Question

How many four-digit numbers, each divisible by 4 can be formed using the digits 1, 2, 3, 4 and 5, repetitions of digits being allowed in any number?

Answer 1

for no. to be divisible by 'y'

last 'z' digit should be divisible by

'y' 6 only combinations are

$12,24,52,32,44,\left(5ways\right)$

no. of ways of filling $1^{st}$ 2 Place

are '5' & '5'

$\therefore$ total ways are $5\times 5\times 5=125$