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digits & decimals problem set, page 21

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jhgordin
 Post subject: digits & decimals problem set, page 21
 Post Posted: Thu Nov 05, 2009 8:28 am 
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Course Students


Posts: 1
2. What is the sum of all possible 3-digit numbers that can be constructed using the digits 3, 4, and 5, if each digit can be used only once in each number?

Solution:

100(24) + 10(24) + (24) = 2400 + 240 + 24 = 2664

Can you please explain how the logic behind 24?

Thanks.
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agha79
 Post subject: Re: digits & decimals problem set, page 21
 Post Posted: Thu Nov 05, 2009 9:12 am 
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Course Students


Posts: 98
what is the offical answer
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rajbir.dadhiyala
 Post subject: Re: digits & decimals problem set, page 21
 Post Posted: Thu Nov 05, 2009 12:54 pm 
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Students


Posts: 6
Quote:
100(24) + 10(24) + (24) = 2400 + 240 + 24 = 2664

Can you please explain how the logic behind 24?


How many numbers can we form using numbers 3,4,5 when we knwo that no number can be repeated?

Six numbers

and on each place ie. units, tens, and hundreds, these nos will appear twice

so 5 + 5 + 4 + 4 + 3 + 3 = 24

and hence

100(24) + 10(24) + (24) = 2400 + 240 + 24 = 2664

Hope u get it

By the way, if you don't get this, just jot those nos

345 + 354 + 435 + 453 + 534 + 543 = 2664
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RonPurewal
 Post subject: Re: digits & decimals problem set, page 21
 Post Posted: Fri Dec 18, 2009 5:24 am 
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ManhattanGMAT Staff


Posts: 7146
yeah - this is definitely one of those problems on which you should just list all the possibilities, if you don't IMMEDIATELY grasp the shorter way of doing the problem.

even if you have to do this - list all the possibilities and add them up - you can still solve the problem in well under one minute.

is this problem actually from GMATPREP? (this is the GMATPREP-only folder.)
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