Olympiad 8 - Test 4

Questions
13. Each face of a regular tetrahedron is painted either red,
Maths white, or blue. Two colorings are considered
1. The number of integral values of m, for which the roots of indistinguishable if two congruent tetrahedral with those
x2 –2mx + m2 –1 = 0 will lie between –2 and 4 is colorings can be rotated so that their appearance are
identical. How many distinguishable colorings are
2. 2
x −y−z possible?
Suppose x . What is the value of
1 2 3
= = =
y+z z+x x+y+z

z−y

x
? 14. Number of positive integers n less than 200 having two
distinct positive divisors with the same distance from n/3.
3. Rishi is four times as old as Minakshi was when Rishi was
old as Minakshi is now. Furthermore, Rishi is twice as old 15. Find the sum of all positive integers n for which 2n + 7n is
Minakshi was when Rishi was six years older than a perfect square
Minakshi is now. How old is Rishi?
16. Into how many regions do 10 circles divide the plane if
4. The integer n is the smallest positive multiple of 15 such each pair of circle intersects at two points and no points
that each digit of n is either 4 or 0. Compute ?
n

1110
lies on the three circles.

5. A rectangular box has width 12 inches, length 16 inches, 17. Parallelogram AECF is inscribed in square ABCD. It is
and height m/n inches, where m and n are relatively prime reflected across diagonal AC to form another
positive integers. Three faces of the box meet at a corner parallelogram AE’CF’. The region common to both
of the box. The center points of those three faces are the parallelograms has area m and perimeter n. Compute the
vertices of a triangle with an area of 30 square inches.
2
n
value of [ m ], if AF:AD = 1:4. Here [x] represents greatest
Find m + n.
integer less than or equal to x.

6. A group of 2021 students sit in a circle, consecutively

18. Find the sum of all positive integer n so that the
counting numbers 1, 2 and 3 and repeating. Starting from
some student A with number 1, and counting clockwise expression 4n + 14n – 5 represents a perfect square?
round the remaining students, students that count
numbers 2 and 3 must leave the circle until only one 19. Find the remainder, when
remains. If nth student counting clockwise from A is the 30 × 30! + 31 × 31!+. . . . . . +99 × 99! divided by 217?
remaining student. Find the sum of digits of n.
20. If P(x) = 1 + x + x2 + x3 + x4 + x5, then remainder when P(x
7. There is a prime number p such that 16p + 1 is the cube of 12)
divide by P(x)
a positive integer. Find sum of digits of p.
21. Six men and 7 women stand in a line in random order.
8. The- phrase "COLORFUL TARTAN" is spelled out with Find the number of ways such that a group of at least four
wooden blocks, where blocks of the same letter are men stand together in the line and every man stand next
indistinguishable. How many ways are there to distribute to at least one other man.
the blocks among two bags of different color such that
neither bag contains more than one of the same letter? 22. For a positive integer n, let p(n) be the product of the
digits of n. Compute the number of positive integers less
9. Three consecutive positive integers raised to the first, than 1000 for which p(p(n)) = 6.
second and third powers respectively, when added make a
perfect square, the square root of which is equal to the 23. The entries in a 3 × 3 array include all the digit from 1
sum of these consecutive integers. Find the sum of these through 9, arranged so that the entries in every row and
integers. column are in increasing order. How many such arrays are
there?
10. The remainder when 2 + [1! + 2(2!) + 3(3!) + . . . . . + 10(10!)]
is divided by 11! is 24. For how many integers n between 1 and 50, inclusive, is
2
(n −1)!

11. If a, b, c, d, e are distinct integers such that n

an integer.
(n!)

