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22 Jan 2024, 02:30
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If n is the product of all the integers from 5 to 20, inclusive, what is the greatest integer k for which 2^k is a factor of n ?
(A) 11
(B) 12
(C) 13
(D) 15
(E) 16
(A) 11
(B) 12
(C) 13
(D) 15
(E) 16
22 Jan 2024, 10:32
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If you count the amount of times that 2 factors into numbers from 5 to 20 you get 15
Posted from my mobile device
Posted from my mobile device
22 Jan 2024, 13:56
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I see two ways to do this.
Option A:
Under timed conditions, it was easiest for me to just write out all the consecutive integers from 5 to 20 and count how many factors of 2 each had, at which point I summed them to get 15. That said, I see two major problems with this. One, if you mess up even one number or don't know how many factors of 2 go into each integer off the top of your head, you'll be over time. Two, if you had a question asking you to find, say, the greatest \(k\) for which \(3^k\) is a factor of the product of consecutive integers from 7 to 63, it can be a lot more time consuming. So I tested out another option.
Option B:
In the spirit of mutual exclusivity of probabilistic events, I tested the hypothesis that you could first find how many powers of 2 go into the product of all consecutive integers from 1 to 20 (i.e. 20!), and subtracted out how many powers of 2 go into the product of all consecutive integers from 1 to 5 (i.e. 5!). Intuitively, this works because you're just removing that last bit of multiplication from 1 to 5. Mathematically, it works because you're dividing 5!, which will always have a constant number of powers of 2, which we can then subtract out. So:
Step One: How Many Powers of 2 Go into 20!?
\(\frac{20}{2^1} = 10\)
\(\frac{20}{2^2} = 5 \)
\(\frac{20}{2^3} = 2\)
\(\frac{20}{2^4} = 1\)
Grand total of 18.
Step Two: How Many Powers of 2 Go into 5!?
\(\frac{5}{2^1} = 2\)
\(\frac{5}{2^2} = 1\)
Grand total of 3.
Step Three: Subtraction
The general formula is # of powers of 2 in 20! - # of powers of 2 in 5!
==> 18 - 3 = 15.
AC D.
The one major caveat is if we're asked to find the number of powers of 2 from another multiple of 2 to 20, say from 6 to 20, removing the factors of 6! will remove all the factors of 6, and we will be left with 14 instead of the correct answer, which is 15. In this case, we simply have to add how many powers of 2 are in 6 (1), to get 15.
Encourage someone to break my hypothesis here.
Option A:
Under timed conditions, it was easiest for me to just write out all the consecutive integers from 5 to 20 and count how many factors of 2 each had, at which point I summed them to get 15. That said, I see two major problems with this. One, if you mess up even one number or don't know how many factors of 2 go into each integer off the top of your head, you'll be over time. Two, if you had a question asking you to find, say, the greatest \(k\) for which \(3^k\) is a factor of the product of consecutive integers from 7 to 63, it can be a lot more time consuming. So I tested out another option.
Option B:
In the spirit of mutual exclusivity of probabilistic events, I tested the hypothesis that you could first find how many powers of 2 go into the product of all consecutive integers from 1 to 20 (i.e. 20!), and subtracted out how many powers of 2 go into the product of all consecutive integers from 1 to 5 (i.e. 5!). Intuitively, this works because you're just removing that last bit of multiplication from 1 to 5. Mathematically, it works because you're dividing 5!, which will always have a constant number of powers of 2, which we can then subtract out. So:
Step One: How Many Powers of 2 Go into 20!?
\(\frac{20}{2^1} = 10\)
\(\frac{20}{2^2} = 5 \)
\(\frac{20}{2^3} = 2\)
\(\frac{20}{2^4} = 1\)
Grand total of 18.
Step Two: How Many Powers of 2 Go into 5!?
\(\frac{5}{2^1} = 2\)
\(\frac{5}{2^2} = 1\)
Grand total of 3.
Step Three: Subtraction
The general formula is # of powers of 2 in 20! - # of powers of 2 in 5!
==> 18 - 3 = 15.
AC D.
The one major caveat is if we're asked to find the number of powers of 2 from another multiple of 2 to 20, say from 6 to 20, removing the factors of 6! will remove all the factors of 6, and we will be left with 14 instead of the correct answer, which is 15. In this case, we simply have to add how many powers of 2 are in 6 (1), to get 15.
Encourage someone to break my hypothesis here.
