Difference between revisions of "1998 CEMC Gauss (Grade 7) Problems/Problem 6"
De-math-wiz (talk | contribs) m (→Solution) |
De-math-wiz (talk | contribs) m (→Solution) |
||
| Line 18: | Line 18: | ||
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 | ||
| − | + | 879 x 492 = 432,468 | |
| − | + | 879 | |
| − | x 492 | + | x 492 |
| − | + | 1,758 | |
| − | + | 79,110 | |
| − | + | + 351,600 | |
| − | +351,600 | + | = 432,468 |
| − | + | Therefore when 1, 9, 1, 2 are added together the total is 13, so (A) 13 is correct answer. | |
| − | =432,468 | ||
| − | |||
| − | Therefore | ||
Revision as of 17:24, 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 when 1, 9, 1, 2 are added together the total is 13, so (A) 13 is correct answer.
