If x is the product of the positive integers from 1 to 8,

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II: Master | Next Rank: 500 Posts; Posts: 400; Joined: Mon Dec 10, 2007 1:35 pm; Location: London, UK; Thanked: 19 times; GMAT Score:680

by II » Wed Jul 02, 2008 3:40 pm

If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = (2^i)(3^k)(5^m)(7^p), then i+k+m+p =

(A) 4 (B) 7 (C) 8 (D) 11 (E) 12

I am very interested getting views and feedback on the following:
- alternative approaches to solving this. Please illustrate your logic.
- guessing strategies you would use on this type of question, if you were running out of time, or if you simply had to guess on this question
- the traps and tricks which are built into this question (if you can see any)
- any key take-aways, strategies to keep in mind for identiying future similar problems etc.

Thanks in advance.
II

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Source: Beat The GMAT — Problem Solving | Wed Jul 02, 2008 3:40 pm

sibbineni: Master | Next Rank: 500 Posts; Posts: 222; Joined: Fri Jan 04, 2008 8:10 pm; Thanked: 15 times

by sibbineni » Wed Jul 02, 2008 4:58 pm

x = (2^i)(3^k)(5^m)(7^p)-----(2)

and x =1*2*3*4*5*6*7*8
=2*3*2^2*5*2*3*7*2^3
=2^9*3^2*5*7----(1)

(1)=(2)
2^9*3^2*5*7=(2^i)(3^k)(5^m)(7^p)

equating the bases and we have

i=9 k=2 m=1p=1

and i+k+m+p=13

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beeparoo: Master | Next Rank: 500 Posts; Posts: 200; Joined: Sun Jun 17, 2007 10:46 am; Location: Canada; Thanked: 9 times

by beeparoo » Wed Jul 02, 2008 5:06 pm

sibbineni wrote:x = (2^i)(3^k)(5^m)(7^p)-----(2)

and x =1*2*3*4*5*6*7*8
=2*3*2^2*5*2*3*7*2^3
=2^7*3^2*5*7----(1)

(1)=(2)
2^7*3^2*5*7=(2^i)(3^k)(5^m)(7^p)

equating the bases and we have

i=7 k=2 m=1p=1

and i+k+m+p=11

Everything looks good, but I think you made a small calculation error. But the logic is a-OK.

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II: Master | Next Rank: 500 Posts; Posts: 400; Joined: Mon Dec 10, 2007 1:35 pm; Location: London, UK; Thanked: 19 times; GMAT Score:680

by II » Thu Jul 03, 2008 1:16 am

Thanks guys ... that is also the approach I used to solve this. 11 is the right answer.

x is the product of the integers from 1 to 8, inclusive: x=1*2*3*4*5*6*7*8

x = (2^i)(3^k)(5^m)(7^p) – this is representing x in terms of it prime factors.

we just need to find out how many of the prime factors, 2, 3, 5, and 7 exist in x. See attachment. We then count the 2's, 3's, 5's. and 7's ... and there you have the values of i, k, m and p !

However, as part of my analysis/review of the question, I am interested in the following:

- alternative (secondary)approaches to solving this. Is there another approach/technique to arrive to the same answer. Please illustrate your logic.
- guessing strategies you would use on this type of question, if you were running out of time, or if you simply had to guess on this question
- the traps and tricks which are built into this question (if you can see any)
- any key take-aways, strategies to keep in mind for identiying future similar problems etc.

Thanks.

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szapiszapo: Junior | Next Rank: 30 Posts; Posts: 28; Joined: Mon Jun 30, 2008 1:45 pm; Thanked: 3 times

by szapiszapo » Thu Jul 03, 2008 3:07 am

the Numbers Theory states that if n is a positive integer and p a prime number then the exponent e of p in the breakdown of n! in prime factors is e = [n/p] + [n/p]^2 + [n/p]^3 + [n/p]^4 + ...

(square brackets meaning that you just keep the integer part of the number)

let me explain with our exemple:

x is the product of the positive integers from 1 to 8, i.e. x =8! (i.e. n=8)
x = (2^i)(3^k)(5^m)(7^p), i.e. 2, 3, 5, 7 are all prime numbers and i, k, m, and p are "prime exponents"

let's apply that theorem for each prime number to find i, k, m, p
8 / 2 = 4
8 / 2^2 = 2
8 / 2^3 = 1
hence i = 7

8 / 3 = 2,66
8 / 3^2 n.a.
hence k = 2

8 / 5 = 1,6 hence m = 1
8 / 7 = 1,14 hence p = 1

although that approach is not really relevant with 8!, I can assure you it's way faster if you have to do the same with 100! or 1000!

if I had to give an advice, it would be to grap a math book and learn the basics of numbers theory, it's always useful

for example Numbers Theory says that n! is divisible by n+1 if n+1>4 and n+1 is not a prime number
it may help one day, who knows

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 amFollowed by: 22 members; Location: New York, NY; Thanked: 39 times

by Jeff@TargetTestPrep » Wed Nov 22, 2017 12:32 pm

II wrote:If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = (2^i)(3^k)(5^m)(7^p), then i+k+m+p =

(A) 4 (B) 7 (C) 8 (D) 11 (E) 12

Let's break 8! into prime factors:

8 x 7 x 6 x 5 x 4 x 3 x 2

2^3 x 7 x 2 x 3 x 5 x 2^2 x 3 x 2

2^7 x 3^2 x 5^1 x 7^1

We see that i = 7, k = 2, m = 1, and p = 1.

Thus, i + k + m + p = 7 + 2 + 1 + 1 = 11.

Answer: D

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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