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How many distinct quotients can be obtained if both the

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How many distinct quotients can be obtained if both the [#permalink] New post 05 Jul 2006, 06:44
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How many distinct quotients can be obtained if both the dividend and the divisor are integers from 1 to 10?

(A) 63 (B) 65 (C) 67 (D) 69 (E) 71

Last edited by kevincan on 05 Jul 2006, 22:30, edited 1 time in total.
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Re: PS. Distinct Quotients [#permalink] New post 05 Jul 2006, 17:04
kevincan wrote:
How many distinct quotients can be obtained if both the dividend and the divisor are positive digits?

(A) 63 (B) 65 (C) 67 (D) 69 (E) 71


i have tried, and I dont' think that I understand the question...
my answer is not one of the five listed above.

If I may ask, is this a GMAT question? I looked up "quotient" and still don't have an answer... :oops:
lets wait for one of the math gurus...
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 [#permalink] New post 05 Jul 2006, 17:26
Is there something missing..??

How many distinct quotients can be obtained if both the dividend and the divisor are positive digits?

Can you clarify the red part.
Thx!
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 [#permalink] New post 06 Jul 2006, 21:53
I tried this. But my method was very time consuming. What is the source of this problem?

All possible combinations - 10*10 = 100.
Now start pairing possible duplicates:
(1/2,2/4,3/6,4/8,5/10) and their reciprocals.
(1/3,2/6,3/9) and their reciprocals.
(1/4,2/8) and their reciprocals.
(1/5,2/10) and their reciprocals.
(2/3,4/6,6/9) and their reciprocals.
(2/5,4/10) and their reciprocals.
(3/4,6/8) and their reciprocals.
(3/5,6/10) and their reciprocals.
(1/1,2/2,....10/10).

Subtract and get the answer. I am feeling too tired to do so!

Hint: You would need to consider at least one member from each group and the corresponding group of its reciprocals.
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 [#permalink] New post 07 Jul 2006, 00:05
Zooroopa wrote:
I tried this. But my method was very time consuming. What is the source of this problem?

All possible combinations - 10*10 = 100.
Now start pairing possible duplicates:
(1/2,2/4,3/6,4/8,5/10) and their reciprocals.
(1/3,2/6,3/9) and their reciprocals.
(1/4,2/8) and their reciprocals.
(1/5,2/10) and their reciprocals.
(2/3,4/6,6/9) and their reciprocals.
(2/5,4/10) and their reciprocals.
(3/4,6/8) and their reciprocals.
(3/5,6/10) and their reciprocals.
(1/1,2/2,....10/10).

Subtract and get the answer. I am feeling too tired to do so!

Hint: You would need to consider at least one member from each group and the corresponding group of its reciprocals.


you are missing 4/5, 8/10 and their reciprocals.
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 [#permalink] New post 07 Jul 2006, 04:59
You are right the list was so long that I forgot noting down a few while I replied to the answer. I have them in my scratchpad!! :)

Is there a shorter method?
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 [#permalink] New post 07 Jul 2006, 17:13
kevin,

do you have OA and OE?

I would really like to know how this works...

thanks...
  [#permalink] 07 Jul 2006, 17:13
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