Question

A Number When Divides 41 And 68 Leaves 5 As The Remainder. What Is The Number?
I Am A 6Grader

Answer 1

Answer:

9

Step-by-step explanation:

41/9 leaves 5 as the remainder

68/9 also leaves 5 as the remainder

Answer 2

Answer:

Let the number be x

41 = 5(modx)

68 = 5(modx)

=> 68 - 41 will be divisible by the number (Since remainder is the same)

=> 27 = kx

Factors of 27 = 1,3,9,27

Going back to the original numbers

There can be no 5 mod (1) number, Remainder is always 0

There can be no 5 mod(3) number, since remainder is always 1 or 2

41 is not 5(mod27) so 27 is out too.

Only option is 9, since there can be numbers that are 9(mod5). We can also quickly check that 41 = 36+5 and 68 = 63 +5.

Hence 9 is the answer.