Question

find the smallest number which when divided by 3, 5 and 7 leaves a remainder 2 in each case, and is divisible by 11

Answer 1

Answer:

Answer is 737

Step-by-step explanation:

According to question,

The smallest no divisible by 5,3 and 7 and leaving reminder 2  and also divisible by 11

The smallest number divisible by 5 , 3 , 7 = LCM ( 5 , 3 , 7 )

LCM of 5 , 3 , 7 = 5 * 3 * 7

= 105

In each of the case remainder is 2

So,

The desired number = 105 + 2 = 107

But 107 is not divisible by 11

Therefore,

The number which when divided by 3, 5 and 7 leaves a remainder 2 and also is divisible by 11 is 737

As,

To be divisible by 11, it is necessary that difference between the digit's sum at odd places and digit's sum at even place is 0 or divisible by 11.

Here 737 = 7 + 7 - 3 = 11, which is divisible by 11

Also,

737 / 3 leaves remainder 2

737 / 5 also leaves remainder 2

737 / 7 also leaves remainder 2

Hence,

Answer is 737