KAPREKAR CONTEST - FINAL - SUB JUNIOR
Classes VII & VIII
AMTI - Saturday, 2nd November_2019.
Instructions:
1. Answer as many questions as possible.
2. Elegant and novel solutions will get extra credits.
3. Diagrams and explanations should be given wherever necessary.
4. Fill in FACE SLIP and your rough working should be in the answer book.
5. Maximum time allowed is THREE hours.
6. All questions carry equal marks.
1. Let an be the units place of 12 + 22 + 32 + …+ n2. Prove that the decimal 0. a1a2a3…an…is a rational
p
number and represent it as , where p and q are natural numbers.
q
Sol. For 12 + 22 + 32 + …+ n2
a1 1 a11 6 a 21 1
a2 5 a12 0 a 22 5
a3 4 a13 9 a 23 4
a4 0 a14 5 a24 0
a5 5 a15 0 a25 5
a6 1 a16 6 a 26 1
a7 0 a17 5 a27 0
a8 4 a18 9 a 28 4
a9 5 a19 0 a29 5
a10 5 a20 0 a30 5
given number 0. a1a2a3…an = 0.1540510455 6095065900
given number is non-terminating and repeating.
p
it is a rational number and can be represent in the form of .
q
1 1 3
2. (a) Find the positive integers m, n such that .
m n 17
1 1 1 3
(b) Find the positive integers m, n, p such that .
m n p 17
(c) Using this idea, prove that we can find for any positive integer k, k distinct integers, n1,n2.....nk such
1 1 1 3
that ... .
n1 n 2 nk 17
Sol.
(a) We know
1 1 1
If
x y a
then, (x – a) (y – a) = a2
NMTC_STAGE-II_PAPER-2019_SUB JUNIOR_PAGE # 1
1 1 3
so,
m n 17
1 1 1
3m 3n 17
(3m – 17) (3n – 17) = 289
= 289 × 1
= 17 × 17
= 1 × 289
If (3m – 17) = 289 and 3n – 17 = 1
m = 102 n=6
so, (m, n) = (102, 6)
If 3m – 17 = 17 and 3n – 17 = 17
34 34
m= n=
3 3
not integer so reject
If (3m – 17) = 1 and 3n – 17 = 289
m=6 n = 102
so, (m, n) = (6, 102)
1 1 3
so, ........(i)
6 102 17
(b) Now,
1 1 1
If
6 x y
(x – 6) (y – 6) = 36
= 1 × 36 or 36 × 1
= 2 × 18 or 18 × 2
= 3 × 12 or 12 × 3
= 4 × 9 or 9 × 4
=6×6
so, (x, y) = (7,42) (8, 24), (9, 18), (10, 15) (12, 12)
so from equation (i)
1 1 1 3
7 42 102 17
1 1 1 3
8 24 102 17
: :
: :
1 1 1
and
102 w z
(w – 102) (z – 102) = (102)2
= 1 × 10404
= 2 × 5202
=: :
=: :
= 102 × 102
Total 27 in which 13 are repeated so total 14 different pais.
so pairs of (w, z) = (103, 10506), (104, 5304).......... (204, 204)
so total 14 pairs
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from equation (i)
1 1 1 3
=
6 103 10506 17
1 1 1 3
6 104 5304 17
: :
: :
Total 5 + 14 = 19 pairs.
1 1 1
(c)
a a 1 a(a 1)
We convert every rational number into definite unit fractions so we can find for any positive integer k.
1 1 1 1 3
Such that ...... .
n1 n2 n3 nk 17
3. Does there exist a positive integer which is a multiple of 2019 and whose sum of the digits is 2019? If
no, prove it. If yes, give one such number.
Sol. Sum of digits
1 × 2019 = 2019 12
2 × 2019 = 4038 15
3 × 2019 = 6057 18
4 × 2019 = 8076 21
5 × 2019 = 10095 16
6 × 2019 = 12114 9
So, sum of digits of number 20196057 = 30
If we take 67 times 20196057 and 1 time 12114 then sum of digits is 2019 and number is also divisible
by 2019.
Number is 20196057
20196057 ......12114
67 times
other number is 4038 4038 .....12114 .
134 times
4. In a triangle XYZ, the medians drawn through X and Y are perpendicular. Then show that XY is the
smallest side of XYZ.
Sol.
