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jamesk486: Master | Next Rank: 500 Posts; Posts: 141; Joined: Wed Mar 28, 2007 9:24 pm; Thanked: 2 times; Followed by:1 members

prep question

by jamesk486 » Mon May 07, 2007 7:00 am

How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3?
A. 10,300
B. 10,030
C. 1353
D. 1352
E. 1339
=>is there an easy way to solve this?

What is the smallest positive integer n for which 324 is a factor of 6^n?
A. 2
B. 3
C. 4
D. 5
E. 6

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Source: Beat The GMAT — Problem Solving | Mon May 07, 2007 7:00 am

Cybermusings: Legendary Member; Posts: 559; Joined: Tue Mar 27, 2007 1:29 am; Thanked: 5 times; Followed by:2 members

by Cybermusings » Mon May 07, 2007 8:27 am

How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3?
A. 10,300
B. 10,030
C. 1353
D. 1352
E. 1339

You must remember that numbers ending with 13 come once in every 100 numbers...
458600 - 324700 = 133900...Thus there must be 133900/100 = 1339 such numbers

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Cybermusings: Legendary Member; Posts: 559; Joined: Tue Mar 27, 2007 1:29 am; Thanked: 5 times; Followed by:2 members

by Cybermusings » Mon May 07, 2007 8:33 am

What is the smallest positive integer n for which 324 is a factor of 6^n?
A. 2
B. 3
C. 4
D. 5
E. 6

6^1 = (2^1)*(3^1)
6^n = (2^n)*(3^n)

324 = (2^2)*(3^4)

So if 324 is a factor it should divide 6^n completely (without any remainder)
6^n / [(2^2)*(3^4)] = An Interger

Thus n's minimum value is 4

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f2001290: Master | Next Rank: 500 Posts; Posts: 400; Joined: Sat Mar 10, 2007 4:04 am; Thanked: 1 times; Followed by:1 members

by f2001290 » Sat May 19, 2007 8:05 am

For the second question

324 = 6^2 * 3^2

so minimum value of n = 2

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abkhan: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Thu May 03, 2007 2:20 am; Location: India

by abkhan » Mon May 21, 2007 1:46 am

Cybermusings wrote:What is the smallest positive integer n for which 324 is a factor of 6^n?
A. 2
B. 3
C. 4
D. 5
E. 6

6^1 = (2^1)*(3^1)
6^n = (2^n)*(3^n)

324 = (2^2)*(3^4)

So if 324 is a factor it should divide 6^n completely (without any remainder)
6^n / [(2^2)*(3^4)] = An Interger

Thus n's minimum value is 4

f2001290 approach is right for the given question.

Cybermusings qould be right if the question were

What is the smallest positive integer n for which 324! is a divisible by 6^n?

I am ..therefore I am..

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Stacey Koprince: GMAT Instructor; Posts: 2228; Joined: Wed Dec 27, 2006 3:28 pm; Location: Montreal, Canada; Thanked: 639 times; Followed by:694 members; GMAT Score:780

by Stacey Koprince » Mon May 21, 2007 9:31 am

Actually, cybermusings is right.

What is the smallest positive integer n for which 324 is a factor of 6^n?

If n = 2, then we'd be talking about 6^2 = 36. 324 is not a factor of 36.
If n = 4, then 6^4 = 324. 324 is a factor of 324.

Let me just clarify something at the end of cybermusing's explanation:

6^1 = (2^1)*(3^1) allows us to split 6 into its primes

6^n = (2^n)*(3^n) showing the split

324 = (2^2)*(3^4) showing 324 split into its primes

Notice that n (the same number) applies to the exponents for BOTH of the terms, so I cannot just use 324's prime split (2^2 * 3^4). So my two choices are either:

(2^2)*(3^2) OR (2^4)*(3^4)

The first one is too small, so it has to be the second one (n=4).

Also, as a reality check, ALWAYS be wary of choosing the smallest given answer choice on a question that asks for the smallest number or the largest given answer choice on a question that asks for the largest number. Those answers are traps (because a decent number of people will pick those answers simply because they are the smallest / largest numbers presented).

Please note: I do not use the Private Messaging system! I will not see any PMs that you send to me!!

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