INDIAN OLYMPIAD QUALIFIER (IOQ) 2023-2024

INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS

(IOQM), 2023

QUESTION PAPER WITH SOLUTION

Sunday, September 03, 2023 |

Duration: 3 Hrs | Time: 10:00 AM to 1:00 PM

Max. Marks : 100

Resonance Eduventures Ltd.

Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN : U80302RJ2007PLC024029
7340010333

This solution was download from Resonance IOQM-2023 Solution portal

 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
Time : 3 hrs September 03, 2023 Total marks : 100

INSTRUCTIONS

1. Use of mobile phones, smartphones, iPads, calculators, programmable wrist watches is

STRICTLY PROHIBITED. Only ordinary pens and pencils are allowed inside the examination hall.

2. The correction is done by machines through scanning. On the OMR Sheet, darken bubbles
completely with a black or blue ball pen. Please DO NOT use a pencil or a gel pen. Darken the
bubbles completely, only after you are sure of your answer; else, erasing may lead to the OMR
sheet getting damaged and the machine may not be able to read the answer. .

3. The name, email address, and date of birth entered on the OMR sheet will be your login
credentials for accessing your score.

4. Incompletely, incorrectly or carelessly filled information may disqualify your candidature.

5. Each question has a one or two digit number as answer. The first diagram below shows improper
and proper way of darkening the bubbles with detailed instructions. The second diagram shows
how to mark a 2-digit number and a 1-digit number.

6. The answer you write on OMR sheet is irrelevant. The darkened bubble will be considered as your
final answer.

7. Questions 1 to 10 carry 2 marks each; questions 11 to 20 carry 3 marks each; questions 21 to 30

carry 5 marks each.

8. All questions are compulsory.

9. There are no negative marks.

10. Do all rough work in the space provided below for it. You also have blank pages at the end of the
question paper to continue with rough work.

11. After the exam, you may take away the Candidate’s copy of the OMR sheet.

12. Preserve your copy of OMR sheet till the end of current Olympiad season. You will need it later for
verification purposes.

13. You may take away the question paper after the examination.

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-1
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023

1. Let n be a positive integer such that 1  n  1000. Let Mn be the number of integers in the set
Xn  4n  1, 4n  2,...., 4n  1000 . Let a = max Mn : 1  n  1000 , and b = min
Mn : 1  n  1000 . Find a – b
Ans. 22
Sol. xn =  4n  1, 4n  2,......., 4n  1000 
for n = 1
5, 6,....... 1004
Number of perfect squares = 29 = a


for n = 1000
4001, 4002,......... 5000
Number of perfect squares = 7 = b
Because when gap is same b/w two number the number of perfect squares b/w two smaller
number will always be greater then or equal to the number of perfect squares b/w two bigger
number.
 a – b = 29 – 7
a – b = 22

2. Find the number of elements in the set a, b  N : 2  a,b  2023,loga b  6 logb a  5
Ans. 54
Sol. logab + 6logba = 5
6
 loga + 5
loga b
 (logab)2 – 5 (logab) + 6 = 0
 (loga b –3) (logab–2) = 0
 (loga b = 2) or logab = 3
 b = a2 or b = a3
Now (a, b)  and 2  a, b  2023
 b = a2 = 22, 32, 42, 52, 62, ……..442
Or b = a3 = 23, 33, 43, 53 , …….., 123
So, number of elements in the set = 43 + 11 = 54

16  7
3. Let  and  be positive integers such that   .
37  16
Find the smallest possible value of .]
Ans. 23
16  7
Sol.  
37  16
16< 37 16 < 7
37
   7  16
16
16
 
7
16 37

7 16
For  = 1, 2, 3, ……,9 ,   I+

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-2
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
at = 10
22.8571 < < 23.125
= 23

4. 
Let x, y be positive integers such that x 4  x  1 y3  23  1 . 
Find the maximum possible value of x + y.
Ans. 07
Sol. x4 – (x–1) (y3–23) – 1
x4 – 1 =(x – 1) (y3–23) – 2
(x – 1) [(x + 1) (x2 + 1) – (y3 –23)] = – 2
= –1 × 2 = 1 x – 2
If x – 1 = –1 x = 0 Rejected
If x – 1 = 1 (x + 1) (x2 + 1) – (y3 – 23) = – 2
x=–2 3× 5 – y3 + 23 = – 2
38 – y3 = – 2 y3 = 40 Rejected
If x – 1 = 2  x = 3
 4.10 – y3 + 23 = –1  y3 = 64
y=4
So maximum possible value of x + y = 3 + 4 = 7

5. In a triangle ABC, let E be the midpoint of AC and F be the midpoint of AB The medians BE and
CF intersect at G. Let Y and Z be the midpoints of BE and CF respectively. If the area of triangle
ABC is 480, find the area of triangle GYZ.
Ans. 10
Sol. By mid point theorem A
1
F || BC & FE = BC  (1)
2
 F ECB is a parallelogram
1 E
ar (BGC) = ar (ABC) F
3 G
1 Y Z
= × 480
3
= 160
B C
In trapezium FECB, since Y & Z are mid– point of diagonal
 YZ || BC +|| EF

& YZ = BC  EF

1
2
1 1 
=  BC  BC 
2 2 
1
YZ = BC
4
 GYZ ~ GBC
ar GYZ   YZ 
2
 
ar GBC   BC 
ar GYZ  1

160 16
ar (GYZ= 10)

