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# number property

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Manager
Joined: 04 Sep 2006
Posts: 113

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27 Jun 2009, 22:33
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When divided by 7, the remainder is 1; when divided by 3, the remainder is 2. How many positive integers less than 500 can satisfy it?

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SVP
Joined: 29 Aug 2007
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27 Jun 2009, 23:19
vcbabu wrote:
When divided by 7, the remainder is 1; when divided by 3, the remainder is 2. How many positive integers less than 500 can satisfy it?

x = 7k+1
x = 3m +2

7k+1 = 3m +2
7k = 3m +1

7k = 3m +1
k = 1, m = 2 and x = 8
k = 4, m = 9 and x = 29
k = 7, m = 16 and x = 50

In the above equation:- The change in k's value is by 3, m's value by 7 and x's value = 21.

x i.e. divisibly by 7 and 3 leveas 1 and 2 reminders respectively are 8+ 21y < 500. Y must be 22 and considering 8 too, then umber of integers below 500 i.e. has 1 and 2 reminders when divided by 7 and 3 respectively are 23.

Alternatively:
using k's value: ((500 - 8)/7)/3 + 1 = 23
using x's value: (500 - 8)/(3/7) = 23
using x's value: (500 - 8)/21 + 1 = 492/21 + 1 = 23

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Re: number property   [#permalink] 27 Jun 2009, 23:19
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