Question

when a particular two digit number is divided by 2 or 4 it leaves a reminder 1 . When it is divided by 5 or 6 ,it leaves a reminder 3 .The number is excatly divisible by 11 .express the number as a product of prime numbers​

Answer 1

Answer:

25201

Suppose we are given that a number when divided by x, y, and z, leaves a remainder of a, b, and c; then the number will be of the format of

LCM(x,y,z)*n + constant

The key in these questions is finding out the value of 'constant'. If all of them leave the same remainder 'r', constant = r. It can also be looked at as the smallest number satisfying the given property.

In this question, the number leaves a remainder of 1 from 2, 3, 4, 5, 6, 7, 8, 9, and 10.

So, the number N = LCM(2,3,4,5,6,7,8,9,10)*n + 1 = 2520n + 1

So, any number which is of the format of 2520n + 1 will satisfy the given conditions.

Some such numbers are 2521, 5041, 7561, ...

We want to find such a number which follows the above rule and is still divisble by 11.

Rem[(2520n+1) / 11] = 0

=> Rem[(n+1) / 11] = 0 {Because 2519 is divisible by 11}

=> n = 10

=> N = 2520*10 + 1 = 25201 is our answer

Let me add 25201 is just one of such numbers. You can also put n as 21, 32,..