Question

A number when divides 41 and 68 leaves 5
as the remainder. What is the number?​

Answer 1

Here is your solution ==>

Let the number be x.

Condition=> When x divided by 41 and 68 leaves remainder 5.

So, ATQ,

x is divisible by 41-5 & 68-5

x is divisible by 36 & 63

So, we have to find HCF of 36 & 63

HCF = 9

Therefore, x= 9.

hopefully it helps!

Answer 2

Answer:

9

Step-by-step explanation:

As the number leaves remainder 5 on dividing 41 and 68 so 5 less from 41 and 68 must be exactly divisible by this number

41-5=36

68-5=63

Now we will find the hcf of 36 and 63

36=2*2*3*3

63= 3*3*7

Hcf=9

therefore 9 Is the required number