The total number of two-digit numbers n such | vedclass

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(A) We want $3^{n} + 7^{n} \equiv 0 \pmod{10}$.Since $7 \equiv -3 \pmod{10}$,we have $7^{n} \equiv (-3)^{n} \pmod{10}$.Thus,$3^{n} + 7^{n} \equiv 3^{n

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(A) We want $3^{n} + 7^{n} \equiv 0 \pmod{10}$.Since $7 \equiv -3 \pmod{10}$,we have $7^{n} \equiv (-3)^{n} \pmod{10}$.Thus,$3^{n} + 7^{n} \equiv 3^{n} + (-1)^{n} 3^{n} \pmod{10}$.If $n$ is even,$3^{n} + 3^{n} = 2 \cdot 3^{n}$,which is not divisible by $10$ (as $3^{n}$ ends in $1$ or $9$,so $2 \cdot 3^{n}$ ends in $2$ or $8$).If $n$ is odd,$3^{n} + (-1) 3^{n} = 3^{n} - 3^{n} = 0$,which is divisible by $10$.Therefore,$n$ must be an odd number.The two-digit numbers are from $10$ to $99$.The total number of two-digit numbers is $99 - 10 + 1 = 90$.Exactly half of these are odd,so the number of odd two-digit numbers is $90 / 2 = 45$.

The total number of two-digit numbers $n$ such that $3^{n} + 7^{n}$ is a multiple of $10$ is ..... .

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