Difference between revisions of "1998 CEMC Gauss (Grade 7) Problems/Problem 6"

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If anyone knows the LaTeX to show long multiplication, any help would be appreciated.
 
If anyone knows the LaTeX to show long multiplication, any help would be appreciated.
  
-edited by De-math-wiz
+
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 = 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.
  
    879
+
-edited by De-math-wiz
x    492
 
________
 
  1,758
 
  79,1100
 
+351,6000
 
_________
 
432,468 therefore add 1+9+1+2 = 13
 
The Answer is (A) 13
 

Latest revision as of 16:30, 12 May 2024

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