Find the least number of five digits which wh | vedclass

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(B) To find the least number of $5$ digits that leaves a remainder of $1$ when divided by $63, 56$,and $42$,we first find the $L.C.M.$ of these number

Vedclass · Accepted Answer

$10081$

Vedclass · Answer

(B) To find the least number of $5$ digits that leaves a remainder of $1$ when divided by $63, 56$,and $42$,we first find the $L.C.M.$ of these numbers.Prime factorization:$63 = 3^2 	imes 7$$56 = 2^3 	imes 7$$42 = 2 	imes 3 	imes 7$$L.C.M. = 2^3 	imes 3^2 	imes 7 = 8 	imes 9 	imes 7 = 504$.The smallest $5$-digit number is $10000$.Divide $10000$ by $504$:$10000 \div 504 = 19$ with a remainder of $424$.The smallest $5$-digit number exactly divisible by $504$ is $10000 + (504 - 424) = 10000 + 80 = 10080$.Since the required number must leave a remainder of $1$ in each case,we add $1$ to the $L.C.M.$ multiple:Required number $= 10080 + 1 = 10081$.

Find the least number of five digits which when divided by $63, 56$ and $42$ leaves remainder $1$ in each case.

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