The greatest number that divides 43, 91, and | vedclass

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Question

(C) Let the greatest number be $x$ and the common remainder be $k$.Then,we have:$43 = nx + k$ $...(1)$$91 = mx + k$ $...(2)$$183 = lx + k$ $...(3)$Sub

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Let the greatest number be $x$ and the common remainder be $k$.Then,we have:$43 = nx + k$ $...(1)$$91 = mx + k$ $...(2)$$183 = lx + k$ $...(3)$Subtracting the equations:$(2) - (1) \Rightarrow 91 - 43 = (m - n)x \Rightarrow 48 = (m - n)x$$(3) - (2) \Rightarrow 183 - 91 = (l - m)x \Rightarrow 92 = (l - m)x$$(3) - (1) \Rightarrow 183 - 43 = (l - n)x \Rightarrow 140 = (l - n)x$The greatest number $x$ is the $H.C.F.$ of the differences $48, 92,$ and $140$.Prime factorization:$48 = 2^4 	imes 3$$92 = 2^2 	imes 23$$140 = 2^2 	imes 5 	imes 7$$H.C.F. = 2^2 = 4$.Thus,the greatest number is $4$.

The greatest number that divides $43, 91,$ and $183$ so as to leave the same remainder in each case is:

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