(9 − a) (9 − b) (9 − c) (9 − d) (9 − e) = 12, then find

a + b + c + d + e ? 25. The center of a circle of radius 2 follows a path around the
edges of a regular hexagon with side length 3. If the area
12. Let P(x) be a polynomial with integer coefficients that of the region the circle sweeps = a + bπ - c√3, find a + b +
satisfies P(17) = 10 and P(24) = 17. Given that P(n) = n + c. Here a, b and c are integers.
3 has two distinct integer solutions n1 and n2, find the
remainder when n1n2 divide by 100.
26. Let A = {1, 2, 3, 4, 5, 6, 7}, and let N be the number of 29. Find the largest positive integer n such that the equations
functions f from set A to set A such that f(f(x)) is a constant (x + 1)
2
+ y
2
= (x + 2)
2
+ y
2
=. . . . . . . = (x + n)
2
2
+ yn
1 2
function. Find the remainder when N is divided by 100. Have an integer solution (x, y1…… yn)

27. In ΔABC, AB = 3, BC = 4, and CA = 5. Circle ω intersects 30. For positive integers n, let m(n) be the number of blocks
at E and B, BC at B and D, and AC at F and G. Given
¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯
AB of consecutive 1’s in binary expansion of n. For example,
that EF = DF and
DG
=
3
, length DE=
a √b
, m(25) = 2 because 25 = 110012 has a block of two 1’s at the
c
EG 4
beginning and a block of one 1 at the end, and m(3) = 1
where a and c are relatively prime positive integers,
because 3 = 112 only has a single block of two 1’s. If x =
and b is a positive integer not divisible by the square of
any prime. Find a + b + c. m(1) + m(2) + m(3) +....+ m(128), find sum of digits of x.

28. Let f : ℕ → ℕ be a function satisfying the following

conditions:
(a) f(l) = 1.
(b) f(a) ≤ f(b) whenever a and b are positive integers with a
≤ b.
(c) f(2a) = f(a) + 1 for all positive integers a.
If n possible values can the 2014-tuple (f(l), f(2), .....,
f(2014)) take, find sum of digits of n.
Answer Key

1 2 3 4 5

3.00 2.00 24.00 4.00 41.00

6 7 8 9 10

22.00 10.00 16.00 12.00 1.00

11 12 13 14 15

42.00 18.00 15.00 33.00 1.00

16 17 18 19 20

92.00 57.00 5.00 63.00 6.00

21 22 23 24 25

64.00 43.00 42.00 34.00 84

26 27 28 29 30

99.00 41.00 8.00 3.00 14.00

Solutions

1. x2 –2mx + m2 –1 = (x – m)2 = 1
X = m+1 or m – 1
m – 1 > -2 and m + 1 < 4
m > -1 and m < 3
m = 0, 1, 2

2.

3. Let Rishi’s age be a and Minakshi’s age be m

ATQ.
a = 4(m – (a – m)) and a = 2(m – (a – m – 6))
5a = 8m and 3a = 4m + 12
Solving these equations
∴ a = 24, m = 15
Hence Rishi is 24

4. According to question
n should be multiple of 5 and 3.
Any multiple of 5 end with 0 or 5.
∴ n should ends with ‘0’.

Also sum of digits of n should be divisible by 3,

Hence 4 + 4 + 4 = 12 is only divisible by 3 as we can use only 4 or 0 as digits in ‘n’
∴ n = 4440

.
n 4440
⇒ = = 4
1110 1110

5. Let the height of the box be x.

2

After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, √( 2 ) ,
x
+ 64

2
x
and √( 2 ) + 36. Since the area of the triangle is 30, the altitude of the triangle from the base with length 10 is 6.

Considering the two triangles created by the altitude, we use the Pythagorean theorem twice to find the lengths of the two
line segments that make up the base of 10.
We find: 10 2
= √(28 + x /4) + x/2
36
Solving for x gives us x = 5
. Since this fraction is simplified:
m + n = 41.

6. If there are 3n students sitting in a circle, then after the first counting there remain 3n–1 students and student X who counts 1
first at the first round will count the same 1 first at the second round. So, X will remain the last one.
We have 2021 students. Since 37 = 729 < 2021 < 37 = 2187.
So, we need to remove 2021 – 729 = 1292 students.
Till 1292+ 1292 /2 = 1938 we remove 1292 students. 1939-th place counting clockwise from A. Sum of digits = 22

7. Since 16p + 1 is odd, let 16p + 1 = (2a + 1)3. Therefore, 16p + 1 = (2a + 1)3 = 8a3 + 12a2 + 6a + 1. From this, we get 8p = a(4a2 +
6a + 3). We know p is a prime number and it is not an even number. Since 4a2 + 6a + 3 is an odd number, we get a = 8.
Therefore, p = 4a2 + 6a + 3 = 4 × 82 + 6 ×8 + 3 = 307.

