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# Magoosh Brain Twister: The Terrible Threes

Welcome to this week’s challenge problem (the third of the series). Though it looks terribly daunting it can be solved relatively quickly. That’s often the case with the GRE: a seemingly inscrutable problem that looks like it’ll take weeks to solve can be unlocked very quickly, if you follow the right path.

To make things even more fun with this question, I’ve made it a Numeric Entry. So no lucky guesses—but good luck anyway!

n = (61!+60!+59!)/3^x, what is greatest value of x where n is an integer?

[ ____________ ]

Check back on Thursday for the explanation!

By the way, students who use Magoosh GRE improve their scores by an average of 8 points on the new scale (150 points on the old scale.) Click here to learn more.

### 42 Responses to Magoosh Brain Twister: The Terrible Threes

1. SD July 22, 2014 at 11:39 am #

I believe the answer is 27.
For “n” to be an integer, each factorial expression must be divisible by 3^x.
Specifically, 59! has to be divisible by 3^x.
Therefore, the problem now boils down to figuring out how many factors of 3 exist in 59!.
By my count, the number is 27.

Good luck!

2. Huan-Lin Chang July 22, 2014 at 9:58 am #

Numerator = 59! + 60! + 61! = 59! * (1 + 60 + 60 * 61) = 59! * (61 + 60 * 61) = 59! * (61 * (1 + 60)) = 59! * 61^2

Since 61^2 cannot be divided by 3, so the problem reduces to n = 59! / 3^x, that is how many 3s are there in 59!.

59/3 = 19 … 2, 19/3 = 6 … 1, 6/3 = 2. So the answer is 19 + 6 + 2 = 27.

3. Ayush jain July 22, 2014 at 9:55 am #

is it 19?

4. Roman July 22, 2014 at 6:05 am #

59

5. Divyanshu July 22, 2014 at 5:49 am #

6. Abhishek July 22, 2014 at 3:38 am #

Sorry, it’s 27

7. Vipin Kumar July 22, 2014 at 3:11 am #

Ans : 27

As we can write (61!+60!+59!) = 59!(61×60 + 60 + 1) = 61x61x59!

And in 59!, the highest power of 3 is 27.

• Chris Lele July 22, 2014 at 11:47 am #

Hmm…I think you got the right answer, but I’m not sure. The question isn’t asking for thie highest power of ‘3’, but how many 3’s can you factor from 61! + 60! + 59!. That also happens to be 27. But maybe that’s what you meant 🙂

8. raj July 22, 2014 at 3:07 am #

For finding the powers of a prime number p, in the n! The formula is: (n/p)+(n/p^2)+….. (n/p^k)
, where k must be chosen such that p^k <= n

Solving for 59!
(59/3)+(59/9)+(59/27)=19+6+2=27
Therfore 59! Will have 27 powers of 3 or 3^27

Similarly solving for 60! And 61! We get 3^29 and 3^29 respectively.

Now 'n' will have (3^26+3^29+3^29)/3^x
So maximum value of x for n to be an integer is 26

• raj July 22, 2014 at 3:09 am #

For finding the powers of a prime number p, in the n! The formula is: (n/p)+(n/p^2)+….. (n/p^k)
, where k must be chosen such that p^k <= n

Solving for 59!
(59/3)+(59/9)+(59/27)=19+6+2=27
Therfore 59! Will have 27 powers of 3 or 3^27

Similarly solving for 60! And 61! We get 3^29 and 3^29 respectively.

Now 'n' will have (3^27+3^29+3^29)/3^x
So maximum value of x for n to be an integer is 27

• Chris Lele July 22, 2014 at 11:48 am #

Yep, that’s it! Perfect 🙂

9. Siddharth Jain July 22, 2014 at 2:52 am #

n = 59!*( 1 + 60 + (60*61) ) / 3^x
n= 59! *(3721) / 3^x

Now, as 3721 is not divisible by 3 , it won’t be divisible by any of its higher powers.
Hence, the answer should be equal to the highest power of 3 in 59!, which is 27.

