The number 1399-1993 must be divisible by

Question

The number 1399-1993 must be divisible by (A) 161 (B) 162 (C) 163 (D) 164

Tardigrade Admin · Accepted Answer

Let, M=1399-1993 = (. 1 + 4 × 3 .)99 - (. 1 + 2 × 32 .)93 = . 1 + 99 C1 × 4 × 3 + 99 C2 × (. 4 × 3 .)2 + ... . - . 1 + 93 C1 × (. 2 × 32 .) + 93 C2 × (. 2 × 32 .)2 + ... . = - 2 × 35 + 2 × 34 × integer = 2 × 34 × an integer Hence, the number is divisible by 162

Q. The number $1 3^{99} - 1 9^{93}$ must be divisible by

$161$

$162$

$163$

$164$

Q. The number 1399−1993 must be divisible by

161

162

163

164

Let, M=1399−1993 =(1+4×3)99−(1+2×32)93 ={1+99C1​×4×3+99C2​×(4×3)2+...}−{1+93C1​×(2×32)+93C2​×(2×32)2+...} =−2×35+2×34× integer =2×34× an integer Hence, the number is divisible by 162

Q. The number $1 3^{99} - 1 9^{93}$ must be divisible by

$161$

$162$

$163$

$164$