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-3
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
6. Let X be the set of all even positive integers n such that the measure of the angle of some regular
polygon is n degrees. Find the number of elements in X.
Ans. 16
180P  2
Sol. n=
P
Where P is number of sides of polygon P  3 {p  I+}
Total number of factors of 180
= (2 + 1) (2 + 1) (1 + 1)
=3×3×2
= 18
Now
For P  3
Total number of factors of 180 = 18 – 2 = 16

7. Unconventional dice are to be designed such that the six faces are marked with numbers from
1 to 6 with 1 and 2 appearing on opposite faces. Further, each face is colored either red or yellow
with opposite faces always of the same color. Two dice are considered to have the same design if
one of them can be rotated to obtain a dice that has the same numbers and colors on the
corresponding faces as the other one. Find the number of distinct dice that can be designed.
Ans. 96
Sol. Arrangement of 3, 4, 5, 6 can be done in 3! ways = 6 (By circular permutation)
Colouring can be done in 2 × 2× 2 = 8 ways
Total design are = 8 × 6 × 2 = 96 ways

8. Given a 2 × 2 tile and seven dominoes (2 × 1 tile), find the number of ways of tiling (that is, cover
without leaving gaps and without overlapping of any two tiles) a 2 × 7 rectangle using some of
these tiles.
Ans. 59
Sol. Case-I:- If we use only dominoes
For 2× n rectangle we get
Recursion formula as f(n) = f(n–1) + f(n–2)
Where f(1) = 1 , f(2) = 2, f(3) = 3, f(4) = 5 , f(5) = 8, f(6) = 13, f(7) = 21
Case-II:- When 2×2 tile is used
2×(f(5) + f(1) ×f(4) + f(2) × f(3))
2×(8 + 1 × 5 +2 × 3)
38
Total = 21 + 38 = 59

9. Find the number of triples (a, b, c) of positive integers such that

(a) ab is a prime;
(b) bc is a product of two primes;
(c) abc is not divisible by square of any prime and
(d) abc  30.
Ans. 14
Sol. (i) ab is a prime
(ii) bc is product of two prime
(iii) abc is not divisible by square of any prime and
(iv) abc  30
Case-1
clearly from (i) & (ii) a = 1 and b is prime
from (i) & (iii) b  c
hence by (i), (ii), (iii) & (iv) we have
a = 1,
If b = 2 then c = 3, 5, 7, 11, 13
Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-4
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
b = 3 then c = 2, 5, 7
b = 5 then c = 2, 3
b = 7 then c = 2, 3
b = 11 then c = 2
b = 13 then c = 2
Hence number of triplets (a, b, c) are  14
Case-2
When b = 1, a – Prime
b=1 (a, b, c) = (2, 1, 15), (3, 1, 10),(5, 1, 6)
so total = 14 + 3 = 17 triples

10. The sequence an n0

is defined by a 0  1, a1  4 and an2  4an1  7an , for n  0. Find the
2
number of positive integer divisors of a50  a49a51.
Ans. 51
Sol. given:-
a0 = 1
a1 = –4
an+2 = – 4an+1 – 7an
x2 + 4x + 7 = 0
Let x1 and x2 are roots
x1 = –2 + 3i x1 + x2 = –4
x2 = –2 – 3i x1 + x2 = 7
n
Let an = p(x1) + q (x2) n

at n = 0 a0 = p + q = 1
p + q = 1 ……. (1)
at n = 1 a1 = p(x1) + q(x2)
–4 = –2(p+q) + 3 i (p-q)
2i
p–q = …..(1)
3
From eqn (1) and (2)
1 i 1 i
p=  and q = 
2 3 2 3
2
a 50 
 a 49 .a 51  px1 
50
 – px   qx  px 
 qx 2 
50 2
1
49
2
49
1
51
 qx 2 
51

= 2pq (x1x2)50 – pq x .x   pqx .x
49
1
51
2
51
1
49
2

 x x  x  x 
7 7 49
=2×  750 1 2
2
2
2
1
12 12
7 51 7
=  749  2
6 12
7 51  7 50
 5 50
6
Number of positive integer divisors = 51

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-5
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
11. A positive integer m has the property that m² is expressible in the form 4n2 – 5n +16 where n is an
integer (of any sign). Find the maximum possible value of Im–n|.
Ans. 14
Sol. m2 = 4n2 – 5n + 16
 16m2 = 64n2 – 80n + 256
 (4m)2 = (8n –5)2 + 231
 (4m)2 – (8n –5)2 = 231
 (4m –8n +5) (4m + 8n –5) = 231 = 3 × 7 × 11
4m  8n  5  231
I   m  29 but n  I
4m  8n  5  1 
4m  8n  5  1 
II   m  29 , n = 15  (m, n) (29, 15)
4m  8n  5  231
4m  8n  5  231
III   m  29 Rej
4m  8n  5  1 
4m  8n  5  1 
IV   m  29 Rej
4m  8n  5  231
4m  8n  5  3 
V   m  10 ; n  I
4m  8n  5  77
4m  8n  5  77
VI   m  10, n  4  (m, n)  (10, – 4)
4m  8n  5  3 
4m  8n  5  3 
VII   m  10, Re j
4m  8n  5  77
4m  8n  5  77
VIII   m  10, Re j
4m  8n  5  3 
4m  8n  5  7
IX   m  5, n  i
4m  8n  5  3
4m  8n  5  33
X   m  5, n  1  (m, n)  (5, – 1)
4m  8n  5  7 
4m  8n  5  33
XI   m  5, Re j
4m  8n  5  7 
4m  8n  5  7 
XII   m  5, Re j
4m  8n  5  33
4m  8n  5  11
XIII   m  4, n  I
4m  8n  5  21
4m  8n  5  21
XIV   m  4, n  0  (m, n)  (4, 0)
4m  8n  5  11
4m  8n  5  21
XV   m  4, Re j
4m  8n  5  11
4m  8n  5  11
XVI   m  4, Re j
4m  8n  5  21
Hence maximum |m – n| = 14

