Maths Olympiad Daily Problems
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BEST OF LUCK EVERYONE FOR THEIR NMTC EXAM
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(Mera toh Adv ka tst h🤡)
Qn 96 is an upgraded daddy version of the 2025 F(n) = remainder left when n^n is divided by 7 qn
Basically a^p(p-1) congruent 1 mod p²
So by concept of order where d is the order then d divides p=> d=p or d | p-1 => a^d congrunt 1 mod p² => (a^d)^k congruent 1 mod p² => a^(dk) congrunt 1 mod p² => a^(p-1) congruent 1 mod p² => a^p congrunt a mod p² meaning it's 1/a mod p²
If d>p then d=p(some prime factor of p-1). = pm for some m
If m = Odd: a^p congruent 1 mod p
If m = Even: a^p congrunt ±1 mod p
Ideally we would need congruent -1 mod p to get the least p
Goal: Parameterise p s.t. we can produce maximum -1 while satisfying the conditions in the qn
This was my approach
You can have a fresh approach or try to parameterise something from here to get the n
For any positive integer n, let τ (n) denote the number of positive divisors of n. If n is a positive integer such that τ(n^2)/τ(n)= 3, compute τ(n^7)/τ(n).
Ari repeatedly rolls a standard, fair, six-sided die. Let R(n) be the n
th number rolled, and let
Q(n) = R(1)· R(2)· . . . · R(n). Let the probability that there exists an n such that Q(n) = 100 and for
all m < n, Q(m) is not a perfect square be p/q
where p, q are relatively prime positive integers. Find
the largest prime factor of p + q.
