Question: When $32^{(32)^{(32)}}$is divided by 7, then the remainder is –...

Question

When $32^{(32)^{(32)}}$ is divided by 7, then the remainder is –

Answer 1

We have,

32³² = (2⁵)³² = 2¹⁶⁰ = (3 – 1)¹⁶⁰

= ¹⁶⁰C₀3¹⁶⁰ – ¹⁶⁰C₁ . 3¹⁵⁹ + ..... – ¹⁶⁰C₁₅₉ . 3 + ¹⁶⁰C₁₆₀3⁰

= 3m + 1, where m Î N

$32^{(32)^{(32)}}$ = (32)^3m+1

= (2⁵) ^3m+1

= 2^15m+5

= (2) ^{3 (5m+1)} . 2²

= (2³)^5m+1 . 2²

= (7 + 1)^5m+1 ×4

= {^5m+1C₀7^5m+1 + ^5m+1C₁ 7^5m + .....+ ^5m+1C_5m

7 + ^5m+1C_5m+1 . 7⁰} × 4= (7n + 1) × 4,

where n = ^5m+1C₀ 7^{5m + 1} +.....+ ^5m+1C_5m , 7

= 28n + 4

Thus, when $32^{(32)^{(32)}}$ is divided by 7, the remainder is 4. Hence (3) is correct.

Question: When \(32^{(32)^{(32)}}\)is divided by 7, then the remainder is –...