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Comment on Listing and Counting
Can you pl explain this - "In
I am confused by the solution.
1) I don't understand the conclusion 'all digits occur the same number of times'
2) is there a different way altogether to solve it, for example if were asked to count the number of zeros, this approach wont work right?
Question link: http:/
Question link: http://gre.myprepclub.com/forum/gre-math-challenge-306.html
1) If we focus on the UNITS digits as we count to 300, we get a units digit of 1 every 10 numbers. So, 1/10 of the 300 numbers will have 1 in the units digit place. This gives us 30 1's so far.
If we focus on the TENS digits as we count to 300, we get a 1 in the TENS location 10 times out of every 100 numbers (e.g., 210, 211, 212, 213..., 219). So, 1/10 of the 300 numbers will have 1 in the TENS digit place. This gives us another 30 1's.
Finally, the numbers from 100 to 199 have a 1 in the HUNDREDS position. So, there are another 100 1's here.
TOTAL = 30 + 30 + 100 = 160
2) We could modify this technique to make it work for counting 0's.
Thank you.
Can you explain the solution in the case we had to find number of zeros used from 1 to 300.
According your above method I got - 30 + 9+ 9 + 1 = 49..
You bet.
You bet.
NEW question: In writing all of the integers from 1 to 300, how many times is the digit 0 used?
0's IN THE UNITS POSITION
We get a 0 in the UNITS position once every 10 values: 10, 20, 30, 40, . . . 280, 290, and 300
In other words, 1/10 of the 300 numbers from 1 to 300 have a 0 in the UNITS position.
1/10 of 300 = 30, so we have 30 zeros so far
0's IN THE TENS POSITION
We get a 0 in the UNITS position in the following formats:
10-
20-
30-
NOTE: the "-" represents a units digit.
For the first case of 10-, the last (units) digit can be 0, 1, 2, 3, 4, ... 9. So, there are 10 zeros used (in the TENS position) in numbers in the 10- format.
Likewise, there are 10 zeros used (in the TENS position) in numbers in the 20- format.
Finally, there's only 1 zero used (in the TENS position) in the 30- format (300).
So, the TOTAL number of 0's used = 30 + 10 + 10 + 1 = 51 (just like you have!)
For the second GRE Prep
I would have thought: 2 slots
First slot: 6 options
Second slot: 4 options
I don't want to double count AZ and ZA, so I divide my total of 24 by 2-factorial.
Is this not a combinations problem?
Hi aseiden,
Hi aseiden,
Question link: http://gre.myprepclub.com/forum/gre-math-challenge-117-on-elm-street-the...
When you're breaking the task into stages (aka slots), you need to be clear what you're actually doing during that stage. For example, what do you mean by "Second slot: 4 options"?
If you state what you're doing in each of those stages, you'll see that there is no duplication. So, we need not divide our answer by 2.
To see what I mean, check out my new post: http://gre.myprepclub.com/forum/gre-math-challenge-117-on-elm-street-the...
Please, regarding the
Am i right?
Actually, 333 and 777 meet
Actually, 333 and 777 meet the condition of using only 3's and 7's
Is it common to get questions
I'm having a harder time figuring out this method.
I'm trying to figure out if I can answer all (or at least most) counting questions on the GRE with other methods later in this section?
There will be times when
There will be times when listing and counting will be the fastest approach. It's also important to mention that, if you're having trouble answer a counting question using traditional counting methods, listing possible outcomes can often help you gain some insight into how to solve the question mathematically.
That said, if you're not having any difficulties answering the counting questions, then you may not need to use the list and count technique.
That question about 1 to 300
It's a difficult question,
It's a difficult question, but if you take a systematic approach to listing and counting, you'll quickly see a pattern.
1-DIGIT NUMBERS WITH 1's
1 (one 1)
Done
2-DIGITS NUMBERS WITH 1's
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31, 41, 51, 61, 71, 81, 91 (19 1's)
Done
3-DIGITS NUMBERS WITH 3's IN THE FORM 2XX
201, 210, 211, 212,... wait! This looks familiar. The number of 1's that appear in the form 2XX = 1 + 19 = 20 (which we got from our two lists above)
Likewise....
3-DIGITS NUMBERS WITH 3's IN THE FORM 1XX
The number of 1's that appear in the form 1XX = 20 + 100 = 120
Aside: There are 100 1's in the hundred digit place from 100 to 199
TOTAL number of 1's = 1 + 19 + 20 + 120 = 160
Cheers,
Brent