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-6
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
12. Let P(x) = x3 + ax2 + bx + c be a polynomial where a, b, c are integers and c is odd. Let pi be the
value of P(x) at x = i. Given that p13  p32  p33  3p1p2p3 , find the value of p 2  2p1  3p 0 .
Ans. 18
Sol. P(x) = x3 + ax2 + 6x + c
P(x1) = x13  ax12  bx1  c  p1
P( x3 )  x32  ax22  bx2  c  p2
P( x3 )  x33  ax32  bx3  c  p3
Now given p13  p32  p33  3p1 p2p3
p1 = p2 = p3 or p1 + p2 + p3 = 0
case-1 p1 = p2 = p3
3a + b + 7 = 0
1+ a +b + c = 8 + 4a + 2b + c = 27 + 9q+ 3b +c
3a+b+7= 0
a = – 6, b = 11
5a + b + 19 = 0
 p(x) = x3 – 6x2 + 11x + c
= (x–1) (x–2) (x–3) + (c + 6)
Hence p2 + 2p1 – 3p0 = (c+6)+ 2(c+6)– 3 (–6+ c+6)
= 3(c+6) + 18 – 3(c+6)= 18
Case 2 If p1+ p2+ p3= 0
36+14a+6b+3c= 0
Which is not possible as c is odd
So Ans = 18

1
13. The ex-radii of a triangle are 10 , 12 and 14. If the sides of the triangle are the roots of the cubic
2
x3  px2  qx  r  0 , where p,q,r are integers, find the integer nearest to pqr .
Ans. 58
  
Sol. ra  ,rb  , rc  p,q, r are integer means
sa sb sc
a+b+c, ab+ bc+ ca, abc are also integer
a, b, cI
1 1 1 1
  
r r1 r2 r3
1 2 1 1
  
r 21 12 14
1 24  21  18 63 1
  
r 252 252 4
r= 4

r 4
S
= 4S
 4S  4S  4S
ra   , rb   , rc  
Sa Sa Sb Sa Sc Sc
21 4S 4S 4S
 , 12  , 14 
2 Sa Sb Sc
21(S–a)= 8S, 12S –12b = 4S, 14S – 14C = 4S
21S – 21a= 85 , 8S= 12b, 10S=14C
Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-7
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
8S 2S 5
13S= 21a, b  , C S
12 3 7
13s
a=
21
a, b, c are integers
13 2 5
a S, b  S, C  S
21 3 7
So S is multiple of LCM (2, 3, 7)so S is take S = 21
13
a  21  13
21
2 2
b  S   21  14
3 3
5 5
c  S   21  15
7 7
So a = 13, b = 14, c = 15
Now, pqr

a  b  c  ab  bc  ca  abc
(1  a)(1  b)(1  c )  1

14  15  16  1
3359 = 57.96
Nearest integers = 58

14. Let ABC be a triangle in the xy plane, where B is at the origin (0,0). Let BC be produced to D such
that BC : CD = 1 : 1, CA be produced to E such that CA: AE = 1: 2 and AB be produced to F such
that AB : BF = 1 :3. Let G(32,24) be the centroid of the triangle ABC and K be the centroid of the
triangle DEF. Find the length GK.
Ans. 40
Sol.
E (x, y)
2
A (x2, y2)

1
G
1 (32, 24) 1
1 C (x1,y1) D (2x1,2y1)

B
3 (0, 0)

F (a, b)
(-3x2,-3y2)
By section formula
Coordinate of point D
D (2x1 , 2y1)
Let E (x , y)
2 x1  x 2y1  y
x2 = y2 =
3 3
Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-8
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
3x2 – 2x1 = x y = 3y2 – 2y1
E (3x2 – 2x1 , 3y2– 2y1)
Let f(a, b)
3x 2  a 3y 2  b
0 0=
4 4
3x2 + a = 0 0 = 3y2= + b
a = –3x2 b = –3y2

 2x  3x2  3x 2  2x1 3y 2  2y1  3y 2  2y1 

centroid of DEF k  1 , 
 3 3 
k (0 , 0)
Distance between GK = 32  02  24  02
= 322  242  1024  576  1600  40

15. Let ABCD be a unit square. Suppose M and N are points on BC and CD respectively such that the
perimeter of triangle MCN is 2. Let O be the circumcentre of triangle MAN, and P be the
2
 OP  m
circumcentre of triangle MON. If    for some relatively prime positive integers m and n,
 OA  n
find the value of m + n.
Ans. 03
Sol. Since perimeter of MCN = 2

E a D b N 1–b C

a+b
1 –a

x y a
x+y
x
B
A
1–b+1–a+MN = 2
2 – a – b + MN = 2
MN = a+b
Extend CD to E
Such that DE = BM = a
ABM  ADE
By SAS congruence criterion
By CPCT
AE = AM
 MAB =  EAD = x
Now EAN and MAN

