MathDB

3

Part of 2001 Cuba MO

Problems(2)

sum of digits of m = n(2n-1) is equal to 2000.

Source: 2001 Cuba MO 2.3

9/15/2024
Prove that there is no natural number n such that the sum of all the digits of the number m, where m=n(2n−1)m = n(2n-1) is equal to 20002000.
number theorysum of digits
(2n+1)^3-2 is sum of 3n-1 perfect squares

Source: 2001 Cuba MO 1.3

9/15/2024
Let nn be a positive integer. a) Prove that the number (2n+1)3−(2n−1)3(2n + 1)^3 - (2n - 1)^3 is the sum of three perfect squares. b) Prove that the number (2n+1)3−2(2n+1)^3-2 is the sum of 3n−13n-1 perfect squares greater than 11.
number theoryPerfect Squares