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17 Aug 2019, 08:54
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B
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Difficulty:
35%
(medium)
Question Stats:
68% (01:24) correct
32%
(01:41)
wrong
based on 142
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What is the highest power of 111 which divides 30! * 31! * 32! * 33! * 34! * 35! * 36! * 37! * 38! * 39! * 40! ?
A. 3
B. 4
C. 5
D. 6
E. 7
A. 3
B. 4
C. 5
D. 6
E. 7
17 Aug 2019, 15:42
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This is a very difficult question in case you don't the shortcut about divisibility for factorial numbers. So firstly let me show you the concept with example numbers:
1) METHDOLOGY TO ANSWER THESE TYPES OF QUESTIONS
To find out k of the factor 2^k in the factorial number 20!, we need to find the TOTAL number of 2s in 20! (which is the same as finding the maximum value of k that makes 2^k divisible by 20!)
I will break down the concept first so that you can understand what we have behind the same "mechanics" we will be applying to solve this kind of questions
20! = 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20
Of the numbers above, these are the numbers that have 2 as a factor: 2, 4 (two of them), 6, 8 (three of them), 10, 12 (two of them), 14, 16 (four of them), 18, 20 (two of them)
This makes 1 + 2 + 1 + 3 + 1 + 2 + 1 + 4 + 1 + 2 = 18
And this is the method to solve it
1º: 20/2 = 10
2º: 10/2 = 5
3º: 5/2 = 2
4º: 2/2 = 1
Stop when quotient would give 0 (next one would be 1/2, whose quotient is zero)
So k = 10 + 5 + 2 + 1 = 18
The greatest possible value of k is 8.
The logic behind this is that each alternate number in 20! will have a 2. Out of 20 numbers, 10 numbers will have a 2 (therefore 1º: 20/2 = 10) (these numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20)
Now, out of these 10 numbers, every alternate number will have another 5 numbers that will have a 2 (therefore 2º: 10/2 = 5)
Basically, same thing with 3º and 4º
Now, to account for all the 2s we add the quotient we've been obtaining 10 + 5 + 2 + 1 = 18 (These are the number of 2s in 20!)
By doing this, you can find the maximum power of any number in any factorial, but we need to know one extra thing for non-prime numbers
If I ask you now if 6^k is a factor of (78!), what is the greatest possible value of k? What would you tell me?
factoring 6 we get 2*3, so which factor should we use to do our method? Easy answer. The one that is more restrictive, in other words, the highest factor. In our case 3. So we would proceed as follows:
Solution:
1º 78/3 = 26
2º 26/3 = 8
3º 8/3 = 2
Greatest possible value of k is 26 + 8 + 2 = 36 (This would be for both 3 and 6)
So now let's tackle the given question
2) ANSWERING GIVEN QUESTION
First impression is that this question looks harder in comparison to what I just showed you because it seems we would need to do it 11 times (for 30!, 31!,...40!). But before making any conclusion, let's factorize 111 to see what we get
This gives 111=3*37, so 37 is the most restrictive number.
Now, wait a second, it seems that 30!, 31!,...36! cannot be divided by 37 cause those factorial numbers would never have this factor as it is higher than the factorial number themselves. So we would need to the method for 37!,...40! (4 times). It would still look like a lot of work, but remember that we need to divide those factorial numbers by 37 so instead of having 4 steps as we had with 2^k in the explanation before, we only have 1 step, so let's do it
- 37! ---> 40/37 = 1 (no more cause 1/37 has quotient zero)
- 38! ---> 40/38 = 1 (no more cause 1/38 has quotient zero)
- 39! ---> 40/39 = 1 (no more cause 1/39 has quotient zero)
- 40! ---> 40/40 = 1 (no more cause 1/40 has quotient zero)
So k for 37 and 111 would be 1 + 1 + 1 + 1 = 4
OPTION B
Please give me kudos if you definitely learnt the way to approach this kind of questions
1) METHDOLOGY TO ANSWER THESE TYPES OF QUESTIONS
To find out k of the factor 2^k in the factorial number 20!, we need to find the TOTAL number of 2s in 20! (which is the same as finding the maximum value of k that makes 2^k divisible by 20!)