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-9
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
D N C

x
M
O
45°

A B
EN = MN (a+b)
AE = MA
AN = AN
By SSS
EAN  MAN
So
ABCD is a square
y + x + y + x = 90
2x+ 2y = 90
x + y = 45
O is circumcentre so
 MON = 2x  MAN = 245 = 90°
 MON = 90°
So MON is a rigid 
  So P(Circumcentre of MON)
P is mid point of MN
OP = PM = PN = X
D N C
x
p
x x
M
O 2x
2x

A B
Only one possibility for quad OMCN
So = OM = 2x
OM = OB = OA = 2x
2 2 2
 OP   x   1  1 m
          = m + n = 1+2 = 3
 OA   2x   2 2 n

16. The six sides of a convex hexagon A1 A2 A3 A4 A5 A6 are colored red. Each of the diagonals of the
hexagon is colored either red or blue. If N is the number of colorings such that every triangle
Ai Aj Ak where 1 i< j < k  6, has at least one red side, find the sum of the squares of the digits
of N.
Ans. 18
nn  3
Sol. Number of diagonals =
2

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-10
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023

A5 A4

A6
A3

A1 A2
66  3
=
2
= 3× 3
=9
Case-1:- when 2 sides red, 1 blue
Number of ways selecting two consecutive red sides = 6
So number of triangles = 6
Case-2:- when 1 sides red, 2 blue
from side A1 A2  (1) A1 A4 A2
(2) A1 A5 A2
From all sides = 6 × 2 = 12 triangles
Total triangle = 15
M –2 6C3 – 2 [A1 A3 A5 and A2 A4 A6] =18

17. Consider the set

S= {(a, b, c, d, e) : 0 <a < b < c < d < e< 100}
where a, b, c, d, e are integers. If D is the average value of the fourth element of such a tuple in
the set, taken over all the elements of S, find the largest integer less than or equal to D.
Ans. 66
Sol. Total selection of a, b, c, d, e are 99C5
When d = 4 then total ways = 3C3 × 95C1
When d = 5 then total ways = 4C3 × 94C1
…………………..…………………………
When d = 98, then total ways = 97C3 × 1C1
4  3C3 95 C1  5 4 C3 94 C1  ......... 98 97 C3 1 C1
Average = 99
C5
97
Number for =  r  1100  r  2 C
r 3
r
3

97
r 1 r r  1 r  2 .r C
97
= 100  4 
r 3
4
C 3  20
r 3
 4 5
3

97 97
= 400 
r 3
r 1
C 4  20 .
r 3
r 2
C5

= 400 99 C5  20  100C6
100
C6 100 1000 200
Average = 400 – 20 × 99
 400  20   400    66.6
C5 6 3 3

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-11
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
18. Let  be a convex polygon with 50 vertices. A set F of diagonals of  is said to be minimally
friendly if any diagonal dF intersects at most one other diagonal in F at a point interior to  . Find
the largest possible number of elements in a minimally friendly set F.
Ans. 71
Sol. Group of nonintersecting diagonals inside polygon are A1A3, A1A4, A1A5,……. A1A49, and A2A4,
A4A6, A6A8, …… A48A50, but diagonal of first group intersect only one diagonal of second group. So
total number of element in set F is 47 + 24 = 71

A1 A2
A50
A49 A3
A48 A4
A5

A6

A7
A8
A9

19. For nN, let P(n) denote the product of the digits in n and S(n) denote the sum of the digits in n.
Consider the set
A= {nN: P(n) is non-zero, square free and S(n) is a proper divisor of P(n)}.
Find the maximum possible number of digits of the numbers in A.
Ans. 92
Sol. A {n N ; p(n)  0 is square free and s(n) is proper divisor of p(n)} p(n) is square free so number
n can containing digit 1, 2, 3,5 7 or, 1, 5, 7, 6 s(n) is proper divisor of p(n)
So maxi possible value of s(n) = 3 × 5 × 7 = 105
For making digit sum 105, n contain digit 2, 3,5 and 7 one time and digit 1, 88 times
s(n) = 2 + 3 + 5 + 7 + 1 × 88 = 105
p(n) = 2 × 3 × 5 × 7 × 1 …… 1 = 210
Maximum number of digits in n = 88 + 4 = 92

20. For any finite non empty set X of integers, let max(X) denote the largest element of X and |X|
denote the number of elements in X. If N is the number of ordered pairs (A, B) of finite non-empty
sets of positive integers, such that max(A)  |B| = 12; and |A|  max(B) = 11 and N can be written
as 100a + b where a, b are positive integers less than 100, find a+ b.
Ans. 43
Sol. max(A) × |B| = 12 = 1 × 12 = 3 × 4 = 2 × 6
|A| × max(B) = 11 = 1 × 11 |A| = 11 4 max (B) = 1
|A| = 1 and max(B) = 11 max (A) = 12 and |B| = 1
Max(A) = 12 |B| =1
=6 =2
=4 =3
=3 =4
=2 =6

Case-1 |A| = 1, max(A) = 12  number of set A = 1

|B| = 1, max (B) = 11  number of set B = 1
Case-2 |A| =1, max(A) = 6  number of set A = 1
|B| =2, max(B) = 11  number of set B = 10C1

