A general had his men line up in 5 rows; no men were left over.
He then had them line up in 8 rows; no men were left over.
He had them line up in 9 rows; no men were left over.
However, when they lined up in 11 rows, one column only had 10 men.
What is the smallest number of men the general could have had under his command?
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Thanks for clarifying the problem statement.
I think you need to reword the question. The Chinese general could have had 3 6 0 solders, which is divisible by 5 , 8 , 9 . Then he could have had 1 0 columns of 3 5 solders deep, and 1 column of 1 0 solders deep, making it 1 1 columns having a total 3 6 0 solders.
There seems to be a disconnect between the wording of your question and the way you've handled the matter of arranging the solders " 1 1 abreast".
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The way you described it they would be lined up 35 abreast, not 11.
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This is exactly what I mean about wording. We can have 1 0 rows of solders 1 1 wide = 1 1 0 solders, plus 2 5 more rows of 1 0 wide = 2 5 0 , making it 3 6 0 in all. What does abreast mean, if not "wide", i.e., "side-by-side"?
In the military, a column of solders in a formation is from front to back, while a line is from side to side, or abreast . You should at least mention in your problem that the total number of solders leaves a remainder of 1 0 when divided by 1 1 . Then 1 4 4 0 is indeed the correct answer. Or, you can simply replace "column" with "line", or "row", although the other way would probably be advisable.
Never thought about it that way, very clever.
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The number of men is an exact multiple of 5, 8, and 9. Since they are mutually coprime, the number of men must be a multiple of 5 * 8 * 9 = 360.
It is also of the form 11n + 10.
360 has a remainder of 8 when divided by 11. The smallest multiple of 8 that is of the form 11n + 10 is 32. So you have to multiply 360 by 4 (32/8) to get the number of men. 360 * 4 = 1440.