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Tough one - Less than 10000

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Tough one - Less than 10000 [#permalink] New post 01 Jul 2009, 06:02
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A
B
C
D
E

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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
How many integers less than 10,000 are there in which sum of digits equals 5?

(A) 31
(B) 51
(C) 56
(D) 62
(E) 93
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Re: Tough one - Less than 10000 [#permalink] New post 01 Jul 2009, 07:09
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And should be C - 56

The possible combinations are

2,3;...... 4,1;...... 5;....... 1,1,3;...... 2,2,1;...... 1,1,1,2

permutations are as follows (using 4 places, and putting zeros in front for 1,2 and 3 digit numbers)

0,0,2,3 = !4/!2 = 12

0,0,4,1 = 12 (same as above)

0,0,0,5 = !4/!3 = 4

0,1,1,3 = !4/!2 = 12

0,2,2,1 = !4/!2 = 12

1,1,1,2 = !4/!3 = 4

Total = 12*4 + 4*2 = 56

Hope the explanation is clear.

btw, thanks for the great question. :)
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Re: Tough one - Less than 10000 [#permalink] New post 01 Jul 2009, 08:37
I wonder if I could have understood this!! :).

Perhaps I need to have initial study on this, then I can take it more easily!!

Thanks for the solution though!!

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Re: Tough one - Less than 10000 [#permalink] New post 01 Jul 2009, 10:42
rashminet84 you are a genius. Not sure what are you doing here....go give ur GMAT... :)
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Re: Tough one - Less than 10000 [#permalink] New post 01 Jul 2009, 10:50
sdrandom1 wrote:
rashminet84 you are a genius. Not sure what are you doing here....go give ur GMAT... :)



Really?? :)

Thanks for your kind words. though i am pretty weak in a lot of areas. I just learnt a few concepts while preparing for the Indian GMAT - Common Admission Test.
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Re: Tough one - Less than 10000 [#permalink] New post 01 Jul 2009, 13:11
So kinda bored at work. And decided to visit the old forums. This one was not too dif.

My approach is not as good or quick as rashmine, but it works.

Ignore all numbers above 5001.

For 1000's Digit We have

5000 nothing above this will work.

4001, nothing above this combo works. Ignore the 4 and we have 3!/2!. Why because 001, 100, 010

3002, Again 3!/2!

3011, again 3!/2!

2003, 3!/2!

2012, 3!/2!

2102, 3!/2!

2201, 3!/2

1013, 3! --> 013 has six combinations

1022, 3!/2!

1121, 3!/2!

1004, 3!/2!

Now to 100's

500,

401, 3!

320, not 3! b/c 0 cannot be in hundreds digit.

113, 3!/2!

10's digit:

50, 41, 14, 32, 23

1's - 5.


Achieved in 4min, so I would def. reccomend Rashmine's approach.
Re: Tough one - Less than 10000   [#permalink] 01 Jul 2009, 13:11
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