Find the greatest number which will divide 21 | vedclass

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(B) To find the greatest number that divides $2112$ and $2792$ leaving a remainder of $4$ in each case,we follow these steps:$1$. Subtract the remaind

Vedclass · Accepted Answer

$68$

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(B) To find the greatest number that divides $2112$ and $2792$ leaving a remainder of $4$ in each case,we follow these steps:$1$. Subtract the remainder from both numbers to make them perfectly divisible by the required number.$2112 - 4 = 2108$$2792 - 4 = 2788$$2$. Find the Highest Common Factor $(HCF)$ of $2108$ and $2788$.Prime factorization of $2108 = 2^2 	imes 17 	imes 31 = 4 	imes 17 	imes 31$Prime factorization of $2788 = 2^2 	imes 17 	imes 41 = 4 	imes 17 	imes 41$$3$. The $HCF$ is the product of the common prime factors:$HCF = 4 	imes 17 = 68$Therefore,the greatest number is $68$.

Find the greatest number which will divide $2112$ and $2792$ leaving the remainder $4$ in each case.

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