1998 CEMC Gauss (Grade 7) Problems/Problem 6

Problem

In the multiplication question, the sum of the digits in the four boxes is:

[Multiply $879 \times 492$ using long multiplication. Find the sum of the four numbers in the thousands place column.]

$\text{(A)}\ 13 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 27 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 22$

Solution

Multiplying, we find that the numbers in the thousands place column are 1, 9, 1, and 2 respectively. Adding yields a sum of 12, so the answer is $\text{(B)}.$

-edited by coolmath34

If anyone knows the LaTeX to show long multiplication, any help would be appreciated.

-edited by De-math-wiz

Solution

While doing the long multiplication the numbers in the thousand place are 1, 9, and 2 respectively. When added the sum is 13, so the answer is (A) 13

879x492 = 432,468

    879
  x 492

________

  1,758
 79,110

+351,600 ________ =432,468

Therefore add 1+9+1+2 = 13, so (A) 13 is correct answer.

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1998 CEMC Gauss (Grade 7) Problems/Problem 6

Problem

Solution

Solution