Difference between revisions of "1998 CEMC Gauss (Grade 7) Problems/Problem 6"
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-edited by De-math-wiz | -edited by De-math-wiz | ||
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| + | == 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 | ||
879 x 492 = 432,468 | 879 x 492 = 432,468 | ||
| Line 19: | Line 22: | ||
879 | 879 | ||
x 492 | x 492 | ||
| − | + | ||
1,758 | 1,758 | ||
| − | 79, | + | 79,110 |
| − | +351, | + | +351,600 |
| − | + | =432,468 | |
| − | 432,468 | + | |
| + | Therefore add 1+9+1+2 = 13 | ||
The Answer is (A) 13 | The Answer is (A) 13 | ||
Revision as of 17:10, 12 May 2024
Problem
In the multiplication question, the sum of the digits in the four boxes is:
[Multiply 
using long multiplication. Find the sum of the four numbers in the thousands place column.]
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 
-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
879 x 492 = 432,468
879
x 492
1,758 79,110
+351,600 =432,468
Therefore add 1+9+1+2 = 13 The Answer is (A) 13
