MathDB

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Part of 2013 India IMO Training Camp

Problems(6)

Exponent of primes

Source: Indian IMOTC 2013, Practice Test 1, Problem 1

5/6/2013
For a prime pp, a natural number nn and an integer aa, we let Sn(a,p)S_n(a,p) denote the exponent of pp in the prime factorisation of apn−1a^{p^n} - 1. For example, S1(4,3)=2S_1(4,3) = 2 and S2(6,2)=0S_2(6,2) = 0. Find all pairs (n,p)(n,p) such that Sn(2013,p)=100S_n(2013,p) = 100.
number theory proposednumber theory
Non homogeneous inequality of degree n-2

Source: Indian IMOTC 2013, Practice Test 2, Problem 1

7/30/2013
Let a,b,ca, b, c be positive real numbers such that a+b+c=1a + b + c = 1. If nn is a positive integer then prove that (3a)n(b+1)(c+1)+(3b)n(c+1)(a+1)+(3c)n(a+1)(b+1)≥2716 . \frac{(3a)^n}{(b + 1)(c + 1)} + \frac{(3b)^n}{(c + 1)(a + 1)} + \frac{(3c)^n}{(a + 1)(b + 1)} \ge \frac{27}{16} \,.
inequalitiesinductionrearrangement inequalityinequalities proposed
Derangements

Source: Indian IMOTC 2013, Team Selection Test 1, Problem 1

7/30/2013
Let n≥2n \ge 2 be an integer. There are nn beads numbered 1,2,…,n1, 2, \ldots, n. Two necklaces made out of some of these beads are considered the same if we can get one by rotating the other (with no flipping allowed). For example, with n≥5n \ge 5, the necklace with four beads 1,5,3,21, 5, 3, 2 in the clockwise order is same as the one with 5,3,2,15, 3, 2, 1 in the clockwise order, but is different from the one with 1,2,3,51, 2, 3, 5 in the clockwise order.
We denote by D0(n)D_0(n) (respectively D1(n)D_1(n)) the number of ways in which we can use all the beads to make an even number (resp. an odd number) of necklaces each of length at least 33. Prove that n−1n - 1 divides D1(n)−D0(n)D_1(n) - D_0(n).
countingderangementfunctioncombinatorics proposedcombinatoricspermutationsPermutation cycles
Functional equation over reals

Source: Indian IMOTC 2013, Team Selection Test 2, Problem 1

7/30/2013
Find all functions ff from the set of real numbers to itself satisfying f(x(1+y))=f(x)(1+f(y)) f(x(1+y)) = f(x)(1 + f(y)) for all real numbers x,yx, y.
functioninductionlimitalgebra proposedalgebraCauchy equation
Double numbers

Source: Indian IMOTC 2013, Team Selection Test 4, Problem 1

7/30/2013
A positive integer aa is called a double number if it has an even number of digits (in base 10) and its base 10 representation has the form a=a1a2⋯aka1a2⋯aka = a_1a_2 \cdots a_k a_1 a_2 \cdots a_k with 0≤ai≤90 \le a_i \le 9 for 1≤i≤k1 \le i \le k, and a1≠0a_1 \ne 0. For example, 283283283283 is a double number. Determine whether or not there are infinitely many double numbers aa such that a+1a + 1 is a square and a+1a + 1 is not a power of 1010.
modular arithmeticDiophantine equationnumber theory proposednumber theory
Sum-friendly odd partitions

Source: Indian IMOTC 2013, Team Selection Test 3, Problem 1

7/30/2013
For a positive integer nn, a sum-friendly odd partition of nn is a sequence (a1,a2,…,ak)(a_1, a_2, \ldots, a_k) of odd positive integers with a1≤a2≤⋯≤aka_1 \le a_2 \le \cdots \le a_k and a1+a2+⋯+ak=na_1 + a_2 + \cdots + a_k = n such that for all positive integers m≤nm \le n, mm can be uniquely written as a subsum m=ai1+ai2+⋯+airm = a_{i_1} + a_{i_2} + \cdots + a_{i_r}. (Two subsums ai1+ai2+⋯+aira_{i_1} + a_{i_2} + \cdots + a_{i_r} and aj1+aj2+⋯+ajsa_{j_1} + a_{j_2} + \cdots + a_{j_s} with i1<i2<⋯<iri_1 < i_2 < \cdots < i_r and j1<j2<⋯<jsj_1 < j_2 < \cdots < j_s are considered the same if r=sr = s and ail=ajla_{i_l} = a_{j_l} for 1≤l≤r1 \le l \le r.) For example, (1,1,3,3)(1, 1, 3, 3) is a sum-friendly odd partition of 88. Find the number of sum-friendly odd partitions of 99999999.
combinatoricsInteger partitions