Difference between revisions of "1998 CEMC Gauss (Grade 7) Problems/Problem 6"
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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 | 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 | 879 | ||
| − | x | + | x 492 |
1,758 | 1,758 | ||
79,110 | 79,110 | ||
+351,600 | +351,600 | ||
| + | |||
=432,468 | =432,468 | ||
| − | Therefore add 1+9+1+2 = 13 | + | Therefore add 1+9+1+2 = 13, so (A) 13 is correct answer. |
| − | |||
Revision as of 16:13, 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
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.
