Find the least number of five digits which wh | vedclass

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Question

(A) Step $1$: Find the $L.C.M.$ of $12, 16, 21, 36, 40$.$12 = 2^2 	imes 3$$16 = 2^4$$21 = 3 	imes 7$$36 = 2^2 	imes 3^2$$40 = 2^3 	imes 5$$L.C.M.

Vedclass · Accepted Answer

$10088$

Vedclass · Answer

(A) Step $1$: Find the $L.C.M.$ of $12, 16, 21, 36, 40$.$12 = 2^2 	imes 3$$16 = 2^4$$21 = 3 	imes 7$$36 = 2^2 	imes 3^2$$40 = 2^3 	imes 5$$L.C.M. = 2^4 	imes 3^2 	imes 5 	imes 7 = 16 	imes 9 	imes 5 	imes 7 = 5040$.Step $2$: The smallest $5$-digit number is $10000$.Divide $10000$ by $5040$: $10000 = 5040 	imes 1 + 4960$.Step $3$: To find the smallest $5$-digit number divisible by $5040$,we add $(5040 - 4960) = 80$ to $10000$.Number $= 10000 + 80 = 10080$.Step $4$: To get a remainder of $8$ in each case,add the remainder to the multiple of $L.C.M.$Required number $= 10080 + 8 = 10088$.

Find the least number of five digits which when divided by $12, 16, 21, 36$ and $40$ leaves a remainder of $8$ in each case.

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