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 February 9th, 2015, 07:34 AM #1 Newbie   Joined: Feb 2015 From: parli Posts: 1 Thanks: 0 math problem 2272 and 875 divde both number by 3 digit number N.such that remainder will same?
 February 9th, 2015, 07:57 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,551 Thanks: 2556 Math Focus: Mainly analysis and algebra $$\begin{array}{c}2272 = an + r \\ 875 = bn + r\end{array} \implies (a-b)n = 1397$$ 1397 is semi-prime with only one of the factors having three digits. Last edited by v8archie; February 9th, 2015 at 08:05 AM.
February 9th, 2015, 12:43 PM   #3
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Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Hello, sabbardabeer!

Quote:
 Both 2272 and 875 are divisible by a 3-digit number $N$ such that the remainders are equal. $\;$Find $N$.

$\;\;\;\begin{Bmatrix}2272 &=& aN \,+\,R \\ 875 &=& bN\,+\,R \end{Bmatrix}\;\;\;\Rightarrow\;\;\; \begin{Bmatrix}R &=& 2272\,-\,aN & [1] \\
R &=& 875\,-\,bN & [2]\end{Bmatrix}$

Equate [1] and [2]: $\:2272\,-\,aN \:=\:875\,-\,bN \;\;\;\Rightarrow \;\;\;aN\,-\,bN \:=\:1397$

$\;\;\;(a\,-\,b)N \:=\:1397 \;\;\;\Rightarrow\;\;\;(a\,-\,b)N \:=\:11\,\cdot\,127$

Therefore: $\:N \:=\:127$

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