Which of the following numbers is prime?

Thanks for the Explanation Bunuel. This questions has many concepts. infact this question cleared my many questions. So I though to share and post so that others could benefit from it.

Thank you for posting.

Bunuel,

..... C. 4^66 + 7^66 --> the units digit of 4^66 is 6 (the units digit of 4 in positive integer power repeats in blocks of two {4, 6})..... 66/2 (number of repetitions of two) is 33, and therefore it should be the first number of the block of 2, meaning the units digits is 4. I think I am missing something

Consider the following example: what is the units digit of 127^124.

First of all, the units digit of 127^124 is the same as that of 7^124 (get rid of all the digits except the units digit).

Next, recall that the units digit of 7 in positive integer power repeats in blocks of four {7, 9, 3, 1}.

Finally, to get the units digit of 7^124, you need to divide the exponent (124) by 4 (cyclicity) and look at the remainder you get:

Remainder = 1 --> the units digit = 1st number from the pattern, so 7.
Remainder = 2 --> the units digit = 2nd number from the pattern, so 9.
Remainder = 3 --> the units digit = 3rd number from the pattern, so 3.
Remainder = 0 --> the units digit = 4th number from the pattern, so 1.

Now, since 124/4 yields the remainder of 0 (124 is divisible by 4), then the units digit of 7^124 is 1.

We can apply the same logic to 4^66: the units digit of 4 in positive integer power repeats in blocks of two {4, 6} --> 66/2 yields the remainder of 0, thus the units do digit of 4^66 is 2nd number from the pattern, so 6. Or another way: 4^odd has the units digit of 4 and 4^even has the units digit of 6.

For more check Number Theory chapter of our Math Book: math-number-theory-88376.html

Hope it's clear.

Additionally, can you help me understand the concept behind the answer you obtained for D
5^82 - 2^82 --> we can factor this as (5^41 - 2^41)(5^41 + 2^41). Not a prime.
I don't fully understand why the factorization helps us deduce that this number is not prime.

Let us say "5^82 - 2^82" results in some number x. If we are able to factor x as x=a*b (a and b are two numbers, can be same or different), this means that x is not a prime. Since any prime number will have 1 and itself as the ONLY factors.

Thanks

Bunuel I would like to understand how if an option can be written as a factor (a-b)(a+b) cannot be Prime?

I don't understand how do we get 5 here. Is there any formula that I don't know? Need help

\(4^{66} + 7^{66}\) --> the units digit of 4^66 is 6 (the units digit of 4 in positive integer power repeats in blocks of two {4, 6}) and the units digit of 7^66 is 9 (the units digit of 7 in positive integer power repeats in blocks of four {7, 9, 3, 1}). Hence, the units digit of 4^66 + 7^66 is 5 (6+9). Thus 4^66 + 7^66 is divisible by 5. Not a prime.

The units digit of 4^(positive integer) repeats in blocks of two - {4, 6}:
4^1 = 4;
4^2 = 16;
4^3 = 8\frac{4[}{fraction];
4^4 = 256;
...

So, the inits digit of 4^256 will be 6 (the odd powers give 4 and even powers give 6).

The units digit of 7^(positive integer) repeats in blocks of four - {7, 9, 3, 1}:
7^1 = 7;
7^2 = 49;
7^3 = ...[fraction]3};
7^4 = ...1;
7^5 = ...7 (7 again)
...

So, the inits digit of 7^66 will be 9. Divide 66 (power) by 4 (cyclisity), remainder is 2. So, the units digit of 7^66 is the same as that of the units digit of 7^2, which is 9.

Hence, the units digit of 4^66 + 7^66 is 5 (6+9).

Theory is here: https://gmatclub.com/forum/math-number- ... 88376.html

Check Units digits, exponents, remainders problems directory in our Special Questions Directory.

Hope it helps.

but why are we multiplying the two units digits by 5 ? why is it 5(6+9)?

We are not multiplying we are adding: 6 + 9. The units digit of a positive number ending with 6 PLUS the units digit of a positive number ending with 9 is a number ending with 5.

Is there any simpler method to solve this

There's no chance you could ever see a similar question on the actual GMAT, because on the GMAT, there's always a way to prove the right answer is right. And there is no reasonable way, in 2 minutes, to prove that 2^16 + 1 is a prime number. To prove a number x is prime, you need to show you cannot divide it by any prime up to √x. So here, to prove 2^16 + 1 is prime, you'd need to prove 2^16 + 1 is not divisible by any prime up to √(2^16 + 1), which is roughly 2^8 = 256. So you have to try dividing 65,537 by 54 different prime numbers, many of them 3-digit primes that most test takers won't be familiar with (and there's no reason to be).

The only reasonable way to do the question is by eliminating wrong answers, but then you still have no way to be sure the right answer is right. And the techniques used earlier in this thread (determining which answers are even, which end in '5', and which have obvious factorizations) are the fastest ones you could use here.

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Useful question and explanation

Not so much because we may see a similar question in the GMAT (we probably won't), more because
Bunuel's techniques here could be useful for other questions. These techniques need to be ready in our armoury.

We don't even need to memorise facts such as [last digits of powers of 7 repeat in blocks of {7, 9, 3, 1}].
We just need to remember that last digits repeat in regular patterns.

It seems miraculous that this question can be done in just a few minutes! Using nothing but simple arithmetic


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I was afraid I have to check the units digits in all options so I went to the replies to see if there's any alternate approach. Is there?

Probably not. Checking the units digits is probably the quickest approach and the only one that's doable.

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I used the units digit approach for all the options and eliminated all but option A and D. Will the actual exam throw a question like this where we necessarily have to use the splitting the term into (a-b)(a+b) approach?

I did the same. But there's no way I was able to eliminate D by the unit digit approach as D has an unit digit of 1. So is there any other way to do this than to factorize D?