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What is the highest power of 6 that can divide 108 * 106! ?
A) 28
B) 53
C) 68
D) 50
E) 52
A) 28
B) 53
C) 68
D) 50
E) 52
Updated on: 12 Aug 2020, 21:55
Originally posted by garcmillan on 19 Aug 2019, 03:33.
Last edited by Bunuel on 12 Aug 2020, 21:55, edited 3 times in total.
Last edited by Bunuel on 12 Aug 2020, 21:55, edited 3 times in total.
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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
So in this questions we have the factor 6 = 2 * 3 and the factorial number 106, for which we will apply the explained method
- 106! --->
106/3 = 35
35/3 = 11
11/3 = 3
3/3 = 1
k = 1 + 3 + 11 + 35 = 50. So the answer would be 50 also for 6, as k is the same for the most restrictive factor than for 6.
But since we have 108 multiplying, we should factorie this number and add the powers of 3 to the 50 obtained i 106!. 108 = 2^2 * 3^3
Now adding 108 to the equation we have:
108 * 106! = 2^2 * 3^3 * 2^102 * 3^50 k = 2^104 * 3^53 k = 6^53 * 2^51 * k
Highest power of 6 in 108 * 106! = 53
OPTION B
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
So in this questions we have the factor 6 = 2 * 3 and the factorial number 106, for which we will apply the explained method
- 106! --->
106/3 = 35
35/3 = 11
11/3 = 3
3/3 = 1
k = 1 + 3 + 11 + 35 = 50. So the answer would be 50 also for 6, as k is the same for the most restrictive factor than for 6.
But since we have 108 multiplying, we should factorie this number and add the powers of 3 to the 50 obtained i 106!. 108 = 2^2 * 3^3
Now adding 108 to the equation we have:
108 * 106! = 2^2 * 3^3 * 2^102 * 3^50 k = 2^104 * 3^53 k = 6^53 * 2^51 * k
Highest power of 6 in 108 * 106! = 53
OPTION B
General Discussion
firas92
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I believe the answer should be \(53\)
\(106! = 2^{102}*3^{50}*x\) [where \(x\) is a positive integer which is not divisible by 2 or 3]
\(108 = 2^2*3^3\)
So \(108*106!=2^2*3^3*2^{102}*3^{50}*x=2^{104}*3^{53}*x=2^{53}*3^{53}*2^{51}*x=6^{53}*2^{51}*x\) [where \(x\) is a positive integer which is not divisible by 2 or 3]
\(106! = 2^{102}*3^{50}*x\) [where \(x\) is a positive integer which is not divisible by 2 or 3]
\(108 = 2^2*3^3\)
So \(108*106!=2^2*3^3*2^{102}*3^{50}*x=2^{104}*3^{53}*x=2^{53}*3^{53}*2^{51}*x=6^{53}*2^{51}*x\) [where \(x\) is a positive integer which is not divisible by 2 or 3]
