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Question

In all numbers, the product of first and last digits is a multiple of the sum of middle two digits.

Thus, $4 	imes 7=35$ is a multiple of $(6+1)$

Vedclass · Accepted Answer

$5412$

Vedclass · Answer

In all numbers, the product of first and last digits is a multiple of the sum of middle two digits.

Thus, $4 	imes 7=35$ is a multiple of $(6+1)$ i.e. $7$ and $\operatorname{so}$ on.

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Given set : $4718, 5617, 6312, 8314$

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Choose the correct alternative for the given relation. Given set : $4718, 5617, 6312, 8314$

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