If P = 1!^2 + 2!^2 + 3!^2 + ... + 10!^2. Then what is the remainder wh

Question

If  P = 1!^2 + 2!^2 + 3!^2 + ... + 10!^2 . Then what is the remainder when  5^{2P}  is divided by 13 ?

A. 0
B. 1
C. 4
D. 8
E. 12

PyjamaScientist · Accepted Answer

5^{2P}  can be written as  25^{P} . Which can be written as  (26-1)^{P} . Now, using the Binomial theorem, the first term would yield no remainder, and it would eventually depend on the even/odd nature of P to determine the remainder.
If, P is even remainder would be 1. If P is odd, the remainder would be -1 = 12.

In the given question,  P = 1!^2 + 2!^2 + 3!^2 + ... + 10!^2 . We can see, that all terms after 1!^2 contain a 2 and thus are even, so the sum of P would be an odd number. Thus, we know P is odd.

So, with this information we can safely conclude that the remainder would be (-1) that is 13*(-1) +  12 .  Option (E)

cat2010 · Answer

Bunuel   chetan2u  What is the value of ! ? As we dont know ! then How can we get !^2 .

Or the question is faulty ? And intended question is (1!)^2

chetan2u · Answer

Hi

5!^2  means  (5!)^2 
Both would mean 5!*5!

chetan2u · Answer

Find a pattern or do it by Binomial theorem.

(1) Pattern

5^{2P}=25^P

Let us see what successive powers leave remainder as when divided by 13.
 25^1=25=26-1=13*2-1  or remainder is -1, that is 13-1 or 12..
 25^2=625=624+1=13k+1  or remainder is 1..
 25^3=15625=15626-1=13x-1  or remainder is -1, that is 13-1 or 12..

So remainder is 12,1,12,1…..
When the power or P is odd, the answer is 12, otherwise 1.
 P=1!^2+2!^2+….. 
All terms are even from 2! onwards, so P is 1+even+even+.. =odd
Hence, remainder is 12.

(2) Easier method - Binomial

5^{2P}=25^P=(26-1)^P , so the remainder is  (-1)^P . 
As P is odd, remainder is  -1^{odd}  = -1
As remainders are considered positive, remainder becomes 13-1 or 12.

Aarya_San · Answer

1! + 2! + .... 10! -> Each term beyond 1! will include a 2 in it, and hence will be even. Therefore
 
P will be an (Odd number (1!) + Even number (Sum of every other factorial term)) -> P is odd.

5^2P will be an even number.

5^1/13 , rem = 5
5^2/13 , rem = 12
5^3/13 , rem = 8
5^4/13 , rem = 1
5^5/13 , rem = 5 (Since rem are cyclical in nature, we are likely to see this pattern repeating). We have also found that it repeats after every 4 power.

So one thing we know id P is even, but ine thing we also know is that 2P is not divisible by 4. So 5^2P will have the same remainder as 5^2 does. Hence 12.

Very calculation intensive, mostly likely won't show up in the real GMAT, but a similar type with easier calc might.

arushi118 · Answer

We can use Euler's Theorem, which states that if we are finding the remainder of (a^x) % b where b is prime, find the remainder of the power of a with b-1 -> and use this remainder as power of a to find the remainder with b.
So here,
we have:
a = 5
x = 2P
b = 13
Remainder of x with b-1: remainder of 2P with 12
2* R  -> as all terms in P will give remainder of 0 after (2!)^2 with -> So R  = 1 + 2^2 = 5

So question gets transformed to:
5^10 % 13
(25)^5 % 13
(-1)^5 % 13
-1 % 13
12

E is correct.

If P = 1!^2 + 2!^2 + 3!^2 + ... + 10!^2. Then what is the remainder wh

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

If P = 1!^2 + 2!^2 + 3!^2 + ... + 10!^2. Then what is the remainder wh

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards