Question

find.the.smallest.5digit.number.which.when.divided.by.5, 7or9leaves2as.the.remainder​

Answer 1

Answer:

LCM of 5, 7 or 9 = 315

The number will be of the form 315k + 2 where k is a natural number.

By inspection the value of k for which 315k + 2 is a five-digit number is 32

Hence, the number = 315 x 32 + 2

= 10082

Step-by-step explanation:

INSPECTION:

[3] [1] [5]

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Multiply 3 with a number such that the product is a two-digit number. Obviously, the number is 32

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315 x 32 = 10080 which is the smallest 5 digit number divisible by 5 , 7 or 9

Hence, k = 32