I will break down the concept first so that you can understand what we have behind the same "mechanics" we will be applying to solve this kind of questions
20! = 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20
Of the numbers above, these are the numbers that have 2 as a factor: 2, 4 (two of them), 6, 8 (three of them), 10, 12 (two of them), 14, 16 (four of them), 18, 20 (two of them)
This makes 1 + 2 + 1 + 3 + 1 + 2 + 1 + 4 + 1 + 2 = 18
And this is the method to solve it
1º: 20/2 = 10
2º: 10/2 = 5
3º: 5/2 = 2
4º: 2/2 = 1
Stop when quotient would give 0 (next one would be 1/2, whose quotient is zero)
So k = 10 + 5 + 2 + 1 = 18
The greatest possible value of k is 8.
The logic behind this is that each alternate number in 20! will have a 2. Out of 20 numbers, 10 numbers will have a 2 (therefore 1º: 20/2 = 10) (these numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20)
Now, out of these 10 numbers, every alternate number will have another 5 numbers that will have a 2 (therefore 2º: 10/2 = 5)
Basically, same thing with 3º and 4º
Now, to account for all the 2s we add the quotient we've been obtaining 10 + 5 + 2 + 1 = 18 (These are the number of 2s in 20!)
By doing this, you can find the maximum power of any number in any factorial, but we need to know one extra thing for non-prime numbers
If I ask you now if 6^k is a factor of (78!), what is the greatest possible value of k? What would you tell me?
factoring 6 we get 2*3, so which factor should we use to do our method? Easy answer. The one that is more restrictive, in other words, the highest factor. In our case 3. So we would proceed as follows:
Solution:
1º 78/3 = 26
2º 26/3 = 8
3º 8/3 = 2
Greatest possible value of k is 26 + 8 + 2 = 36 (This would be for both 3 and 6)
So now let's tackle the given question
2) ANSWERING GIVEN QUESTION
First impression is that this question looks harder in comparison to what I just showed you because it seems we would need to do it 11 times (for 30!, 31!,...40!). But before making any conclusion, let's factorize 111 to see what we get
This gives 111=3*37, so 37 is the most restrictive number.
Now, wait a second, it seems that 30!, 31!,...36! cannot be divided by 37 cause those factorial numbers would never have this factor as it is higher than the factorial number themselves. So we would need to the method for 37!,...40! (4 times). It would still look like a lot of work, but remember that we need to divide those factorial numbers by 37 so instead of having 4 steps as we had with 2^k in the explanation before, we only have 1 step, so let's do it
- 37! ---> 40/37 = 1 (no more cause 1/37 has quotient zero)
- 38! ---> 40/38 = 1 (no more cause 1/38 has quotient zero)
- 39! ---> 40/39 = 1 (no more cause 1/39 has quotient zero)
- 40! ---> 40/40 = 1 (no more cause 1/40 has quotient zero)
So k for 37 and 111 would be 1 + 1 + 1 + 1 = 4
OPTION B
Please give me kudos if you definitely learnt the way to approach this kind of questions
17 Aug 2019, 17:02
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\(\frac{(30! * 31! * 32! * 33! * 34! * 35! * 36! * 37! * 38! * 39! * 40!)}{111^{k}}\)-- what is the highest value of k???
111=3*37
--> 37 (prime number) is involved only in the last 4 factorial:
37!=1*2*3...*36*37
38!=1*2*3...*36*37*38
39!=1*2*3...*36*37*38*39
40!=1*2*3...*36*37*38*39*40
So, the highest value of k will be 4.
The answer is B.
111=3*37
--> 37 (prime number) is involved only in the last 4 factorial:
37!=1*2*3...*36*37
38!=1*2*3...*36*37*38
39!=1*2*3...*36*37*38*39
40!=1*2*3...*36*37*38*39*40
So, the highest value of k will be 4.
The answer is B.
