Please ignore people saying it's "divisible by anything." Everyone should realize that we're talking about being evenly divisible (divisible, to anyone but an uninformed high school algebra student, always means "evenly" divisible).
Call your number n. We know 2 divides n, meaning n is even.
n must also be divisible by the following:
n
1
n/2
Giving four factors of n that we know:
1, 2, n/2, n
Keep in mind that if n=2, there are not actually four distinct factors (since n=2 and n/2=1 in this case).
2007-07-09 03:58:11
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answer #1
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answered by сhееsеr1 7
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Since the problem doesn't say what the number is, it could be 2 or 4 or 6 or 8, etc. If you consider 0 and negative numbers, then it could also be any one of 0, -2, -4, -6, etc. Look at 2, 4, 6, 8, and 10 2 is divisible by 1 and 2 (every integer is divisible by 1) 4 is divisible by 1, 2, and 4 6 is divisible by 1, 2, 3, and 6 8 is divisible by 1, 2, 4, and 8 10 is divisible by 1, 2, 5, and 10 All we know for sure is that the number is divisible by two, but we DON'T KNOW which one of the numbers it is. If you look at the sample above, what is true about ALL of those numbers? The answer is that all of them are divisible by both 1 and 2. If you allow for negative numbers, then -1 and -2 would also be divisors.
2016-05-17 17:27:02
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answer #2
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answered by Anonymous
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1
2007-07-09 03:22:39
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answer #3
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answered by Bruce O 3
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By 1, by that number itself, by half of that number and maybe by some other numbers.
2007-07-09 03:22:16
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answer #4
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answered by oregfiu 7
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you need to be more specific, the maximum is every number if the number being divided by is zero, since zero divided by anything is zero.
2007-07-09 03:22:47
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answer #5
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answered by grompfet 5
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All other numbers (except zero).
2007-07-09 03:28:53
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answer #6
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answered by Runa 7
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