M32-01

Question

How many zeros are at the end of  49! + 50! ?

A. 9
B. 10
C. 11
D. 12
E. 22

Bunuel · Accepted Answer

Official Solution:

How many zeros are at the end of  49! + 50! ?

A. 9
B. 10
C. 11
D. 12
E. 22

Factor out 49! from the expression:  49! + 50! = 49!(1 + 50) = 49! * 51 .
 
51 doesn't contribute to the number of zeros at the end of the number, so all zeros will come from 49!.
 
To find the number of zeros at the end of 49!, we can use this calculation:  \frac{49}{5}+\frac{49}{5^2}=9+1=10 
 
 Some background on trailing zeros: 
 
Trailing zeros refer to a sequence of 0's in a decimal representation of a number, after which no other digits are present.
 
For instance, 125,000 has 3 trailing zeros.
 
The number of trailing zeros in the factorial of a non-negative integer,  n! , can be determined using this formula:
 
 \frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k} , where  k  must be selected such that   5^k \leq n 
 
Let's consider an example:
 
How many zeros are at the end of 32!?
 
 \frac{32}{5}+\frac{32}{5^2}=6+1=7 . Observe that the last denominator ( 5^2 ) must be less than 32. Also, note that we only consider the quotient of the division, that is  \frac{32}{5}=6 . 
 
So, there are 7 zeros at the end of 32!.
 
Another example: how many trailing zeros does 125! have?
 
 \frac{125}{5}+\frac{125}{5^2}+\frac{125}{5^3}=25+5+1=31 ,
 
The formula essentially counts the number of factors of 5 in n!, but since there are at least as many factors of 2, this is equivalent to counting the number of factors of 10, each of which contributes one more trailing zero.

Answer: B

Nikhil · Answer

Hello,

I solved this question in slightly different manner.

To find the number of trailing zeroes in 49!

we know if we write 49! as product from 49 to 1 like 49*48*47*...*1, each 10(or 2*5) in this product contributes a trailing zero

So to find the number of 10's in 49! we need to find how many (2*5) pairs in 49! expansion

We know since each even number contributes a 2 there are more 2's than 5's in the product, so 5's are the limiting factor. Hence we need to find number of 5's in the product.

So counting the number of 5's in 49! expansion

Integer ------ Number of 5's in that integer

5 (5*1) ------> 1
10 (5*2) ------> 1
15 (5*3) ------> 1
20 (5*4) ------> 1
25 (5*5) ------> 2
30 (5*6) ------> 1
35 (5*7) ------> 1
40 (5*8) ------> 1
45 (5*9) ------> 1

Total 10

C3BISHT · Answer

I think this is a  high-quality  question.

Andinds1991 · Answer

What I don't understand is why Excel gives me the following number when I use "=FACT(49)*51" expression:

31 022 375 065 747 700 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000

The number has 48 zeroes.

Can someone please explain?

Bunuel · Answer

Using Excel to perform calculations involving large numbers is not recommended. Use better calculations: https://www.wolframalpha.com/

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

renzosm93 · Answer

Thank you Bunuel. My doubt is why 51 alone doesn't contribute to any zero?

Bunuel · Answer

To get a trailing zero, we need both a 2 and a 5. Since 53 is a prime number, it doesn't contain any 2's or 5's.

mellowmev · Answer

If you mean 50!, its because in this case 49! will always be lower than 50! and the trailing zeroes willl be dependent on the smaller numbers.
Also notice that 50! = 49! x 50, so in their sum 49! will contribute fewer zeroes

sssdd1234 · Answer

I think this is a  high-quality  question and I agree with explanation.

juliarajkovic · Answer

Why do we factor out 49! and not count the zeros in 50! then add them together?

Bunuel · Answer

Factoring out 49! simplifies the expression to 49!(1 + 50), and since 51 (1 + 50) does not contribute any additional factors of 5, the number of zeros is determined entirely by 49!.

vedvar · Answer

" additional factors of 5" is this a rule that is there? Factors of 5 contribute zeros to the final number?

According to the explanation above 51/5 + 51/5^2 = 10 + 2 = 12 additional zeros??
This is obviously wrong, but it seems to be consistent with the initial explanation above?

Bunuel · Answer

1. We don't have 51!, just 51.

2. 49! ends with 10 zeros, as explained above.

3. 49! obviously contains more factors of 2 than factors of 5.

4. If we had say 5 * 49!, 5 would pair with an extra 2 in 49!, adding one more zero, making it 11 zeros.

5. However, since we have 51 * 49!, and 51 does not have 5, it does not contribute an extra factor of 5 to pair with the 2’s in 49!, meaning the number of zeros remains 10.

olivierosacchet · Answer

I don’t quite agree with the solution. 49/5 + 49/25 is not equal to 10

Bunuel · Answer

Note that we only consider the quotient of the division, that is 49/5 = 9 and 49/25 = 1. For similar questions check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory . Theory on Trailing Zeros: https://gmatclub.com/forum/everything-a ... 85592.html Hope it helps.

shubhamagarwal44 · Answer

Why do we not add the factors individually here? So basically for 49! it would be 10 zeroes (dividing by 5) and for 50! it would be 12 zeroes so total 10+12 + 22 zeroes. Why is this method wrong?

Bunuel · Answer

10 has one trailing zeros, and 100 has two. Does 10 + 100 = 110 have three?

dradi123 · Answer

I like the solution - it’s helpful.

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