31M5 is a multiple of 9 and two values are obtained by M. Why? Where M

1.	31M5 is a multiple of 9 and two values are obtained by M. Why? Where M is a digit?
Answer» Given Numbers is 31M5 ∴Sum of digits = 3 + 1 + M + 5 = 9 + M But it is given that 9 + M is a multiple of 9 which is only possible if 9 + M is either 9 or 18. If 9 + M = 9 Then M = 9 – 9 = 0 If 9 + M = 18 Then M = 18 – 9 = 9 Therefore, M has two values 0 and 9

31M5 is a multiple of 9 and two values are obtained by M. Why? Where M is a digit?

Answer»

Given Numbers is 31M5

∴Sum of digits = 3 + 1 + M + 5 = 9 + M

But it is given that 9 + M is a multiple of 9 which is only possible if 9 + M is either 9 or 18.

If 9 + M = 9

Then M = 9 – 9 = 0

If 9 + M = 18

Then M = 18 – 9 = 9

Therefore, M has two values 0 and 9