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m is a three-digit integer such that when it is divided by

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m is a three-digit integer such that when it is divided by [#permalink] New post 12 Aug 2017, 10:13
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m is a three-digit integer such that when it is divided by 5, the remainder is y, and when it is divided by 7, the remainder is also y. If y is a positive integer, what is the smallest possible value of m?

enter your value

[Reveal] Spoiler: OA
106

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Re: m is a three-digit integer such that when it is divided by [#permalink] New post 12 Aug 2017, 13:44
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Carcass wrote:
m is a three-digit integer such that when it is divided by 5, the remainder is y, and when it is divided by 7, the remainder is also y. If y is a positive integer, what is the smallest possible value of m?

enter your value

[Reveal] Spoiler: OA
106


There's a nice rule that say, "If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3

When m is divided by 5, the remainder is y
So, m = 5k + y for some integer k

When m is divided by 7, the remainder is y
So, m = 7j + y for some integer j

Since both equations are set to equal m, we can write: 5k + y = 7j + y
Subtract y from both sides to get: 5k = 7j
Well, 5k represents a multiple of 5, and 7j represents a multiple of 7
So, what's the smallest 3-digit number that is a multiple of 5 AND a multiple of 7?

The smallest 3-digit number is 100
100 is a multiple of 5, but it's NOT a multiple of 7

Next we have 105
105 is a multiple of 5, AND it's a multiple of 7
Now be careful. This does NOT mean that m = 105

When we divide 105 by 5 we get a remainder of 0, but we're told that the remainder (y) is a POSITIVE INTEGER.
To MINIMIZE the value of m, we need a super small remainder.
The smallest possible non-zero remainder is 1.
105 + 1 = 106

So, 106 is the smallest possible 3-digit value of m.

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Re: m is a three-digit integer such that when it is divided by   [#permalink] 12 Aug 2017, 13:44
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