P = 1^1 + 2^2 + 3^3 + 4^4 + 5^5 +............+48^{48}+49^{49}+50^{50}

Disclaimer: This is a slightly lengthy answer. Be patient initially and you'll be able to make sense of it.

This is, by far, one of the better questions on remainders. It’s not a surprise that it’s classified as a 700-level question, because, you’ll have to be at your best to be able to solve this in 2 minutes.
Honestly speaking, 2 mins is actually insufficient to solve this question. A better estimate would be 3 minutes.

Using the divisibility rule of 8 might help, but only to a certain extent. This is because the divisibility rule relies on the last 3 digits of a number, and as numbers become bigger, it’s virtually impossible to calculate the last 3 digits. So, the best approach in this question is to eliminate certain values and then establish a pattern for the remaining numbers.

\(1^1\) leaves a remainder of 1 when divided by 8.

\(2^2\) leaves a remainder of 4 when divided by 8.

\(3^3\) leaves a remainder of 3 when divided by 8.

\(4^4\) is a multiple of \(2^3\)(i.e. 8), so it gets completely divided by 8 and does not leave any remainder.

Similarly, all even numbers greater than 4, will be divisible by 8, in our case. For example, \(6^6\) definitely has a \(2^3\) in it. So does \(8^8\), \(10^{10}\), \(12^{12}\) and so on. Therefore, we can eliminate the numbers where the bases are even numbers.

There are 25 such numbers, of which \(2^2\) is the exception. So, we can eliminate the remaining 24 numbers. We will be left with 26 numbers, of which we have already found out the remainders for \(1^1\), \(2^2\) and \(3^3\).

All of the remaining 23 numbers are odd numbers. Of these 23 odd numbers, 13 will be prime (there are a total of 15 prime numbers from 1 to 50).

Any prime number greater than 3 can be written in the form of (6k+1) or (6k-1). Therefore, when we square these numbers, we get (36\(k^2\) + 12k + 1) or (36\(k^2\) – 12k +1). For any value of k, when we divide the above expressions by 8, we see that the remainder will be 1.

For example, if k = 1, (36 + 12 + 1) divided by 8 leaves a remainder of 1; (36-12 + 1) also leaves a remainder of 1. We can use this as a basis for calculating the remainders for prime numbers.

\(5^5\) = \((5^2)^2\) * 5. The first part, as per the above discussion, will give a remainder of 1; the second part will give a remainder of 5. Therefore, the remainder when \(5^5\) is divided by 8 will be 5.

\(7^7\) = \((7^2)^3\) * 7. Remainder = 1 * 7 = 7.

We can apply the above concept to all prime numbers. So, essentially, now, we just have to calculate the remainder of that number divided by 8 (like 5 by 8 or 7 by 8 or 11 by 8 and so on).


Now, coming to the next part of odd composite numbers, like 9, 15, 21 and so on.

\(9^9\) will leave a remainder of 1 (because 9 = 8+1)

\(15^{15}\) = \((15^2)^7\) * 15 = \((225)^7\) * 15. The first part leaves a remainder of 1 and the second part, 7. So the remainder is 7.

\(21^{21}\) = \((21^2)^{10}\) * 21 = \((441)^{10}\) * 21. Again, the first part leaves remainder of 1 and the second part, 5. So, the remainder is 5.
Hope this pattern is clear.


Now, let’s combine all of these into one and see what pattern emerges.
\(9^9\) => remainder 1
\(11^{11}\)=> remainder 3
\(13^{13}\) => remainder 5
\(15^{15}\) => remainder 7.
Notice that 9, 11, 13 and 15 are 1,3, 5 and 7 away from 8, respectively.

Similarly,
\(17^{17}\) => remainder 1
\(19^{19}\) => remainder 3
\(21^{21}\)=> remainder 5
\(23^{23}\) => remainder 7.

If \(25^{25}\) gives us a remainder of 1,we can say that we have got the pattern

\(25^{25}\) = \((25^2)^{12}\) * 25. Both the parts leave a remainder of 1. Therefore, we can continue with the pattern for every four numbers.