Case-3 |A| =1, max(A) = 4  number of set A = 1

|B| = 3, max(B) = 11  number of set B = 10C2

Case-4 |A| =1, max(A) = 3  number of set A = 1

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-12
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
|B| = 4, max(B) = 11  number of set B = 10C3

Case-5 |A| =1, max(A) = 2  number of set A = 1

|B| = 6, max(B) = 11  number of set B = 10C5

Case-5 |A| = 11, max(A) = 12  number of set A = 11C10

|B| = 1, max(B) = 1  number of set B = 1
N = 1 × 1 + 1 × 10C7 + 1 × 10C2 + 1 + 10C3 + 1 × 10C5 + 1 × 11C10
= 1 + 10 + 45 + 120 + 252 + 11 = 439
a = 4 b = 39
a + b = 43

21. For n N, consider non-negative integer-valued functions f on {1,2………., n } satisfying f(i)  f(j)


n
for i > j and ni 1(i  f (i))  2023 . Choose n such that f (i) is the least. How many such
i 1
functions exist in that case?
Ans. 15
n
Sol.  i  f i  2023
i1
n n

 i   f i  2023
i1 i 1

nn  1  
n

2

  f i  2023
 i1 
  nn  1 
n


 f i
 
  2023   7
 i1 least  2 least
for n = 63
n
So  f i = 7
i1
So number of such functions = 15 f(i)  0, f(i)  f(i)
(6, 1), (5, 2) (4, 3)
(1, 2, 5), (1, 2, 4), (1, 3, 3), (2, 2, 3)
(1, 1, 1, 4), (1, 1, 2, 3), (1, 2, 2, 2)
(1, 1, 1, 1, 3), (1, 1, 1, 2, 2)
(1, 1, 1, 1, 1, 2), (1, 1, 1, 1, 1, 1, 1)

22. In an equilateral triangle of side length 6, pegs are placed at the vertices and also evenly along
each side at a distance of 1 from each other. Four distinct pegs are chosen from the 15 interior
pegs on the sides (that is, the chosen ones are not vertices of the triangle) and each peg is joined
to the respective opposite vertex by a line segment. If N denotes the number of ways we can
choose the pegs such that the drawn line segments divide the interior of the triangle into exactly
nine regions, find the sum of the squares of the digits of N.
Ans. 77
Sol. Case-I

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-13
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023

A
P11 P10
P12 P9
P13 P8
P14 P7
P15 P6

B P 1 P2 P3 P4 P5 C
As per given condition we need to divide triangle into exactly I region, and for flu's to happen 3 line
must be concurrent as shown in the above figure.
(i.e. in a way we are choosing 3 points on those sides, such that three lines from those points are
concurrent)
So basically this is ideal solution of ova's theorem in which product of three different ratio leads to
1.
m n
Possible ration on side AB, BC and CA will be of the form , and 1.
n m
m n
i.e  1  1
n m
Now, we can choose the ration 1:1 in 3 ways for all three sides and other ration can be choose in 4
ways other two sides.
i.e three are a 3 × 4 + 1 = 13 ways
Now, fourth point can be choose in 12C ways
 Total such possibilities = 12× 13 = 156 ways
Case-II:-

A
P11 P10
P12 P9
P13 P8
P14 P7
P15 P6

B P 1 P2 P3 P4 P5 C
Seating any two points on any two sides, no of ways: = 3 × 5C2 × 5C2 = 300
Total case possible = 300 + 156 = 456
Some of square of digit = 42+ 52 + 62 = 16 + 25 + 36 = 77

23. In the coordinate plane, a point is called a lattice point if both of its coordinates are integers. Let A
be the point (12,84). Find the number of right angled triangles ABC in the coordinate plane where
B and C are lattice points, having a right angle at the vertex A and whose in center is at the origin
(0,0).
Ans. 18
Sol.

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-14
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023

45°
 -45°

m = 3/4 m1 = –4/3
I

B
(3t1,4t1) C
(-4t2,3t2)

84
Using concept of shifting co-ordinate slope of AI = 7
12
7 1 3
Slope of BC = tan(– 45) = 
1 7 4
AI 122  84 2
Radius of in circle is   60
2 2
r = 60
AB  AC  BC
Now r
2
(BC)2 = (AC)2 + (AB)2
(BC)2 = (AC + AB – 2r)2

(5t1 + 5t2 – 2 × 5 × 12)2 = 25 t12  t 22 
t1  t 2  2  122  t12  t 22
Put t1 = x + 12
t2 = y + 12
Equation become
(x – 12) (y– 12) = 2 × 122
0 0
Total number of triangle is equal to pair of (x, y)
(x – 12) (y – 12) = 25 × 32
number of pair (x, y) = 6 × 3 = 18

24. A trapezium in the plane is a quadrilateral in which a pair of opposite sides are parallel.
A trapezium is said to be non-degenerate if it has positive area. Find the number of mutually
non-congruent, non-degenerate trapeziums whose sides are four distinct integers from the set
(5,6,7,8,9,10).
Ans. 31
Sol.

