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May 9th, 2016, 10:34 PM   #1
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when a number is successively divided by 7,5 and 4 it leaves remainders of 4,2 and 3 respectively what will be the respective remainders when the smallest such number is successively divided by 8,5,6?

Can you please help me out?
Prudhvi raj k is offline  
 
May 9th, 2016, 11:29 PM   #2
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I get 3, 2 and 1 for the remainders. I did this by brute force.
To speed up the brute force, I can see that because:
1) dividing by 5 gives remainder 2, the last digit of the number must be a 2 or 7
2) dividing by 4 gives remainder 3, the number must be odd
The last digit must be a 7
Now, if we take the remaining piece of information:
3) dividing by 7 gives remainder 4
If we start with a 4 and keep adding 7's until we get a number ending in 7, that is our answer.
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May 10th, 2016, 07:07 AM   #3
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Quote:
Originally Posted by Prudhvi raj k View Post
when a number is successively divided by 7,5 and 4 it leaves remainders of 4,2 and 3 respectively what will be the respective remainders when the smallest such number is successively divided by 8,5,6?

Can you please help me out?
The three conditions translate to

$\displaystyle x \equiv 4$ mod 7
$\displaystyle x \equiv 2$ mod 5
$\displaystyle x \equiv 3$ mod 4

This can be solved by the Chinese Remainder Theorem.

The solution is $\displaystyle x \equiv 67$ mod 140

and the answer to your question is 3, 2, 1
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May 14th, 2016, 01:42 PM   #4
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Quote:
Originally Posted by Prudhvi raj k View Post
when a number is successively divided by 7,5 and 4 it leaves remainders of 4,2 and 3 respectively what will be the respective remainders when the smallest such number is successively divided by 8,5,6?

Can you please help me out?
Call the "number" x. As mrtwhs said, the conditions are equivalent to
x= 4 (mod 7), x= 2 (mod 5), and x= 3 (mod 4)

x= 4 (mod 7) is the same as x= 7i+ 4 for some integer i. Taking that equation "mod 5", x= 2i+ 4= 2 (mod 5) or 2i= -2 and then i= -1= 4 (mod 5). So i= 5j+ 4 for some integer j and then x= 7(5j+ 4)+ 4= 35j+ 28+ 4= 35j+ 32. Taking that equation "mod 4", since 32 is a multiple of 4 and 35= 32+ 3, we have 3j= 3 (mod 4). So j= 1 (mod 4), j= 4k+ 1 for some integer k and x= 35j+ 32= 35(4k+ 1)+ 32= 140k+ 35+ 32= 140k+ 67.

Since we were asked to use "the smallest such number", take k= 0 and x= 67. 67 divided by 8 has remainder 3, divided by 5 has remainder 2, and divided by 6 has remainder 1.
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