• Chris Lele July 22, 2014 at 11:52 am #

Hi Siddharth,

You got the right answer, but an even quicker way is to see that (1 + 60 + 60*61) = 61^2, and 61 is not divisible by ‘3’.

Good job!

10. Aroushi July 22, 2014 at 1:29 am #

• Chris Lele July 22, 2014 at 11:53 am #

Almost, but make sure to include the 3’s in 27 and 54.

11. Urvish Parikh July 21, 2014 at 7:40 pm #

22
If we simplify this it looks something like
(61*60*59! + 60*59! + 59!) factor out 59!:
59!(61*60*1 +60 + 1) = 59!(61*60 + 61) factor out 61:
59!*61*(60+1) so that’s 59!*61*61

the 61^2 will not play a roll in divisibility by 3^x because 61 is not divisible 3.
so we have to focus on 59!

lets expand to see if we can find a pattern.

59! = 1*2*(3)*4*5*(3*2)*7*8*(3*3)*10*11*(3*4)

notice that every third number has atleast one 3. so how many 3’s are in 59!?

59/3 =19 so thats 19 positions with atleast one 3. and then we have to look for the actual exponents of 3 so for 9 theres 1 extra 3 in that position and for 27 there are 2 extra 3s. so total of 22 threes will factor in to 59!61^2

• Chris Lele July 22, 2014 at 11:55 am #

Hi Urvish,

Almost, but don’t forget the multiples of ‘9’ and ’27’ that are less than 59 🙂

12. Urvish July 21, 2014 at 7:32 pm #

22
If we simplify this it looks something like
(61*60*59! + 60*59! + 59!) factor out 59!:
59!(61*60*1 +60 + 1) = 59!(61*60 + 61) factor out 61:
59!*61*(60+1) so that’s 59!*61*61

the 61^2 will not play a roll in divisibility by 3^x because 61 is not divisible 3.
so we have to focus on 59!

lets expand to see if we can find a pattern.

59! = 1*2*(3)*4*5*(3*2)*7*8*(3*3)*10*11*(3*4)

notice that every third number has atleast one 3. so how many 3’s are in 59!?

59/3 =19 so thats 19 positions with atleast one 3. and then we have to look for the actual exponents of 3 so for 9 theres 1 extra 3 in that position and for 27 there are 2 extra 3s. so total of 22 threes will factor in to 59!61^2

13. Tarun July 21, 2014 at 7:30 pm #

edit-
explaination:-

simplifying. the numerator
59!(61X60+60+1)=59!X61X61=59X58X57!X61X61
now only term 57! is divisible by 3
now the no.of 3’s needed can be calculated as
greatest no. in 57! which is divisible by 3 is 57 itself
3. ,19’s are 57 therefore, there are total 19 no.s that are divisible by 3 of which 6 no.s are divisible by 9(as 9X6=54 is largest in it ) , and in these no’s one no. is 27 .Now total no.of 3’s from 3,9 and 27 are (19-6)X1+(5X2)+(1X3) = 26. as 3^1=3 ,3^2=9 and 3^3=27 Hence 3^26 is the larget required no. [may be]

p.s- I know procedure is little long , im looking forward to your short explaination

• Chris Lele July 22, 2014 at 11:57 am #

Hi Tarun,

Sooo close. I think you missed one of the 3’s. The answer is 27. I think you forgot the three 3’s in 54. I’ll have a quick answer up later this week 🙂

• Tarun July 22, 2014 at 8:17 pm #

haha yeah dang!! I missed 27 in 54 that would make it 27

14. Ankur July 21, 2014 at 7:20 pm #

15. Vasanth July 21, 2014 at 7:18 pm #

59! (60*61 + 60 + 1) = 59!*61*61. number of 3’s in this is = number of multiples of 3 (19) + extra 3’s in multiples of 9 and 27(6+2=8) .Total 27 3’s. x= 27

16. Ram July 21, 2014 at 5:07 pm #

Hi Chris,

I think x=27.

• Chris Lele July 22, 2014 at 1:38 pm #

Yep, that’s it!