Sum of digits = 10

8. Observe that, there are five pairs of letters and four singletons. It is not necessary to care about
the pairs, since each pair must have one letter in bag. It then remains to distribute four distinct letters among two
distinguishable bags; this can be done in 24 = 16 ways.
9. Let (n – 1), n, n + 1 be the three consecutive integers.
Then, (n – 1)1 + n2 + (n + 1)3 = (3n)2 = 9n2
⇒ n – 1 + n2 + n3 + 3n2 + 3n + 1 = 9n2
⇒ n3 – 5n2 + 4n = 0 ⇒ n(n – 1)(n – 4) = 0
⇒ n = 0 or n = 1 or n = 4
But n = 0 and n = 1 will make the consecutive integers – 1, 0, 1 and 0, 1, 2 which contradicts the hypothesis that the
consecutive integers are all greater than zero.
Hence, n = 4, corresponding to which the consecutive integers are 3, 4, and 5.

10. n(n!) = (n + 1)! − n! and proceed

11. The prime factorization of 12 is 22 × 3.

∴ The 5 distinct integer factors must have some negative numbers in them.

There are two 2’s in the prime factorization, one of them must be negative and other positive
Distinct integer factors must be
−2, 2, 1, −1, 3

Taking a = 6
b=7
c=8
d = 10
e = 11.

12. We define Q(x) = P(x) – x + 7, noting that it has roots x = 17 and 24. Hence P(x) – x + 7 = A (x–17) (x–24). In particular, this
means that P(x) – x – 3 = A (x–17) (x–24) – 10. Therefore, x = n1, n2 satisfy A (x–17) (x–24) = 10, where A, (x–17), and (x–24) are
integers. This cannot occur if x ≤ 17 or x ≥ 24 because the product (x - 17)(x - 24) will either be too large or not be a divisor of
10. We find that x = 19 and x = 22 are the only values that allow (x–17) (x–24) to be a factor of 10. Hence the answer is 19.22 =
418.

13. A tetrahedron has 4 sides. The ratio of the number of faces with each color must be one of the following 4 : 0: 0, 3 : 1 : 0, 2 : 2 :
0, or 2 : 1 : 1
The first ratio yields 3 appearances, one of each color.
The second ratio yields 3.2 = 6 appearance, three choices for the first color, and two choices for the second.
3
The third ratio yields ( ) = 3 appearances since the two colors are interchangeable.
2

The fourth ratio yields 3 appearances. There are three choices for the first color, and since the second two colors are
interchangeable, there is only distinguishable pair that fits them.
The total is 3 + 6 + 3 + 3 = 15 appearances

14. Since the smallest possible divisors of an integer n are 1,2 and 3 the greatest possible divisors are n, n/2 and n/3. Hence a
divisor that s bigger than n/3 can only be n or n/2. Since there is no positive divisor of an having the same distance from n/3
as n, the bigger one of the two divisors must be n/2. The distance from n/2 and n/3 and since n/3 – n/6 = n/6 the smaller
divisor must be n/6. Hence n is a multiple or 6. On the other hand it is clear that all positive multiples of 6 have the desired
property. There are 33 numbers which are multiple of 6.

15. Since 21 + 71 = 9 = 32, n = 1 is a solution. We will now show that it is the only solution. For n > 1, we have 2n ≡0 (mod 4). We
also have 7n ≡ (−1)n (mod 4). Since all perfect square are either congruent to 0 or 1 modulo 4, 2n + 7n cannot be a perfect
square if n is odd and greater than 1. So write n = 2m, where m is a positive integer.
Considering this expression modulo 5, we have 2n + 7n = 4m + 49m ≡2 ×(−1)m (mod 5). Therefore 2n + 7n is congruent to 2 or
3 modulo 5. On the other hand, all perfect squares are congruent to 0, 1 or 4 modulo 5.
Therefore, n = 1 is indeed the only solution to the problem.
16. Denote by P(n), the number of regions divided by n circles. We have P(1) = 2 P(2) = 4, P(3) = 8, P(4) = 14,…and from this we
notice that
P(1) = 2,
P(2) = P(1) + 2
P(3) = P(2) + 4
P(4) = P(3) + 6
…..
P(n) = P(n-1) + 2(n-1)
Summing up these equations we obtain
P (n) = 2 + 2 + 4+. . . . . . +2 (n − 1)

