A Mersenne number is a positive integer that is one less tha

The smallest Mersenne prime is 3. Now we know that any power of 2 ends with a even integer. Thus the units digit of (2^[43112609])-1) will be an odd integer. This when multiplied by 3 will still be an odd integer. Thus the answer is the only odd integer.

B.
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Good solution.

The smallest Mersenne prime = 2^2 - 1 = 3 = Odd;
2^(43112609) - 1 = Even - Odd = Odd;

Odd*Odd = Odd --> the units digit must be odd. Only B fit.

Answer: B.
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First of all, think which is the smallest Mersenne prime?
The smallest prime 1 less than a power of 2 is 3.

We need the last digit of 2^{43112609} - 1. What will be the last digit of 2^{43112609}?
Recall cyclicity of last digits. After every 4 powers, the last digit repeats itself. Let's look at the last digits of powers of 2.
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 6
2^5 = 2
2^6 = 4
and so on...
When 43112609 is divided by 4, you get remainder 1 (divide just the last two digits by 4. Whatever remainder you get will be the remainder when you divide the actual number by 4)
So the last digit of 2^{43112609} is 2
Last digit of 2^{43112609} - 1 is 1

Product of the last digits is 1*3 = 3

Answer (B)
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The smallest Mersenne Prime is \(2^2 - 1 = 3\)
We know that any prime other than '2' is odd.
Hence the product of these two primes should be odd.

From the options, only option B is possible.

Hence option B is the answer.

That is exactly how I did it but I was worried it I was missing something just in case. You explained it really well, thank you

Yay, a reason>math solution...I love it!

We could have even gone a step further. All primes aside from 2 are odd. We are multiplying two such primes. Odd.

Answer choice B.