2025 AMC 12B Problem 5

Attempt Problem 5 of the 2025 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

Positive integers

x

and

y

satisfy the equation

57x + 22y = 400.

What is the least possible value of

x + y?

10

11

13

14

15

Reduce

57x + 22y = 400

modulo

22

to pin down

x

Since

x

is positive and

57x \lt 400,

only a few values of

x

are possible

Modulo

22,

the equation gives

13x \equiv 4,

x \equiv 2 \pmod{22}.

With

57x \lt 400,

the only option is

x = 2,

which gives

22y = 286,

y = 13.

Then

x + y = 15.

Thus, the correct answer is E.

Problem 5 in other years