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# When an integer Z is divided by 3, it leaves 1 as a

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SVP
Joined: 03 Feb 2003
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When an integer Z is divided by 3, it leaves 1 as a [#permalink]  15 Jul 2003, 08:54
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When an integer Z is divided by 3, it leaves 1 as a remainder. When Z is divided by 5, it leaves 3 as a remainder. When Z is divided by 7, it leaves 5 as a remainder.

1. Find the smallest positive Z.
2. Find the next smallest positive Z.
Manager
Joined: 08 Apr 2003
Posts: 149
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Kudos [?]: 8 [0], given: 0

This one was dificult indeed.
I got the first one..
it's 103.

I shall try to post my soltion again after some thought
Manager
Joined: 07 Jul 2003
Posts: 56
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Please submit the solution too (Explaination) which book do I get such problems.

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Attain Moksha

SVP
Joined: 03 Feb 2003
Posts: 1608
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Kudos [?]: 76 [0], given: 0

satgates wrote:
Please submit the solution too (Explaination) which book do I get such problems.

The question is mine; probably you will not find it in books.
Meanwhile, the solution is not that difficult.

Z=3n+1
Z=5k+3
Z=7p+5 the trick is to see that each time a quotinent is 2 more than a remainder. Therefore, add 2 to both parts and get what you need.

Z+2=3n+1+2=3n+3, divisible by 3
Z+2=5k+3+2=5k+5, divisible by 5
Z+2=7p+5+2=7k+7, divisible by 7

so, Z+2 is divisible by 3, 5, and 7.

To find the smallest positive Z, we have to subtract 2 from the LCM of 3, 5, 7.

LCM = 3*5*7, since they all are primes = 105

To find the second smallest number: 2LCM-2=210-2=208 and so on.

Do you like the question? It is one of my best.
Manager
Joined: 08 Apr 2003
Posts: 149
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Kudos [?]: 8 [0], given: 0

Excellent!!! Can i say anything better than that... Well i logged to post my second answer but saw your post and was pleased to see it. The idea behind the question is simply brilliant.
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
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Kudos [?]: 41 [0], given: 0

This is one of the best solutions I have ever seen.
For these kinds of problems I simply used to mutiply the divisors and get a product. The desired number should lie in the visinity of this product. I used to search around this number using trial and error.
stolyar's solution has made it eaven easier.
Manager
Joined: 28 Jan 2004
Posts: 203
Location: India
Followers: 2

Kudos [?]: 14 [0], given: 4

What a great logic stolyar.........i use to beat my head with such questions,but looks to me that it will not happen any more.........Kudos man for such a nice solution.
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