document.write( "Question 763573: find the greastest number that will divide 43,91,and 183 so as to leave the same remainder in each case \n" ); document.write( "

Algebra.Com's Answer #466517 by MaartenRU(13) $\"\"$ $\"About$

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The idea here is to look at the differences between the given numbers. If two numbers give the same remainder when divided by some other number, then their difference must give a remainder of zero when divided by that number:\r
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\n" ); document.write( "\n" ); document.write( "For example: 51 and 93 both give a remainder of 2 when divided by 7. Therefore the difference is divisible by 7: $\"93-51=42=6%2A7\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now, let's look at our numbers here. 91-43=48, 183-91=92, 183-43=140\r
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\n" ); document.write( "\n" ); document.write( "So we have the set of numbers {48, 92, 140}, and we want to know the biggest number that divides all these numbers. It's obvious that all these numbers are even, so they are at least divisible by 2. By doing that, we leave {24, 46, 70}. These are still all even, so we can do it again. The new set, {12,23,35}, has no common divisors. \r
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\n" ); document.write( "\n" ); document.write( "We got this set by dividing the original numbers by 4. So the greatest common divisor of {48, 92, 140} is 4. And that's your answer.\r
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\n" ); document.write( "\n" ); document.write( "Let's check:
\n" ); document.write( " $\"43%2F4=10\"$ with remainder 3.
\n" ); document.write( " $\"91%2F4=22\"$ with remainder 3.
\n" ); document.write( " $\"183%2F4=45\"$ with remainder 3. \n" ); document.write( "

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