2011 AMC 12B Problem 4

Attempt Problem 4 of the 2011 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 2011 AMC 12B solutions, or check the answer key.

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

In multiplying two positive integers

a

and

b,

Ron reversed the digits of the two-digit number

a.

His erroneous product was

161.

What is the correct value of the product of

a

and

b?

116

161

204

214

224

Factor

161

into primes

The reversed

a

is the two-digit factor; reverse it back before multiplying

Since

161=7\cdot23,

the only two-digit factor is

23.

This must be the reversed value of

a,

so the true value of

a

32,

and

b=7.

The correct product is

32\cdot7=224.

Thus, the correct answer is E.

Problem 4 in other years