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(A) Let the number be $x$. According to the problem,$7 	imes x = 555...5$ (a number consisting only of fives).To find the least such number,we start

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$79365$

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(A) Let the number be $x$. According to the problem,$7 	imes x = 555...5$ (a number consisting only of fives).To find the least such number,we start testing numbers consisting of $n$ fives,where $n = 1, 2, 3, ...$We need to find the smallest $n$ such that $55...5$ (n times) is divisible by $7$.$5/7$ (Remainder $5$)$55/7 = 7$ remainder $6$$555/7 = 79$ remainder $2$$5555/7 = 793$ remainder $4$$55555/7 = 7936$ remainder $3$$555555/7 = 79365$ remainder $0$.Since $555555$ is the smallest number consisting only of fives that is divisible by $7$,the quotient $79365$ is the least such number.Therefore,the least number is $79365$.

When a certain number is multiplied by $7$,the product consists entirely of fives; find the least value of such a number.

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