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by Prasanna » Sat Apr 28, 2007 1:52 pm
The question is asking us to determine how many numbers such as 324713,324813, 324913 .....are there between 324,700 and 458,600.

For every hundered there is a 13(a integer with 1 in tens digit and 3 in units digit). Between 324,700 and 458,600, there are 1339 hunderds. (458600-324700=133900). Hence the answer would be (E).
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by dunkin77 » Sat Apr 28, 2007 2:08 pm
Thank you!

yes the answer is E)
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by Cybermusings » Mon Apr 30, 2007 1:16 am
Thanks for the simple solution! I was going nuts with permutation!
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by Scott@TargetTestPrep » Mon May 20, 2019 6:14 pm
dunkin77 wrote:. How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3?
(A) 10,300
(B) 10,030
(C) 1,353
(D) 1,352
(E) 1,339
The first few integers that have this property are: 324,713, 324,813, 324,913, .... As we can see, this is an arithmetic sequence with common difference of 100. Since the last integer that has this property is 458,513, the number of integers that have this property is:

(458,513 - 324,713)/100 + 1 = 133,800/100 + 1 = 1,338 + 1 = 1,339

Answer: E

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by Brent@GMATPrepNow » Tue May 21, 2019 6:17 am
dunkin77 wrote:. How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3?
(A) 10,300
(B) 10,030
(C) 1,353
(D) 1,352
(E) 1,339

Hi,

Can anyone help pls..?
We want to determine how many integers end in 13
For example, 324713, 324813, 324913, 325013, . . . etc.

Recognize that, for every 100 consecutive integers, only 1 will end in 13.

There are 133,900 integers between 324,700 and 458,600 (since 458,600 - 324,700 = 133,900)
Of those 133,900 integers, 1/100 of them end in 13.

1/10 of 133,900 = 1339, so the correct answer is E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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