Find the greatest number of four digits which | vedclass

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(A) Step $1$: Observe the difference between the divisors and their respective remainders:$10 - 4 = 6$$15 - 9 = 6$$21 - 15 = 6$$28 - 22 = 6$Since the

Vedclass · Accepted Answer

$9654$

Vedclass · Answer

(A) Step $1$: Observe the difference between the divisors and their respective remainders:$10 - 4 = 6$$15 - 9 = 6$$21 - 15 = 6$$28 - 22 = 6$Since the difference is constant $(6)$,the required number is $(L.C.M. 	ext{ of } 10, 15, 21, 28) 	imes k - 6$.Step $2$: Find the $L.C.M.$ of $10, 15, 21, 28$.$10 = 2 	imes 5$$15 = 3 	imes 5$$21 = 3 	imes 7$$28 = 2^2 	imes 7$$L.C.M. = 2^2 	imes 3 	imes 5 	imes 7 = 420$.Step $3$: Find the greatest $4$-digit number divisible by $420$.The greatest $4$-digit number is $9999$.$9999 \div 420 = 23$ with a remainder of $339$.So,$9999 - 339 = 9660$.Step $4$: Subtract the constant difference $(6)$ from this number.$9660 - 6 = 9654$.

Find the greatest number of four digits which when divided by $10, 15, 21$ and $28$ leaves $4, 9, 15$ and $22$ as remainders,respectively.

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