Question

Find the smallest 5digit number which when divided by 5,7 or 9 leaves 2 as the remainder

Answer 1

Answer:

First we need to find out a 5 digit number which would be divisible by 5,7,9

So, (5-2) = 3

(7-2) = 5

(9-2) = 7

LCM = 3x5x7 = 105

Smallest 5 digit number would be 10000

SO, 10000/105 = we get 9975 as quotient and 25 as remainder.

So 9975 is the number completely divisible by 105 but its a 4 digit number so we need to add 105 to it so, 9975 + 105 = 10080 is the least 5 digit number completely divisible by 105.

But we need a number which leaves remainder 2 so,

10080 + 2 = 10082 is the number we want which when divided by 5,7,9 will leave remainder 2.

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