D b C

c d
c
h

A b E B
a

ha  b ……….. (1)

1
[BEC] =
2
total area = bh + [BEC]
Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-15
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
by egn
(1)
2BEC 
h=
ab
 2BEC 
So total area =  b  BEC
 ab 
ab
= [BEC]  
ab
So total area is non-zero if [BEC] has non-zero area we want to find a, b, c, d  {5, 6, 7, 8, 9, 10}
such that [BEC] has non-zero area, and are pairwise non-congruent.
note that {c, d, a-b} for a non-degenerate triangle if semi perimeter > all side
c  d  a  b
i.e  c, d, a – b
2
Notice that since a, b, c, d 
c, d  5  a – b, so we need to cheek
c  dab
 c, d this beams |c –d| < a– b
2
Since exchanging c, d lead to a congruent trapezium.
Let c > d, since they are distinct , so 0 < c – d < a – b is are condition
Now a–b can range from 1 to 5
Case–I:
a–b=1 0< c–d<1 no solutions
Case–II:
a–b=2 0< c–d<2 c–d=1
c=d+1
(a, b) = (7, 5) d = 8 or 9
(a, b) = (8, 6) d = 9
(a, b) = (9, 7) d = 5
(a, b) = (10, 8) d = 5 or 6 6 Solutions
Case–III:
a–b=3 0< c–d<3
c–d=1 or c–d=2
c=d+1 or c=d+2
(a, b) = (8, 3) d = 6, c = 7
d = 7, c = 9
d = 8, c = 10
(a, b) = (9, 6) d = 5, c = 7
d = 7, c = 8
d = 8, c = 10
(a, b) = (10, 7) d = 5, c = 6
d = 6, c = 8
d = 8, c = 9
Total = 9 solutions
Case–IV:
a–b=4 0<c–d<4
c–d=1 c–d=2 c–d=3
c=d+1 c=d+2 c=d+3
(a, b) = (9, 5) d = 6, c = 7, 8
d = 7, c = 8, 9
d = 8, c = 10
(a, b) = (10, 6) d = 5, c = 7, 8
d = 7, c = 8, 9
d = 8, c = 9

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-16
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
Total = 10 solutions
Case–V:
a–b=5 0<c–d<5
c–d=1 c–d=2 c–d=3 c–d=4
c=d+1 c=d+2 c=d+3 c=d+4
(a, b) = (10, 5) d = 6, c = 7, 8, 9
d = 7, c = 8, 9
d = 8, c = 9
Total = 6 solutions
Total = 0 + 6 + 9 +10 + 6= 31 trapezium

25. Find the least positive integer n such that there are at least 1000 unordered pairs of diagonals in a
regular polygon with n vertices that intersect at a right angle in the interior of the polygon.
Ans. 30
Sol.
A1
A4k A2

Case-I
Let n= 4k
[1 + 3 + 5 + …… + (2k–1)+…… + 3 +1]k
[k2 + (k–1)2k] 1000
k (k2 + (k–1)2] 1000
k  9 n  36
Case-II
n= 4k + 2
(1 + 3 + 5 + …… + (2k–1) × 2 × (2k + 1)
2k2 (2k + 1) 1000
k2 (2k + 1) 500
k2  7
n  30
Final Answer = 30

26. In the land of Binary, the unit of currency is called Ben and currency notes are available in
denominations 1, 2, 22, 23.... Bens. The rules of the Government of Binary stipulate that one can
not use more than two notes of any one denomination in any transaction. For example, one can
give a change for 2 Bens in two ways: 2 one Ben notes or 1 two Ben note. For 5 Ben one can give
1 one Ben note and 1 four Ben note or 1 one Ben note and 2 two Ben notes. Using 5 one Ben
notes or 3 one Ben notes and 1 two Ben notes for a 5 Ben transaction is prohibited. Find the
number of ways in which one can give change for 100 Bens, following the rules of the
Government.
Ans. 19
Sol. 20(1), 21, 22, 23 , …... = 1, 2, 4, 8, 16, 32, 64
64 + 32 + 4
64 + 32 + 2 + 2
64 + 32 + 2 + 1 + 1
Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-17
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
64 + 16 + 16 + 4 = 32 + 32 + 16 + 16 + 4
64 + 16+ 16 + 2 + 2 = 32 + 32 + 16+ 16 + 2 + 2
64 + 16 + 16 + 2 + 1 + 1 = 32 + 32 + 16 + 16 + 2 + 1 + 1
64 + 16 + 8 + 8 + 4 = 32 + 32 + 16 + 8 + 8 + 4
64 + 16 + 8 + 8 + 2 + 2 = 32 + 32 + 16 + 8 + 8 + 2 + 2
64 + 16 + 8 + 8 + 2 + 1 + 1 = 32 + 32 + 16 + 8 + 8 + 2 + 1 + 1
64 + 16 + 8 + 4 + 4 + 2 + 2 = 32 + 32 + 16 + 8 + 4 + 4 + 2 + 2
64 + 16 + 8 + 4 + 4 + 2 + 1 + 1 = 32 +32 + 16 + 8 + 4 + 4 + 2 + 1 + 1
Sol. (Incomplete)
20(1), 21, 22, 23
100B
100 = 26 + 25 + 22 (64 + 32+ 4) 64 + 16 + 8 + 8 + 4
= 26 + 25 + 21 + 21 26 + 24 + 2 × 23 + 22
= 26 + 25 + 21 + 20 + 20 26 + 24 + 2 × 23 + 2. 21
= 26 + 2 × 24 + 2.21 2 × 25 + 24 + 2 × 23 + 2 × 22
= 2 × 25 + 2 × 24 + 22 2 × 25 + 24 + 2 × 23 + 2 × 21
= 2 × 25 + 2.24 × 2.21 64 + 16 + 8 + 4 + 4 + 2 + 2
= 26 + 25 + 2 + 2.20 26 + 24 + 23 + 2 × 22 + 2 × 21
= 26 + 2.24 + 21 +2.20 2 × 25 + 24 + 23 + 2 × 22 + 221
= 2 × 25 + 2 × 24 + 21 + 2 × 20
Total cases = 15