X
C
2a 2y Q
x
2x G C
y
Y b P b Z
XP YQ, XP and YQ intersect at G
Let XY = 2a
YZ = 2b
XZ = 2c & XG = 2y
GP = y and YG = 2x
NMTC_STAGE-II_PAPER-2019_SUB JUNIOR_PAGE # 3
GQ = x
In XGQ
c2 = 4y2 + x2
c= 4 y 2 x 2 ….(1)
In YGP
b2 = 4x2 + y2
b= 4 x 2 y 2 ….(2)
In XGY
2 2 2
4a = 4x +4y
2 2 2
a =x +y
a= x 2 y 2 …..(3)
from eq.(1) & (3)
a < c 2a < 2c XY < XZ …..(4)
From eq. (2) & (3)
a<b
2a < 2b
XY < YZ ….(5)
From eq. (4) & (5) we can say that XY is the smallest side.
5. Let PQR be a triangle of area 1cm2. Extend QR to X such that QR = RX ; RP to Y such that RP = PY
and PQ to Z such that PQ = QZ. Find the area of XYZ.
X
P 1sq.cm
Q
Z
Y
Sol.
X
P 1sq.cm
Q
Z
Y
area PRQ = area PXR (PR is a median)
1 = area PXR
area PXR = area PXY (PX is a median)
1 = area PXY
area PQY = area PRQ (PQ is a median)
area PQY = 1
area QYZ = area PQY (YQ is a median)
NMTC_STAGE-II_PAPER-2019_SUB JUNIOR_PAGE # 4
area RQZ = area PRQ (RQ is a median)
area RQZ = 1
area RQZ = area RZX (RZ is a median)
1 = area RZX
area XYZ = 7 cm2
3 2657
6. Find the real numbers x and y given that x – y = and x4 + y4 = .
2 16
4 4 2657
Sol. x +y =
16
2657
(x2)2 + (y2)2 =
16
2 2 2 2 2 2657
(x + y ) – 2x y =
16
2657
(x2 + y2)2 – 2(xy)2 =
16
2657
((x – y)2 + 2xy)2 – 2(xy)2 =
16
xy = t
3
x– y =
2
2
3 2
2t – 2t2 = 2657
2 16
81 2657
+ 4t2 + 9t – 2t2 =
16 16
2576
2t2 + 9t =
16
2t2 + 9t – 161 = 0
2t2 + 23t – 14t – 161 = 0
t(2t + 23) – 7(2t + 23) = 0
(2t + 23) (t – 7) = 0
– 23
t=7,t=
2
– 23
xy = 7 or xy =
2
when xy = 7
7
y=
x
3
x–y=
2
7 3
x– =
x 2
x2 – 7 3
x 2
2
2x – 14 = 3x
2x2 – 3x – 14 = 0
2x2 – 7x + 4x – 14 = 0
NMTC_STAGE-II_PAPER-2019_SUB JUNIOR_PAGE # 5
x(2x – 7) + 2(2x – 7) = 0
(2x – 7) (x + 2) = 0
7
x= ,x=–2
2
7
when x =
2
y=2
when x = – 2
–7
y=
2
– 23
when xy =
2
– 23
y=
2x
3
x–y=
2
x 23 3
1 2x 2
2x 2 23 3
2x 2
2
2x – 3x + 23 = 0
D = –ve
No real value
7. The difference of the eight digit number ABCDEFGH and the eight digit number GHEFCDAB is divisible
by 481. Prove that C = E and D = F.
Sol. Difference of ABCDEFGH – GHEFCDAB is k.
k = 1000000 AB + 10000 CD + 100EF + GH – 1000000GH – 10000EF – 100CD – AB
k = 999999 (AB – GH) + 9900(CD – EF)
here 999999 is divisible by 481 so 9900(CD – EF) should be divisible by 481.
9900(10C + D – 10E – F) = 481x
99000 (C – E) + 9900 (D – F) = 481 x
It is possible when C – E or D – F should be multiple of 37 or 0.
C–E=0 ,D–F=0
C=E D=F
8. ABCD is a parallelogram with area 36cm2. O is the intersection point of the diagonals of the
parallelogram. M is a point on DC. The intersection point of AM and BD is E and the intersection point
of BM and AC is F. The sum of the areas of triangles AED and BFC is 12cm2. What is the area of the
quadrilateral EOFM ?
D M C
E
F
O
A B
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Sol. area of parallelogram ABCD = 36
1 1
area AOD = area AOB = area BOC = area DOC = area ABCD = × 36 = 9
4 4
1 1
area AMB = area ABCD = × 36 = 18
2 2
let area AED = x
area BFC = 12 – x
area BOF = area BOC – area BFC = 9 – (12 – x) = x – 3
area AOE = area AOD – area AED = 9 – x
area AMB = area AEO + area AOB + area BOF + area quadrilateral EOFM
18 = 9 – x + 9 + x – 3 + area quadrilateral EOFM or quadrilateral EOFM
area of quadrilateral EOFM = 3 cm2.
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