What is the highest integral power of 100 in the expression 100!+101!?

Hi Yashika,

This looks really interesting, but unfortunately I couldn't understand it.

If you could explain your method in detail, it'll be really helpful.?


Thank you.

P.S:- I got the answer by fluke as I used estimation after a failed attempt. 7! is 5040 so option A and B were out. 10^24 or 10^48 would be too much for 100!+101!, so 12.

This is not a realistic GMAT problem, so I'll write up a solution quickly, but:

101! + 100! = 100! (101 + 1) = (102)(100!)

We want to know the largest power of 100 that will divide this, so we want to know how many times we can divide this by (2^2)(5^2). If you think of what 100! is equal to, the product of all integers from 1 through 100, there are tons and tons of even numbers in that product. So we'll have no shortage of 2s. We'll have many fewer 5's if we prime factorize 100!. So the 5's are the limitation we need to focus on (and since the '102' doesn't give us any 5's, we can just ignore it).

If you were to write out 100! in full, you'd find there are twenty multiples of 5 in the product (5, 10, 15, ... 95, 100). So 100! is divisible by at least 5^20. But there are also four multiples of 5^2 in 100! (25, 50, 75, 100), and each of those gives us one additional '5'. So 100! is divisible by 5^24. We don't find 125 = 5^3 if we write out 100! in full, so we have just shown that 100! is divisible by 5^24, but by no higher power of 5. It's also divisible by 2^24 for sure (it's divisible by a much higher power of 2 than that), so 100! is divisible by 10^24, and thus by 100^12, but by no higher power of 100. So the answer is 12.

Damn! This is really interesting.

Yes it is quite a speculation, but I would've had no other option on the real test.


But, thanks to you, Bunuel, Chetan2u, GMATNinja and many others, that people like me can learn and not just speculate.

I'll make note of this explanation and the concept of trailing zeroes.


Thank you so much Yashika!

Posted from my mobile device

100 = 10^2nd = 2^2nd * 5^2nd

When looking for Trailing Zeroes or Factor Pairs of (2 * 5), there will always be fewer 5 Prime Factors in the Factorial! < than 2 Prime Factors.

Thus, if we find the Count of 5 Prime Factors that exist in the Prime Factorization of a Factorial!, that will give us the Count of (2*5) Factor Pairs.

In this case, we need TWO 2's and TWO 5's to make 1 Composite Number of 100. So we will look for the Highest No. of 5 Prime Factors that exist in the Expression.

1st) need to simplify the Expression.

100! + 101! = 1 * (100!) + 101 * (100!)

take 100! as Common Factor

= 102 * (100!) = 2 * 51 * 100!


Concept: to find the Max No. of Prime Factors that Divide into a Factorial!, keep dividing the Prime Factor into the Quotient until you can no longer get a (+)positive Quotient.

100 / 5 = 20

20 / 5 = 4

4 / 5 = NO + Quotient

There is a MAX 24 Prime Factors of 5 that evenly divide into the Expression. Since we need a Pair of TWO 5's to make up ONE 100, the Highest Integral Power of 100 that we can make is (24/2) = 12

Answer 12 C

To find the highest power of '100', first, find the highest power of '10'.

10 = 2 * 5

The number of '2's will be always more than those of '5'. But we can only pair the same numbers of power. So, find the highest power of '5' in 100!.

\(\frac{100!}{ 5}\) + \(\frac{100!}{ 5^2}\) + \(\frac{100!}{ 5^3 }\)+ .....

=> 20 + 4 + 0.8 ( As we consider only the integer part, rest all results will be '0')

So, the highest power of 10 in 100 ! is 24.

=> \(10 ^{24}\) = \((10^2)^{12}\)

=> \(100 ^{12}.\)

Answer C