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kiran120680
Moderator - Masters Forum
Joined: 18 Feb 2019
Last visit: 27 Jul 2022
Posts: 706
Given Kudos: 276
Location: India
Schools: AGSM '20 EDHEC Sept 2019 Intake Great Lakes '21 IE
GMAT 1: 460 Q42 V13
GPA: 3.6
16 May 2019, 01:04
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
76% (01:14) correct
24%
(01:34)
wrong
based on 223
sessions
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Not Attempted Yet
For any integer x greater than 1, x! denotes the product of all integers from 1 to x, inclusive. What is the hundred-thousands digit of 25!?
A. 0
B. 2
C. 4
D. 5
E. 8
A. 0
B. 2
C. 4
D. 5
E. 8
16 May 2019, 05:14
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This is a problem which is based on the concept of maximum power of a number in a factorial.
To start off, know that all numbers till 9! end with a ZERO. 10! is the first number which ends with 2 zeroes since it has 2 fives in it. Therefore, an intelligent estimate will tell us that the last few digits of 25! may be Zeroes. This, in turn will lead us on to calculating the highest power of 5, because the number of zeroes at the end of a number depends on the highest power of 5 in that number.
The procedure to calculate the highest power of a prime number in a factorial is shown below.
16th May 2019 - Reply 1.JPG [ 19.65 KiB | Viewed 5287 times ]
As is evident from the calculation, the highest power of 5 in 25! is 6. This means that there are 6 zeroes at the end of 25!
The sixth digit, when counted from the right, is the hundred-thousand’s digit. So, the correct answer option is A.
In a question like this, where the start point for the solution is not defined clearly, it’s a good idea to use your knowledge of smaller factorials to establish a pattern and then extrapolate from there to a large number like 25!
Hope this helps!
To start off, know that all numbers till 9! end with a ZERO. 10! is the first number which ends with 2 zeroes since it has 2 fives in it. Therefore, an intelligent estimate will tell us that the last few digits of 25! may be Zeroes. This, in turn will lead us on to calculating the highest power of 5, because the number of zeroes at the end of a number depends on the highest power of 5 in that number.
The procedure to calculate the highest power of a prime number in a factorial is shown below.
Attachment:
16th May 2019 - Reply 1.JPG [ 19.65 KiB | Viewed 5287 times ]
As is evident from the calculation, the highest power of 5 in 25! is 6. This means that there are 6 zeroes at the end of 25!
The sixth digit, when counted from the right, is the hundred-thousand’s digit. So, the correct answer option is A.
In a question like this, where the start point for the solution is not defined clearly, it’s a good idea to use your knowledge of smaller factorials to establish a pattern and then extrapolate from there to a large number like 25!
Hope this helps!
19 May 2019, 19:34
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kiran120680
Since the multiples of 5’s in 25! are 25, 20, 15, 10, 5, for a total of 6 factors of 5, which equate to 6 trailing zeros, we see that the last 6 digits of 25! are zero and thus the hundred-thousands digit is 0.
Alternate Solution:
Any time we multiply a number by 10, we create a trailing zero. In a factorial, we look for 5-and-2 pairs, each of which makes 10, and thus every 5-and-2 pair makes a trailing zero. Let’s determine how many 5-and-2 pairs are in 25!. First, we see that there are more 2s than 5s, so we will determine the number of factors of 5 in 25!, and this will also determine the number of 5-and-2 pairs.
In 25!, we obtain a 5 from each multiple of 5. Thus, from 5, 10, 15, and 20, we have 4 fives, and from 25 we have 2 additional fives. Thus, in 25! There are six factors of 5, and so there are six 5-and-2 pairs. This will create 6 trailing zeros, and so we see that the hundred-thousands digit of 25! Is 0.
Answer: A
