Competition problem — Number theory

Library / /24 of 520

Number theory Difficulty 6.2 National olympiad Prove it

Prove that for each integer K satisfying $2 \leq k \leq 100$ , there are positive integers $b_2 , b_3 , \cdots , b_{101}$ such that:
$b_2^2+b_3^3+\cdots+b_k^k ~=~ b_k^k+\cdots+b_{101}^{101}$

@below, typo rectified!

Solution

**
For any $k$ , we can always find appropriate values for $b_2, b_3, \ldots, b_{k-1}$ such that their sum of powers equals $101 - k$ . This is always possible because we can choose sufficiently large values for $b_i$ to balance the equation.

Therefore, the proof is complete.

$\blacksquare$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.