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

Question

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

A. 1​
B. 2​
C. 12​
D. 24​
E. 48

yashikaaggarwal · Accepted Answer

101 don't have any prime factorization so trailing zeroes of 100! Will be same in 100!+101!

For the highest power of 100 integer in 100!+101! we need to find the total no. Of 10 power in the given equation.
10 = 2*5 (prime factorization)

No. Of 2s = 
100/2 = 50
50/2 = 25
25/2 = 12 (ignore remainder)
12/2 = 6
6/2 = 3
3/2 = 1 (ignore remainder)

Total no. Of 2s in 100! = 50+25+12+6+3+1 = 97

No. Of 5s = 
100/5 = 20
20/5 = 4
Total no. Of 5s = 20+4 = 24
But 10 is pair of 2&5 so for 24 5s there will be only 24 2s
So total no. Of 10 in 100! = 24

10 raise to the power 24
100 = 10^2
So 10^24 = 100^12

Answer is C.

Posted from my mobile device

Nups1324 · Answer

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.

IanStewart · Answer

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.

yashikaaggarwal · Answer

Well that is quiet a speculation

but i guess you need a more specified answer. the method i use is of trailing zeroes, we are required to determine 100 power in 100! + 101! so let's say if there are no 10 or 100 in 100! there will be no zero after the digits xxxxxxxxxxxxxxxxxx.............. so in similar term we just have to find zero after the product of rest digit. let say the no. of trailing zeroes in 9! = 1 (362880) which is also the product of (3*4*6*7*8*9*(2*5)) so every pair of 2 and 5 will make a ten and will add in trailing zeroes. now you can find the no. of 2 in 9! as 9/2 = 4 4/2 = 2 2/2 = 1 so there are 4+2+1 = 7 2s in 9! or (2*3*4*5*6*7*8*9 have 2 *3* 2*2 *5 *2 *3*7* 2*2*2 *9 = 7 2s, however this method is way long and require time or we go by the trailing zeroes method) and 9/5 = 1 1 5s in 9! which resulted into one 10. similarly 100! have 24 5's and 97 2's but we get only 24 pair of 2 and 5 so only have 24 trailing zero after the digits of 100! (1 to 100 product) And since we want 100 power and 10^2 = 100 there will be only 12 hundred in the whole 100!+101! product. Hopefully this is helpful.

Nups1324 · Answer

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

Fdambro294 · Answer

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

MathRevolution · Answer

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

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

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

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