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.

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

879 x 492 = 432,468

       879
    x  492
   _______
     1,758
    79,110
 + 351,600
 _________
 = 432,468

Therefore when 1, 9, 1, 2 are added together the total is 13, so (A) 13 is correct answer.

-edited by De-math-wiz

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1998 CEMC Gauss (Grade 7) Problems/Problem 6

Problem

Solution