= 2 + 2 (1 + 2+. . . . + (n − 1))

n(n−1)
= 2 + 2.
2

= 2 + n (n − 1)
P(10) = 92

17.

18.
n n
4 + 14n − 5 > 4
n n
√4 + 14n − 5 > 2
n n
√4 + 14n − 5 ⩾ 2 + 1
n n n
4 + 14n − 5 ⩾ 4 + 2 × 2 + 1
n+1
2 − 14n ⩽, −6

n ⩽ 5

Check for n = 1, 2, 3, 4 and 5

Only n = 5 give perfect square.

19. 217 = 31 x 7
All term Divisible by 217 except 30x30!
30! ≡ - 1 mod (31) and 30! ≡ 0 mod (7)
⇒ 31k – 1 ≡ 0 mod7
3k ≡ 1 mod7
k=5
31k -1 = 154
⇒ 30! = 217m + 154
30 x 30! = 217m x 30 + 30 x 154 = 217(30m + 21) + 63
Remainder is 63

20. P(x 12) = 1 + x12 + x24 + x36 + x48 + x60 = 6 + (x12 - 1 ) + (x24 - 1 ) + (x36 - 1 ) + (x48 - 1 ) + (x60 – 1)
= 6 + (x6 - 1)Q(x)
Remainer is 6.

21. Let n be the number of women present, and let _ be some positive number of women between groups of men. Since the
problem states that every man stands next to another man, there cannot be isolated men. Thus, there are five cases to
consider, where (k) refers to a consecutive group of k men:
_(2)_(2)_(2)_
_(3)_ (3)_
_(2)_ (4)_
_(4)_ (2)_
_(6)_
For the first case, we can place the three groups of men in between women. We can think of the groups of men as dividers
splitting up the n women. Since there are n + 1 possible places to insert the dividers, and we need to choose any three of
n + 1
these locations, we have ( ) ways.
3

The second, third, and fourth cases are like the first, only that we need to insert two dividers among the n + 1 possible
n + 1 n + 1
locations. Each gives us 3 ( ) ways, for a total of ( ) = n + 1 ways.
2 1

The total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case,
or
n + 1
2( ) + (n + 1) = 28C2 + 8 = 64
2
22.

23. Case 1: Center 4

3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So there
are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. 2 × 6 =
12.
Case 2: Center 5

Here, no 3’s or 7’s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above.
Remembering that 4 < 5, logically see that the numbers of cases are then 2, 3, 3, 1 respectively. By symmetry, 2 × 9 = 18.
Case 3: Center 6
By inspection, realize that this is symmetrical to case 1 except that the 7’s instead of the 3’s are assured. 2 × 6 = 12.
12 + 18 + 12 = 42.
24.

25.

26. Any such function can be constructed by distributing the elements of A on three tiers.
The bottom tier contains the constant value, c = f(f(x)) for any x. (Obviously f(c) = c)
The middle tier contains k elements x ≠ c such that f(x) = c, where 1 ≤ k ≤ 6..
The top tier contains 6 – k elements such that f(x) equals an element on the middle tier.

ways to choose the elements on the middle tier, and then k6-
6
There are 7 choices for c. Then for a given k, there are ( )
k

k ways to draw arrows down from elements on the top tier to elements on the middle tier.
6
6
Thus N = 7 ⋅ ∑ ( ) ⋅ k
6−k
, giving the answer 99.
= 7399
k
k=1
27.