27. A quadruple (a, b, c, d) of distinct integers is said to be balanced if a + c = b + d. Let S be any set
of quadruples (a, b, c, d) where 1  a < b < d < c  20 and where the cardinality of S is 4411. Find
the least number of balanced quadruples in S.
Ans. 91
Sol. At first find maximum cardinality of quadruple (a, b, c, d) ignoring the balanced one, the that is
simply 20C4 = 4845
But cardinality of s {(a, b, c, d)} is given to be 4411.
So leaving out maximum possible balanced quadruple in set S = {(a, b, c, d)} with 1 a < b < d < c
 20
Now, counting balanced quadruples is s, we have
a + c = b + d = 5  (1, 4), (2, 3)  2C2
a + c = b + d = 6  (1, 5), (2, 4)  2C2
a + c = b + d = 7  (1, 6), (2, 5), (3, 4)  3C2
a + c = b + d = 8  (1, 7), (2, 6), (3, 5)  3C2
a + c = b + d = 9  (1, 8), (2, 8), (3, 7), (4, 5)  4C2
a + c = b + d = 10  (1, 9), (2, 8), (3, 7), (4, 6)  4C2


a + c = b + d = 36  (16, 20), (17, 19),  2C2
a + c = b + d = 37  (17, 20), (18, 19),  2C2
Total such case are = 4(2C2 + 3C2 + ……. + 9C2) + 10C2
= 480 + 45
= 525 balanced quadruples
Remaining will be 525 – 434 = 91

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-18
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
28. On each side of an equilateral triangle with side length n units, where n is an integer, 1  n  100,
consider n –1 points that divide the side into n equal segments. Through these points, draw lines
parallel to the sides of the triangle, obtaining a net of equilateral triangles of side length one unit.
On each of the vertices of these small triangles, place a coin head up. Two coins are said to be
adjacent if the distance between them is 1 unit. A move consists of flipping over any three mutually
adjacent coins. Find the number of values of n for which it is possible to turn all coins tail up after a
finite number of moves.
Ans. 67
Sol. This can be done for n  0 , 2 mod 3. below by a triangle , we will mean three coins which are
mutually adjacent for n = 2, clearly it can be done.
H
T

H H
T T and
for n = 3 , flip each of the four triangle

H T T T T

H H T T H T T T H H

H H H H H H T T H T T TT H T
Original (I) steps (II) steps (III) steps (IV) steps
for n  0 , 2 mod 3
and n > 3 flip every triangle. Then the coins at the corners are flipped one. The coins on the sides
(not comers) are flipped three times each. So all these coins will have tails up. The interior coins
flipped six times each and have heads up. Since the interior coins have side length n – 3 by the
induction step all of them (an be flipped so to have tails up.
n  0 mat 3
Black
Red

n=4

n=6

reduces to n = 6  reduces n = 3 point

Next suppose r = 1 mod 3 colour the heads of each coin red, white and blue 80 that adjacent coins

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-19
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
different colour and any three coins is a raw have different colours. then the coins in the corner
have the same colour sag red [ n = 3 k +1] A simple count shown that three are one more red
coins then white or blue coins, so the (odd or even) parities of the red and white coins increase by
1 or (b) both by 1 or (c) one increases by 1 and other decreases by 1. So the particles of the red
and white coins stay different. In the case all coins are fails up the number of red and white coins
could be zero and the parities would be the same so this cannot happens n  1 mod 3 case
number of coins

R W step
4 3 original
3 2 (I) – 1
2 3 (II) ± 1
1 2 (III) – 1

T T
T
R W T
T T W T
W B B R W T B W
T W
B R T
W R R T R
W B R W B R B
R R
W B Ist step IInd step IIIrd step

R
W B
B W
R R
W B
B W
R R
W B R W B
so it is possible for r = 33
n = 1, 2, 3, 5, 6, 8, 9, ……, 98, 99
for n = 67
possible value it can be done

29. A positive integer n > 1 is called beautiful if n can be written in one and only one way as
n = a1 + a2 + ……..+ ak = a1. a2 …..ak for some positive integers a1,a2,……ak , where k > 1 and
a1  a2  ak. (For example 6 is beautiful since 6 = 3. 2. 1 = 3 + 2 + 1, and this is unique. But
8 is not beautiful since 8 = 4 + 2 + 1 + 1 = 4. 2. 1.1 as well as 8 = 2 + 2 + 2 + 1 + 1 = 2.2.2.1.1,
so uniqueness is lost.) Find the largest beautiful number less than 100.
Ans. 95
Sol. 95 = 19 × 5 × (1 × 1 × 1 ……. 71 times)
95 = 19 + 5 + (1 + 1 + 1 …… 71 times)

Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-20
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 INDIAN OLYMPIAD QUALIFIER IN MATHEMATICS (IOQM) | 03-09-2023
30. Let d(m) denote the number of positive integer divisors of a positive integer m. If r is the number of

i1d(i) is odd, find the sum of the digits of r

n
integers n 2023 for which
Ans. 18
Sol. As m is a positive integer and If m is perfect square then number of its divisor is odd otherwise
n
number of divisor is even and  di = odd
i1
So for sum to be odd, number of perfect square to be odd number of times
Hence number of interfere n 2023 for which d(i) = odd is equal to
r = (22– 12) + (42–32) + (62–52) + ………. +(442–432)
 r = 1 + 2 + 3 +4 + ……. + 44
44  45
=  22  45  990
2
Sum of digits of r is 18


Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
IOQM2023-21
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
 JEE (ADVANCED) 2023 Result
#PlanningSafaltaKi

From Excellence to Prominence...

