Funny exponent (Part 2)

What are the last 7 7 digits of 3 11 , 000 , 001 3^{11,000,001} ?

Note: Ignore initial zeroes of the answer for example if the answer is 0002456 0002456 , submit 2456 2456 .


Funny exponent (Part 1)


The answer is 3.

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2 solutions

Chew-Seong Cheong
Jan 21, 2021

3 11 , 000 , 001 3 × 3 2 × 5 , 500 , 000 (mod 1 0 7 ) 3 × 9 5 , 500 , 000 (mod 1 0 7 ) 3 ( 10 1 ) 5 , 500 , 000 (mod 1 0 7 ) 3 ( + 5 , 500 , 000 ( 5 , 499 , 999 ) ( 50 ) 55 , 000 , 000 + 1 ) (mod 1 0 7 ) 3 ( + 55 , 000 , 000 ( 27 , 499 , 995 1 ) + 1 ) (mod 1 0 7 ) 3 ( + 55 , 000 , 000 ( 27 , 499 , 994 ) + 1 ) (mod 1 0 7 ) 3 ( + 110 , 000 , 000 ( 13 , 749 , 997 ) + 1 ) (mod 1 0 7 ) 3 (mod 1 0 7 ) \begin{aligned} 3^{11,000,001} & \equiv 3 \times 3^{2 \times 5,500,000} \text{(mod }10^7) \\ & \equiv 3 \times 9^{5,500,000} \text{(mod }10^7) \\ & \equiv 3 (10-1)^{5,500,000} \text{(mod }10^7) \\ & \equiv 3 ( \cdots + 5,500,000(5,499,999)(50) - 55,000,000+ 1) \text{(mod }10^7) \\ & \equiv 3 ( \cdots + 55,000,000(27,499,995-1)+ 1) \text{(mod }10^7) \\ & \equiv 3 ( \cdots + 55,000,000(27,499,994)+ 1) \text{(mod }10^7) \\ & \equiv 3 ( \cdots + 110,000,000(13,749,997)+ 1) \text{(mod }10^7) \\ & \equiv \boxed 3 \text{(mod }10^7) \end{aligned}

Pi Han Goh
Jan 22, 2021

The Carmichael's lambda function of 1 0 7 10^7 is 5 × 1 0 5 5\times 10^5 .

Thus, 3 5 × 1 0 5 1 ( m o d 1 0 7 ) 3^{5\times10^5} \equiv 1 \pmod{10^7} . Raise both sides to the power of 22 yields 3 11 × 1 0 6 1 ( m o d 1 0 7 ) . 3^{11\times10^6} \equiv 1 \pmod{10^7}. Now multiply both sides by 3, and the answer follows.

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