Difference between revisions of "1998 CEMC Gauss (Grade 7) Problems/Problem 6"
Coolmath34 (talk | contribs) (Created page with "== Problem == In the multiplication question, the sum of the digits in the four boxes is: [Multiply <math>879 \times 492</math> using long multiplication. Find the sum of the...") |
De-math-wiz (talk | contribs) m (→Solution) |
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-edited by coolmath34 | -edited by coolmath34 | ||
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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. | ||
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| + | 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 | ||
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 
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.
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
