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# Is there an integer that leaves a remainder of 11 when

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SVP
Joined: 03 Feb 2003
Posts: 1607
Followers: 6

Kudos [?]: 84 [0], given: 0

Is there an integer that leaves a remainder of 11 when [#permalink]  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?
Manager
Joined: 26 Aug 2003
Posts: 233
Location: United States
Followers: 1

Kudos [?]: 2 [0], given: 0

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?
SVP
Joined: 03 Feb 2003
Posts: 1607
Followers: 6

Kudos [?]: 84 [0], given: 0

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.
SVP
Joined: 30 Oct 2003
Posts: 1794
Location: NewJersey USA
Followers: 5

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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.
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