QJEEM21Ma2d5e: JEE Main 2021 Mathematics - Class 11 JEEM21Ma2d5e Solution: The total number of two digit numbers 'n', such that 3 n + 7 n…

$\because$ ${7^n} = {(10 - 3)^n} = 10k + {( - 3)^n}$

${7^n} + {3^n} = 10k + {( - 3)^n} + {3^n}$



$\therefore$ 3ⁿ = 3^2t = (10 $-$ 1)^t

= 10p + ( $-$ 1)^t

= 10p $\pm$ 1

$\therefore$ if n = even then 7ⁿ + 3ⁿ will not be multiply of 10

So if n is odd then only 7ⁿ + 3ⁿ will be multiply of 10

$\therefore$ n = 11, 13, 15, ..........., 99

$\therefore$ Ans : 45