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

879 x 492 = 432,468

x 492 ________

  1,758
 79,1100

+351,6000 _________ 432,468 therefore add 1+9+1+2 = 13 The Answer is (A) 13

Page

Toolbox

Search

1998 CEMC Gauss (Grade 7) Problems/Problem 6

Problem

Solution