Since ∠DBE = 90O, DE is a diameter of ω. Then ∠DFE = ∠DGE = 900. But DF = FE, so ΔDEF is a 45-45-90 triangle. Letting DG
5x
= 3x, We have that EG = 4x, DE = 5x, and DF = EF = .
√2

Note that ΔDGE~ΔABC by SAS similarity, so ∠BAC = ∠GDE and ∠ACB = ∠DEG. Since DEFG is cyclic quadrilateral,
∠BAC = ∠GDE = 1800 −∠EFG = ∠AFE and ∠ACB = ∠DEG = ∠GFD, implying that ΔAFE and ΔCDD are isosceles. As a result, AE
= CD = , so BE .
5x 5x 5x
= 3 − and BD = 4 −
√2 √2 √2

2 2

Finally, using the Pythagorean Theorem on ΔBDE, (3 −

5x 5x 2
) + (4 − ) = (5x)
√2 √2

5 √2 25√2
Solving for x, we get that x = , so DE = 5x = . Thus, the answer is
√14 14

25 + 2 + 14 = 41.

28.

29. We prove that the given system has no integer solution if n = 4, and therefore, it has no integer solution of all n ≥ 4.
Note that if a k ∈ Z then
⎧ 1 ( mod 8), if k ≡ ±1 (mod 4),
⎪
2
k ≡ ⎨ 0 (mod 8), if k ≡ 0 (mod 4),
⎩
⎪
4 (mod 8), if k ≡ 2 ( mod 4).

This implies that for any two integers k, ℓ we have

⎧ 2,1,5 ( mod 8), if k ≡ ±1 (mod 4),
⎪
2 2
k + ℓ ≡ ⎨ 1,0,4 (mod 8), if k ≡ 0 (mod 4),
⎩
⎪
5,4,0 (mod 8), if k ≡ 2 ( mod 4).

Assume that there are integers x, y1, y2, y3, y4 satisfying

2 2 2 2 2 2 2 2
(x + 1) + y = (x + 2) + y =. (x + 3) + y = (x + 4) + y
1 2 3 4

Then, due to the fact x + 1, x + 2, x + 3, x + 4 form a complete congruent

System modulo 4, there exits an integer m such that
m ∈ {2; 1; 5} ∩ {1; 0; 4} ∩ {5; 4; 0} = φ

Which is impossible. Thus there is no integer solution for n ≥ 4.

When n = 3 we can see that (− 2, 0, 1,0) is a solution. Thus nmax = 3.
30. We prove that m(1) + m(2) +….+ m(2n) = 1 + 2n-2 (n+1) for all n ≥ 1, giving an answer of 1+25. 8 = 257. First note that m(2n) = 1,
and that we can view 0, 1, …..,2n-1 as n digit binary sequences by appending leading zeros as necessary. (Then m(0) = 0.) Then
for 0 ≤ x ≤ 2n-1, x and 2n –x are complementary n digit binary sequences (0’s and 1’s), with x’s strings of 1’s (0’s)
corresponding to 2n–x strings of 0’s (resp.1’s). It follows that m(x) + m(2n – x) is simply 1 more than the number of digit
changes in x (or 2n–x), i.e. the total number of 01 and 10 occurrences in x. Finally, because exactly half of all n – digit binary
sequences have 0,1 or 1,0 at positions k, k +1(for 1 ≤ k ≤ n –1 fixed), we conclude that the average values of m(x) + m(2n–x) is
n−1 n+1 n+1
, and thus that the total value of m(x) is ad desired.
1 n 2n−2
1 + = .2 . = (n + 1) ,
2 2 2 2

We prove that m(1) + m(2) +…..+ m(2n–1)

= 2n–2
(n+1). Identify each block of 1’s with its rightmost 1. Then it suffices to count
the number of these ‘’rightmost 1’s’’. For each 1 ≤ k ≤ n –1, the kth digit from the left is a rightmost 1 is and only if the k and k
+ 1 digits from the left 1 and 0 respectively. Thus, there are 2n-2 possible numbers. For the rightmost digit, it is a rightmost 1
if and only if it is a 1, so there are 2n-1 possible numbers. Sum up we have (n–1)2n-2 +2n-1 = 2n-2 (n+1), as desired.