Ana has $22$ coins. She can take from her friends either $6$ coins or $18$ coins, or she can give $12$ coins to her friends. She can do these operations many times she wants. Find the least number of coins Ana can have.
Problem
Source: Kosovo Mathematical Olympiad 2022, Grade 9, Problem 1
Tags: combinatorics
StarLex1
06.03.2022 23:10
$22+18k+6l-12d = 0$ $22+18k+6l-12d=1$(not possible since all even) . . . . . $22+18k+6l-12d=n$ $2m=n$ we can also say that $m = -6d+9k+3l+11$ $0 = -6d+9k+3l+11$
$l=2d-3k-\frac{11}{3}$ obviously non int since d and k are integer
$1=-6d+9k+3l+11$
$l=2d-3k-\frac{10}{3}$ obviously non int since d and k are integer
$2 = -6d+9k+3l+11$ $l =2d-3k-3$ for $2d \geq 3k+3$ hence the least number that ana could have is 4 take example $(l,d,k) = (0,3,1)$ $22+18-36 = 4$
hhh1234
07.03.2022 06:59
mod 6 and ez game
Rijadinho
28.04.2023 00:10
Since $6 \equiv 12 \equiv 18 \equiv 0 \pmod{6}$ And $22 \equiv 4 \pmod{6}$ It is clear that 4 is the least amount of coins Ana can have $\blacksquare$