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prep question

Expert replies
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: — Problem Solving |

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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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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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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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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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!!

Stacey Koprince
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Manhattan GMAT

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by abkhan » Tue May 22, 2007 4:21 am
so sorry .. i misunderstood the question . Was looking for 6^n in 324 where as we should be looking for 324 in 6^n..
I am ..therefore I am..
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