AIR
22 AIR
26 AIR
29 AIR
32 AIR
33

AIR
SC- 1
DESHANK PRATAP SINGH MAYANK SONI TANISHQ M MANDHANE KRITIN GUPTA NAMAN GOYAL
Reso Roll No.: 22235068 Reso Roll No.: 22741545 Reso Roll No.: 22235023 Reso Roll No.: 22235049 Reso Roll No.: 22235036

AIR
99 AIR
37

7
STUDENTS IN
AIR
AIR
OBC-
NCL-
PWD 1
TOP-100
MD Sahil Akhtar All India Ranks (AIRs)

15
S S SUMEDH
Reso Roll No.: 22235069 Reso Roll No.: 22235059

AIR
92 AIR
44
BIKKINA A. CHOWDARY
All AIRs are in Common Rank List (CRL) Reso Roll No.: 22235070

All Students are from

Apurva samota Our Offline/Online Classroom Programs KAUSHAL VIJAYVERGIYA
Reso Roll No.: 22235055 Reso Roll No.: 22741546

AIR
69 AIR
66 AIR
63 AIR
61 AIR
54

AIR: All India Rank

AIR
PWD-1
KRISH GUPTA keyaan K rajesh dipen sojitra ridam jain bhavya bansal
Reso Roll No.: 22235051 Reso Roll No.: 21214207 Reso Roll No.: 22235054 Reso Roll No.: 22235057 Reso Roll No.: 22235065

Special Achievers Total Selections

1200
The Power of SHE fgUnh ek/;e ds
The Power of SHE Distance Learning
ADITI SINGH fotsrk
Reso Roll No.: 22235053 ANUSHKA fcfiu eh.kk sanjay p
AIR
GIRLS’ Topper SINGHAL Reso Roll No.: 21218129 mallar
IIT-Bombay Zone
17
Reso Roll No.: 21153204 Reso Roll No.: 23405290

(OBC-NCL) AIR
(CRL)
104 AIR
(CRL)
291 AIR
(ST)
61 AIR
(CRL) 86 (Classroom: 920 | Distance: 280)
Eligible for Counselling

Congratulations...!!!
To all the Selected Students & their Proud Parents
REGISTERED & CORPORATE OFFICE (CIN: U80302RJ2007PLC024029): CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Rajasthan) - 324005
0744-2777777 73400 10345 contact@resonance.ac.in www.resonance.ac.in Follow Us: @ResonanceEdu @Resonance_Edu
 #PlanningSafaltaKi
JEE (Main) 2023 RESULT

22 o”kkZs ls yxkrkj--- Js”B f’k{k.k] Js”B ifj.kke---

ERGIYA | C
AYV ou
VIJ r

AL

se
: iV
H
KAUS

IJETA
N | Course: SOHAM
O BI DL DA
ER P S
RG

|C
5
AIR
FREDIE GEO

our
se: DLP
98 26
AIR AIR

Course: VIJETA

7
ASHIK STE
| NN
OD 10

50 29
S

Y|
AIR AIR
0%
LA

Cou
100%ile
ile (Maths)
HARSHAL

rse: DLP
1 00 % % i le
il 00 AIR

34 31
AIR
e

RESULT HIGHLIGHTS

20 9228 19 3 3
Subject Wise 100%ile

AIRs (Category) in Top - 100 Students Qualified for JEE (Advanced) 2023 Physics Chemistry Maths

Classroom Students (O ine/Online) in Top-1000 AIRs (CRL)

AIR 115 AIR 258 AIR 378 AIR 390 AIR 407 AIR 454 AIR 465 AIR 481 AIR 538 AIR 551 AIR 564 AIR 584

RAMKRISHNA GENA ABHISHEK GUPTA KAVYA AGRAWAL APOORV SHARMA KRITIN GUPTA ARHAN M. VORA ARJUN KRISHNASWAMY SANYAM AGRAWAL ARYA SAMEER JOSHI KEYAAN K. RAJESH SUJIT ADIGA AYUSH GUPTA

AIR 641 AIR 642 AIR 657 AIR 668 AIR 716 AIR 757 AIR 765 AIR 768 AIR 853 AIR 854 AIR 860 AIR 867

UTKARSH GUPTA CEZAN V. DAMANIA ANUSHKA SINGHAL DEVANSH JAIN HARSH AGRAWAL DARSHIT MITTAL PRAJJWAL YADAV PRATYUSH DASH MOHIT AGARWAL SHAURYA SINGHAL ANIRUDH BHARDWAJ PARTH GUPTA

*Result Received so far

Congratulations to all the Qualified Students & their Parents

REGISTERED & CORPORATE OFFICE (CIN: U80302RJ2007PLC024029): CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Rajasthan) - 324005
0744-2777777 73400 10345 contact@resonance.ac.in www.resonance.ac.in Follow Us: @ResonanceEdu @Resonance_Edu