
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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May 9th, 2016, 11:34 PM  #1 
Newbie Joined: Mar 2016 From: India Posts: 14 Thanks: 0  HCF and LCM related
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? 
May 10th, 2016, 12:29 AM  #2 
Senior Member Joined: Apr 2014 From: UK Posts: 895 Thanks: 328 
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. 
May 10th, 2016, 08:07 AM  #3  
Senior Member Joined: Feb 2010 Posts: 702 Thanks: 137  Quote:
$\displaystyle x \equiv 4$ 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  
May 14th, 2016, 02:42 PM  #4  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896  Quote:
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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