17. Shahadat Hasan REZA July 21, 2014 at 4:29 pm #

27 should be the answer. First set of multiples of 3 in 59!*61*61 are 3 through 57, 19 in total, then from 3 to 18, 6 in total and then 3 and 6, 2 in total. Adding them results in 27

18. Shahadat Hasan REZA July 21, 2014 at 4:14 pm #

I think it’s 22. The factorial can be simplified as 59!*61*61. Now since 59! Has 19 multiples of 3 in it and 3 more for 27 and 9. Thus 22 in total.

19. Shahadat Hasan REZA July 21, 2014 at 4:09 pm #

Is it 19? I think the factorial can be simplified into the form 59!*61^2. Now since 59! Has 20 numbers in it that are multiples of 3, the answer should be 19.

20. Sriram July 21, 2014 at 1:40 pm #

I presume that the standard formula to determine prime decomposition should help. Using the same, amongst (61!+60!+59!) for n to be an integer, its suffices to consider the smallest amongst the 3 values, that is, 59!/3^x which works out to be 59/3 + 59/3^2 + 59/3^3 = 19+6+2 = 27 (taking integer values)

• Chris Lele July 22, 2014 at 1:46 pm #

Yep, that’s it! Good job 🙂

21. priya July 21, 2014 at 12:33 pm #

hi Chris,

value of x = 27.

59/3 + 59/9 + 59/27 =27.

• Chris Lele July 22, 2014 at 1:47 pm #

Good–you got it! Of course factoring out of 61 is important as well 🙂

22. Ankit July 21, 2014 at 11:33 am #

Hello Chris,
you are doing a great job with your posts and helping thousands of students

i think the answer is 26

23. max July 21, 2014 at 11:06 am #

(61!+60!+59!)/3^x,

The numerator is 59! (61*61*61)

59! will consist of 19 multiples of 3 that are starting from 57 till 3
So, x will be 1+3+……19= 19*20/2=190

Greatest value can be 190

24. Tarun July 21, 2014 at 10:45 am #

explaination:-

simplifying. the numerator
59!(61X60+60+1)=59!X61X61=59X58X57!X60X61
now only term 57! is divisible by 3
now the no.of 3’s needed can be calculated as
greatest no. in 57! which is divisible by 3 is 57 itself
3. ,19’s are 57 therefore, there are total 19 no.s that are divisible by 3 of which 6 no.s are divisible by 9(as 9X6=54 is largest in it ) , and in these no’s one no. is 27 .Now total no.of 3’s from 3,9 and 27 are (19-6)X1+(5X2)+(1X3) = 26. as 3^1=3 ,3^2=9 and 3^3=27 Hence 3^26 is the larget required no. [may be]
p.s- I know procedure is little long , im looking forward to your short explaination

25. Utsav Akhoury July 21, 2014 at 10:29 am #

x should be 19

My logic was that 59! needs to be dived by a 3^x such that there no remainders. 57 is the largest number in 59! that is divisible by 3 which is 3*19.

This is applicable for 60! and 61! Cant have 20 3’s as that may satisfy 60! and 61! but not 59! so n would not remain and integer.

26. Piyush Chaudhary July 21, 2014 at 10:00 am #

Is it 27 ?

27. Ankur July 21, 2014 at 9:57 am #

The greatest value of x is 27.
For n to be integer, 61!/3^x, 60!/3^x and 59!/3^x all the three terms should be integers.
And we can find the greatest value of x for which 59!/3^x be an integer then obviously the other two terms will also be integers..
And there are 27 threes in 59!.
So the ans should be 27

28. Ankur July 21, 2014 at 9:45 am #

For n to be integer, 61!/3^x , 60!/3^x & 59!/3^x …all the three terms should be an integer.
And if we can the greatest value of x for 59!/3^x to be integer , obviously the other two terms will also be integer…
And there are 27 threes in 59!

29. Piyush Chaudhary July 21, 2014 at 9:39 am #

Is it 27 ??
If it is right then I will tell the explanation.

30. Guri Kejriwal July 21, 2014 at 9:39 am #