Question

If $(67^{67}+67)$ is divided by $68$, the remainder is: 

• A. 61
• B. 67
• C. 63
• D. 66

Hint:

When a base is “one less than the modulus” replace it by $-1$ (or $-k$) to simplify powers quickly.

Answer 1

The Correct Option is D

Work modulo $68$. Since $67\equiv -1\pmod{68}$ and the exponent is odd, \[ 67^{67}\equiv (-1)^{67}\equiv -1\pmod{68}. \] Therefore, \[ 67^{67}+67\equiv (-1)+(-1)\equiv -2\equiv 68-2=\boxed{66}\pmod{68}. \] 

Answer 2

Expand \( 67^{67} \) as \( (68-1)^{67} \) using the binomial theorem: every term of the expansion except the very last carries a factor of \( 68 \) and vanishes under modulo \( 68 \).

1. Option A (61): The binomial expansion gives \[ (68-1)^{67}\equiv(-1)^{67}\equiv-1\pmod{68}, \] so \( 67^{67}+67\equiv-1+67=66\pmod{68} \), which is not \( 61 \); this option is rejected.
2. Option B (67): The same computation gives \( 66 \), not \( 67 \), so this option is rejected.
3. Option C (63): Again the result is \( 66 \), not \( 63 \), so this option is rejected.
4. Option D (66): As shown, \( 67^{67}+67\equiv-1+67=66\pmod{68} \), which matches exactly.

The binomial expansion of \( (68-1)^{67} \) confirms the remainder is \( 66 \).

Hence, the correct answer is 66.