If p is the product of the integers from 1 to 30,
inclusive, what is the greatest integer k for which 3k is
a factor of p ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
If p is the product of the integers from 1 to 30,
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Hi vikkimba17,vikkimba17 wrote:If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
Pl. post the correct question. It's not '3k,' it's '3^k.' Moreover, do mention the correct answer and the source of the question.
We have to find the greatest value of k such that 3^k is a factor of 30! (1*2*3...*30).
Thus, we have to count the total number of 3s in 30!.
It would be given by quotients of [30/3 + 30/(3^2) + 30/(3^3) + ......]; we must not add '30/(3^4)' since 3^4 = 81 > 30.
Thus the greatest value of k = quotients of [30/3 + 30/(3^2) + 30/(3^3)] = quotients of [30/3 + 30/9 + 30/27] = 10 + 3 + 1 = 14.
The correct answer: C
Hope this helps!
Relevant book: Manhattan Review GMAT Math Essentials Guide
-Jay
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The method Jay described is the most efficient way to solve this problem, but because the number of multiples of 3 is reasonably small, it's also feasible to employ brute force, and just list them out and count the number of threes in each number's prime factorization.vikkimba17 wrote:If p is the product of the integers from 1 to 30,
inclusive, what is the greatest integer k for which 3k is
a factor of p ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
3 --> 1 three
6 --> 1 three
9 --> 2 threes (9 = 3^2)
12--> 1 three
15 --> 1 three
18 --> 2 threes (18 = 2 * 3^2)
21---> 1 three
24 -->1 three
27 --> 3 three's (27 = 3^3)
30 ---> 1 three
1 + 1 + 2 + 1 + 1 + 2 + 1 + 1 + 3+ 1 = 14; The answer is C
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For a lengthier post describing the mechanics of such problems, see here: https://www.beatthegmat.com/mba/2011/07/ ... r-problems
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vikkimba17 wrote:If p is the product of the integers from 1 to 30,
inclusive, what is the greatest integer k for which 3^k is
a factor of p ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
This question is really asking us to determine the number of 3's "hiding" in the prime factorization of p.
p = (1)(2)(3)(4)(5)(6)(7)(8)(9) . . . (27)(28)(29)(30)
= (1)(2)(3)(4)(5)(2)(3)(7)(8)(3)(3)(10)(11)(3)(4)(13)(14)(3)(5)(16)(17)(3)(3)(2)(19)(20)(3)(7)(22)(23)(3)(8)(25)(26). . . (3)(3)(3)(28)(29)(3)(10)
= (3)^14(other non-3 stuff)
Answer: C
Cheers,
Brent
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I posted a solution here:
https://www.beatthegmat.com/product-of-t ... 94247.html
https://www.beatthegmat.com/product-of-t ... 94247.html
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The product of the integers from 1 to 30 inclusive is 30!. Thus, we need to determine the number of factors of 3 in 30!. To do so, we can use the following shortcut in which we divide 30 by 3 then divide the quotient (ignore any nonzero remainder) by 3, and then continue this process until we no longer get a nonzero quotient.vikkimba17 wrote:If p is the product of the integers from 1 to 30,
inclusive, what is the greatest integer k for which 3k is
a factor of p ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
30/3 = 10
10/3 = 3 (we can ignore the remainder)
3/3 = 1
Since 1/3 does not produce a nonzero quotient, we can stop.
The final step is to add up our quotients; that sum represents the number of factors of 3 within 30!.
Thus, there are 10 + 3 + 1 = 14 factors of 3 within 30!.
Answer: C
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Hi vikkimba17,
These types of questions are based on a math concept called "prime factorization", which basically means that any integer greater than 1 is either prime OR the product of a bunch of primes.
Here's a simple example:
24 = (2)(2)(2)(3)
Now, when it comes to this question, we're asked to multiply all the integers from 1 to 30, inclusive and find the greatest integer K for which 3^K is a factor of this really big number.
Here's a simple example with a smaller product:
1 to 6, inclusive...
(1)(2)(3)(4)(5)(6)
Then numbers 1, 2, 4 and 5 do NOT have any 3's in them, so we can essentially ignore them:
3 = one 3
6 = (2)(3) = one 3
Total = two 3's
So 3^2 is the biggest "power of 3" that goes into the product of 1 to 6, inclusive.
Using that same idea, we need to find all of the 3's in the product of 1 to 30, inclusive. Here though, you have to be careful, since there are probably MORE 3's than immediately realize:
3 = one 3
6 = one 3
9 = (3)(3) = two 3s
12 = one 3
15 = one 3
18 = (2)(3)(3) = two 3s
21 = one 3
24 = one 3
27 = (3)(3)(3) = three 3s
30 = one 3
Total = 14 3's
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
These types of questions are based on a math concept called "prime factorization", which basically means that any integer greater than 1 is either prime OR the product of a bunch of primes.
Here's a simple example:
24 = (2)(2)(2)(3)
Now, when it comes to this question, we're asked to multiply all the integers from 1 to 30, inclusive and find the greatest integer K for which 3^K is a factor of this really big number.
Here's a simple example with a smaller product:
1 to 6, inclusive...
(1)(2)(3)(4)(5)(6)
Then numbers 1, 2, 4 and 5 do NOT have any 3's in them, so we can essentially ignore them:
3 = one 3
6 = (2)(3) = one 3
Total = two 3's
So 3^2 is the biggest "power of 3" that goes into the product of 1 to 6, inclusive.
Using that same idea, we need to find all of the 3's in the product of 1 to 30, inclusive. Here though, you have to be careful, since there are probably MORE 3's than immediately realize:
3 = one 3
6 = one 3
9 = (3)(3) = two 3s
12 = one 3
15 = one 3
18 = (2)(3)(3) = two 3s
21 = one 3
24 = one 3
27 = (3)(3)(3) = three 3s
30 = one 3
Total = 14 3's
Final Answer: C
GMAT assassins aren't born, they're made,
Rich