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Question

The total number of two digit numbers n, such that 3n+7n is a multiple of 10, is

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Solution

7n=(103)n=10k+(3)n
f=7n+3n=10k+(3)n+3n
So, if n is odd, f=10k
if n is even, f=10k+23n
Let n=2t;t N
3n=32t=(101)t
=10p+(1)t
=10p±1
if n = even then 7n+3n wil not be multiple of 10
So if n is odd then only 7n+3n will be multiple of 10
n=11,13,15,............,99
n=45

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