Last Digit In A Huge Product

Number Theory Level 2

Find the last digit of the number below.

$(2^{2^0} + 1) ( 2^{2^1} +1 ) ( 2^{2^2} + 1)(2^{2^3} + 1) \cdots (2^{2^{1024}} + 1 )$

The answer is 5.

2 solutions

Raj Magesh
Jan 2, 2016

Evaluate the second term in the product: $2^{2^1} +1=5$ .
All the terms are odd numbers.
When odd numbers are multiplied, the result is odd.
When an odd number is multiplied by $5$ , the final digit will be $\boxed{5}$ .

I'm thinking what would be the last two digits ?

Akshat Sharda - 5 years, 5 months ago

The last two digits are 35.

The last two digits of the factors are $03, 05, 17, 57, \underbrace{37, 97, 17, 57}_{255\ \text{times}}.$ Now $37\cdot97\cdot17\cdot 57 \equiv 41$ mod 100, and $41^5 \equiv 1$ mod 100, so that the product is $3 \times 5\times 17\times 57\times (41^5)^{51}$ mod 100, which simplifies to $3 \times 5\times 17\times 57 \equiv 35$ mod 100.

Arjen Vreugdenhil - 5 years, 5 months ago

Yes ... Thank You !!

Akshat Sharda - 5 years, 5 months ago

Yeah, that was actually the original problem. :D

Raj Magesh - 5 years, 5 months ago

Got em lol

Bryan Gonzalez - 5 years, 4 months ago

Saswata Banerjee
Jan 2, 2016

Multiply this by ( $2^{2^0} - 1$ ) to simplify the whole expression to $2^{2^{1025}} - 1$ .

The last digit of any power of 2 follows the cyclic fashion 2, 4, 8, 6, 2, 4, etc... As $2 ^ {1025}$ is divisible by 4, the last digit of this number will be 6.

$6 - 1 = 5$

Hence, final answer is $\boxed{5}$

That was my solution too! Just replace the 1025 for 1024 (;

Dudu Bello - 5 years, 4 months ago

@Saswata Banerjee How was INMO?

Upamanyu Mukharji - 4 years, 3 months ago

@Saswata Banerjee Are you planning on giving RMO this year(for 11th)?

Upamanyu Mukharji - 5 years, 2 months ago

I gave it in 10th. Couldn't go to camp, but got the merit certificate in INMO. And yeah, I'll give it this year too. Sorry for the extremely late reply.

Saswata Banerjee - 5 years ago

Last Digit In A Huge Product

The answer is 5.

2 solutions

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