chesstitans wrote:

If A, B and C represent different digits in the multiplication, AAB * B = CB5B, then A + B + C = ?

9

12

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17

It does take time to solve this problem.

Questions like these are best approached with a mix of trial and error and logic.

This combines Alternative and Logical approaches.

First, the ones digit of AAB*B is B. This means that B*B is B (because the ones digit of the product is determined only by the ones digits of the factors!).

Only three numbers fit the pattern - 0,1, 5 and 6. 0 and 1 are clearly irrelevant so B must be 5 or 6.

Next, the tens digits AAB*B is 5 meaning that AB*B has a tens digit of 5.

If B=5 then A5*5 has a tens digit of 5 --> 50*A + 25 has a tens of 5 --> 5*A +2 has a ones digit of 5 --> 5*A has a ones digit of 3.

This is impossible so B must equal 6.

Let's verify:

If B=6 then A6*6 has a tens digit of 5 --> 60*A + 36 has a tens of 5 --> 6*A + 3 has a ones digit of 5 --> 6*A has a ones digit of 2 --> A equals 2 or 7.

So, our problem is now 226*6 = C656 or 776*6 = C656

Let's solve the first problem as it looks easier. 226*6 = 1200+120+36= 1356. But then B is both 3 and 6. Impossible! Then A must equal 7.

776*6 = 4200 + 420 + 36 = 4656 and C = 4. In this case A+B+C = 7+6+4=17 and (E) is our answer.

_________________

David

Senior tutor at examPAL

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