The next four numbers are 25, 27, 29 and 31.
The next four are 33, 35, 37 and 39.
The next four are 41, 43, 45 and 47.

When we add up the remainders 1,3, 5 and 7, the total is 16. There are a total of 5 cycles. The total of all these remainders will be 80, which is equivalent to a remainder of 0 (because 80 is divisible by 8).

The only number left out is \(49^{49}\) which will leave a remainder of 1 (because 49 = 48 + 1).

Now, we only have to add the remainders of
\(1^1\), \(2^2\), \(3^3\), \(5^5\), \(7^7\) and \(49^{49}\).

The remainders were 1, 4, 3, 5 , 7 and 1 respectively. The sum of these values is 21.
21 leaves a remainder of 5, when divided by 8. The correct answer option is C.

This might appear a lengthy approach. But, that is only because, I have had to type out the entire thing to show you the patterns. When you actually solve this on a scratch paper, you’ll be able to establish the pattern much faster. Establishing the pattern is of paramount importance in questions like these, which look overwhelming owing to their sheer monstrosity.

Hope this helped!

Can we really do this in 2-3 min?

Posted from my mobile device

We can write any odd number as 2k+1

(2k+1)^2= 4k^2+4k+1= [4(k)(k+1)]+1
[4(k)(k+1)] is always divisible by 8
Hence odd^{even power} will give you remainder 1 when divided by 8

(2n)^2n is divisible by 8, if 2n>2

If you apply above concepts, you can solve it under 2 minutes. You might not see such question in GMAT, but this question will hone your basics and might be helpful to solve many other questions.

Hello Sach,

If you have practiced sufficient questions on numbers concepts and remainders, I don't see any reason why you cannot do this in 3 minutes ( i know 2 minutes is not realistic).

Let me break it down for you:

First 15 to 30 seconds - read the question, the options and plan your approach (eliminate any if you can)

30 seconds to 120 seconds - do the actual calculations on numbers beyond 7^7 and establish a pattern. Note that this is a good 1.5 minutes you have to do all the calculations you want (obviously, your calculation skills need to be very good and you need to have a good smattering of remainder concepts and tricks)

120 seconds to 180 seconds - based on the pattern, calculate the actual remainder and mark your answer.

Also, these kind of questions are very very rare on the GMAT. You need to use these questions to build stamina (as Nick mentioned) , so much so that, the actual remainder questions on the GMAT should look trivial to you.

Hope this helps!

R = remainder
1/8 gives R.1
2^2 = 4/8 gives R.4

---------------------------
for 4^4
4*4*4*4 = (2*2*2*....)/8 gives R.0
for 6^6
6*6*6*6*6*6= (2*2*2....)/8 gives R.0
holds so for all higher even number on the series
------------------
for 3^3
3^1 / 8 gives R.3 (odd power)
3^2 =9/8 gives R.1 (even power)
(cycle repeats as R.1 achieved)
3^3 (odd) repeats to give 3
3^4 (even) gives R.1
------------------
for 5^5
5^1/8 gives R.5
5^2/8 gives R.1 (cycle repeats from R.1)
odd power R.5
-----------------
for 7^7
7^1 / 8 gives R.7
7^2 / 8 gives R.1 (cycle repeats from R.1)
odd power of 7 so, R.7
--------
for 9^9
always R.1
-----------------
doing so for 11^11;
For 9^9 R.1 is already achieved. Cycle for complete series repeats so,

11^11 (R.3), for 13 (R.5), for 15 (R.7) and for 17 (R.1)
same for all other odd numbers
-----------------------------------------------------
from 1 to 10 (except 1 and 2^2)
R=(3^3+5^5+7^7+9^9)=(3+5+7+1)=16
so, up to 50 this repeats 5 times; 16*5 --- anyway remainder is 0. as 16*5/8

Only remaining:
1^1 and 2^2 = 1+4 = 5