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(B) To find the number that is exactly divisible by $20, 28, 32$ and $35$,we must first calculate their Least Common Multiple $(L.C.M.)$.Prime factori

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$4714$

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(B) To find the number that is exactly divisible by $20, 28, 32$ and $35$,we must first calculate their Least Common Multiple $(L.C.M.)$.Prime factorization of the numbers:$20 = 2^2 	imes 5$$28 = 2^2 	imes 7$$32 = 2^5$$35 = 5 	imes 7$$L.C.M. = 2^5 	imes 5 	imes 7 = 32 	imes 35 = 1120$.We need to find a number $x$ such that $(5834 - x)$ is exactly divisible by $1120$. To make $(5834 - x)$ the largest possible multiple of $1120$ that is less than $5834$,we divide $5834$ by $1120$:$5834 \div 1120 = 5$ with a remainder of $234$.So,$5834 = (1120 	imes 5) + 234$.To get a number exactly divisible by $1120$,we must subtract the remainder $234$ from $5834$:$5834 - 234 = 5600$.However,the question asks for the number which when subtracted from $5834$ gives a result divisible by $1120$. If we subtract $234$ from $5834$,we get $5600$,which is divisible by $1120$. Thus,the number to be subtracted is $234$. Wait,re-evaluating the question: "The greatest number,which when subtracted from $5834$,gives a number exactly divisible by each of $20, 28, 32$ and $35$". This implies we want the result $(5834 - x)$ to be a multiple of $1120$. The multiples of $1120$ are $1120, 2240, 3360, 4480, 5600$. To make $x$ the greatest,$(5834 - x)$ must be the smallest multiple,which is $1120$.$5834 - x = 1120 \implies x = 5834 - 1120 = 4714$.

The greatest number,which when subtracted from $5834$,gives a number exactly divisible by each of $20, 28, 32$ and $35$,is

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