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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]  05 Jul 2006, 05:44
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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, 21:30, edited 1 time in total.
Manager
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Re: PS. Distinct Quotients [#permalink]  05 Jul 2006, 16:04
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...
lets wait for one of the math gurus...
VP
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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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"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

Manager
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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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Thanks,
Zooroopa

VP
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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.
Manager
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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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Thanks,
Zooroopa

Manager
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kevin,

do you have OA and OE?

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

thanks...
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