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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?


Solution

for no. to be divisible by 'y' 
last 'z' digit should be divisible by 
'y' 6 only combinations are
$$ 12,24,52,32,44,(5\,ways) $$
no. of ways of filling $$ 1^{st} $$ 2 Place
are '5' & '5' 
$$ \therefore $$ total ways are $$ 5\times 5 \times 5 = \boxed{125} $$ 

1108126_518549_ans_870d97177a9d498aadc3cbcd889ba42d.jpg

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