CMI Entrance Exam 2022 Question Paper Solution B.Sc Maths & Computer 23 May

CMI BSc entrance make-up exam on May 23, 2022
Part A, Correct answers
1. True
2. True
3. True
4. False
5. True
6. False
7. True
8. True
9. True
10. False
11. False
12. False
13. True
14. True
15. False
16. False
17. False
18. True
19. False
20. True
21. False
22. False
23. False
24. False
25. True
26. False
27. True
28. False
29. True
30. False
31. False
32. True
33. True
34. False
35. True
36. True
37. False
38. False
39. False
40. False

CHENNAI MATHEMATICAL INSTITUTE
Undergraduate Programme in Mathematics and Computer Science/Physics
Solutions of the 23rd May 2022 exam

Note: The solutions below consist only of main steps and strategies and do not contain all
the details expected in the exam.

B1. [12 points] Let N = {1, 2, 3, 4, 5, 6, 7, 8, 9} and L = {a, b, c}.

1. Suppose we arrange 12 elements of L ∪ N in a line such that no two of the three letters
occur consecutively. If the order of the letters among themselves does not matter, find
the number of such arrangements.

2. Find the number of functions from N to L such that exactly 3 numbers are mapped
to a, b, c.

3. Find the number of onto functions from N to L.

Solution:

1. Imagine that the numbers are placed with a gap between two consecutive numbers.
Moreover there is a gap before 1 and one after 9. So there are 10 gaps, where letters
can be placed. In order to find the number of such arrangement
first choose 3 gaps
from 10 and then permute the 9 numbers. The final answer is 10 3
· 9!.

2. Nine digits have to be placed in 3 groups. The answer is 93 · 63 · 33 .





3. By inclusion-exclusion principal the answer is

39 − 3 · 29 + 3 · 19 .

The first summand is the total number of functions, the second summand is the number
of function which miss 2 values and last summand is the number of functions which
miss 1 value.

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B2. [12 points] Let f function from natural numbers to natural numbers that satisfies

f (n) = n − 2, for n > 3000;
f (n) = f (f (n + 5)) for n ≤ 3000.

Show that f (2022) is uniquely determined and find its value.
Solution: The important step is to discover the following closed form expression for the
function:
f (n) = 2999 + (n + 2)mod(3).
The equality can be derived in a number of ways, including induction. Once this is established
it is straightforward to conclude that f (2022) = 3001.

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B3. [14 points] In △ABC, ∠BAC = 2∠ACB and 0◦ < ∠BAC < 120◦ . A point M
is chosen in the interior of △ABC such that BA = BM and M A = M C. Prove that
∠M CB = 30◦ . See the schematic figure below (not to scale).
C

M

A B

Solution: There are various ways to solve this problem. Let us start with a geometric
approach.
Construct a line segment CD such that it is equal to AB and ∠DCA = ∠BAC. Denote the
intersection of AD and BC by E. First step, show that △BAC ≡ △DCA. Second step,
show that △CDE ≡ △ABE. Next, show that △EM A ≡ △EM C. Now conclude that
∠CAB = 90◦ , this will lead to the solution.
Another construction is to draw a line segment BD such that it is equal to CD and is parallel
to AC. First show that ABDC is an isosceles trapezium. Use it prove that △ABM ≡
△CM D. Conclude that △BM D is equilateral. Use this along with the usual angle relations
to find the exact value.
One can also use trigonometric techniques, like sine rule, to solve this problem.

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B4. We want to find a nonzero polynomial p(x) with integer coefficients having the following
property.  
p(x) 1
Lettingq(x) := , q(x) = q .
x(1 − x) 1−x

Solution: First derive the following relationship
 
3 1
p(1 − x) = −x p .
x

This implies that the degree of the polynomial is at most 3. An easy check shows that linear
polynomials can’t satisfy this relation. Substituting p(x) = ax2 + bx + c in the above relation
gives us that
p(x) = ax(x − 1).
Let the cubic be of the form

p(x) = ax3 + bx2 + cx + d.

Then any cubic whose coefficients satisfy the following is a candidate.

a + d = 0, c − b − 3a = 0, 3b + c + 3a = 0, b + c + d = 0.

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B5. [14 points] Let R+ denote the set of positive real numbers. A bijection f : R+ → R+
is called golden if f ′ (x) = f −1 (x) for every x ∈ R+ .

1. Find all golden functions (if any) of the form f (x) = axb . Find all golden functions (if
any) of the form f (x) = abx .

2. Show that there is no bijection f : R → R such that f ′ (x) = f −1 (x).

Solution:

1. When f (x) = axb equate the derivative with the inverse to get b2 − b − 1 = 0. Whose
roots are the golden ration, ϕ, and its negative inverse, 1 − ϕ. The answer is
x 1−ϕ
f (x) = (x(ϕ − 1))ϕ OR f (x) = ( ) .
ϕ−1

On the other hand f (x) = abx is not a candidate for the golden function since the
derivative is an exponential and the inverse is a logarithmic function.

2. If f is a bijection defined on the entire real line then the derivative doesn’t change its
sign. However, the inverse does change sign. Hence it is impossible that the derivative
is the inverse of such a function.

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B6. [14 points] Suppose n > 1 is a natural number which is not congruent to 3 modulo 4.
Prove that there exist 1 ≤ i < j ≤ n such that the following is a perfect square
1!2! · · · n!
.
i!j!

Solution: The important observaion needed is that k!(k −1)! = k((k −1)!)2 . Case 1: n = 4k
for some k. In that case the numerator simplifies as follows:

1!2! · · · (4k)! = 4k((4k − 1)!)2 · · · 2 · 1
= (4k)(4k − 2) · · · 2 · M 2
= 22k · 2k · (2k − 1) · · · 1 · M 2
= (2k)! · N 2 .

Here M 2 is the product of the terms of the form ((4k − 1)!)2 and N 2 includes the factor
(2k )2 . Hence one can choose i = 1 and j = 2k. The other two remaining cases of n can be
dealt in a similar fashion.

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