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Is there an integer that leaves a remainder of 11 when

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Is there an integer that leaves a remainder of 11 when [#permalink] New post 04 Oct 2003, 07:24
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Is there an integer that leaves a remainder of 11 when divisible by 12 and a remainder of 1 when divisible by 8? If yes, why yes—if not, why not?
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 [#permalink] New post 04 Oct 2003, 19:29
I'll so no.

Because 12 and 8 have common factor 4. They differ by additional factors 3 and 2, respectively. And all numbers, when divided by, 2 and/or 3 leave either 0 or 1 as remainder. So if the number if divided by 8 leaves remainder 1, it cannot leave 11 when divided by 12. It can only leave multiple of 3 and 5. And 11 is not a multiple of either. If you reverse the situation then only multiples of 3 and 7 remain. So when you put all the remainders (from both remainder calculations) in a line and divide by 2, you get a remainder 1.

Is that confusing?
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 [#permalink] New post 04 Oct 2003, 23:05
I would say that it is confusing.
consider 25. When divided by 8, it leaves 1. When divided by 12, it leaves 1 again. You write that such a number (25) can leave multiples of 3 or 5 only.

My approach:
N=12n+11 and N=8k+1; equal them:
12n+11=8k+1, or
10=8k–12n, or
5=4k–6n, or
5=2*(2k–3n)—a clear contradiction. The rigth part is always even, but the left one is odd. Therefore, there is no such N.
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 [#permalink] New post 09 Jan 2004, 09:15
I did it in a slightly different way

12K+11 = 8P+1
so P = 12K/8 + 10/8 = 6K/4+5/4
Since 5/4 has a remainder of 1 when divided by 4 6K/4 has to have a remainder of 3 to make P an integer. There no such value of K for which 6K/4 yeilds remainder of 3. The remainder is either 0 or 2
So no such number exists.
  [#permalink] 09 Jan 2004, 09:15
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