Competition problem — Algebra — Maths Olympiad Prep

Library / /61 of 86

Algebra Difficulty 6.6 National olympiad Prove it Estonia

There are three consecutive positive integers on a blackboard. A move consists of erasing the smallest number on the blackboard and replacing it by the sum of itself and the greatest number on the blackboard. Is it possible that the sum of the numbers on the blackboard will be a power of 10:

a) after the 6th move?

b) after the 8th move?

Solution

Answer: (a) Yes; (b) No.

Let $x$ be the smallest of the three numbers initially on the blackboard. We will find the numbers on the blackboard after having made 0, 1, 2, 3, 4, 5, 6, 7 and 8 moves, and their sums:

Number of moves	Numbers on the blackboard	Sum of numbers
0	$x$ , $x+1$ , $x+2$	$3x + 3$
1	$x+1$ , $x+2$ , $x+3$	$4x + 5$
2	$x+2$ , $2x+2$ , $3x+3$	$6x + 7$
3	$2x+2$ , $3x+3$ , $4x+5$	$9x + 10$
4	$3x+3$ , $4x+5$ , $6x+7$	$13x + 15$
5	$4x+5$ , $6x+7$ , $9x+10$	$19x + 22$
6	$6x+7$ , $9x+10$ , $13x+15$	$28x + 32$
7	$9x+10$ , $13x+15$ , $19x+22$	$41x + 47$
8	$13x+15$ , $19x+22$ , $28x+32$	$60x + 69$

(a) The total sum of the numbers on the blackboard after 6 moves will be

28x + 32

. Either by direct computation or by considerations modulo 4 and 7, we observe that the equation

28x + 32 = 10^4

has an integer solution. Therefore it is possible that after the 6th move, the sum of the numbers on the blackboard will be a power of 10.

(b) The total sum of the numbers on the blackboard after 8 moves will be $60x + 69$ . As $60x$ and $69$ end with 0 and 9, respectively, their sum will also end with the digit 9. But such a number cannot be equal to a power of 10.

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: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.