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by weena82 » Thu Dec 11, 2008 4:50 pm
I can't solve it within 2 mins. There must be some tricks to solve this. Anyone can tell?

How many positive intergers less than 10,000 are there in which the sum of the digits equal 5?
weena
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by earth@work » Thu Dec 11, 2008 6:27 pm
if the OA is 16 maybe you cud try this method, took me 1.50 minutes
first pick two digit nos which have total =5
05,50,14,41,23,32
now: now to make them three digit keeping the sum 5 just add zero to the end till u get number below 10000
05, 50-but we already have 50 so leave it as just 05
50,500,5000
14,140,1400
41,410,4100
23,230,2300
32,320,3200
gives us total 16 numbers.
infact even faster than this will be =1+5*3 =16 - just skip the arrangement above!
pls do confirm the OA
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by anayeri » Thu Dec 11, 2008 6:35 pm
if the OA is 16 maybe you cud try this method, took me 1.50 minutes
first pick two digit nos which have total =5
05,50,14,41,23,32
now: now to make them three digit keeping the sum 5 just add zero to the end till u get number below 10000
05, 50-but we already have 50 so leave it as just 05
50,500,5000
14,140,1400
41,410,4100
23,230,2300
32,320,3200
gives us total 16 numbers.
infact even faster than this will be =1+5*3 =16 - just skip the arrangement above!
pls do confirm the OA
What about numbers like 131, 311, or 4010? There's a ton missing from you list.

I feel like there's gotta be a shortcut. But I can't tell what it is either!
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by dmateer25 » Thu Dec 11, 2008 6:41 pm
I think the OA is 56 if I have all of the numbers.

But I can't see any patterns or shortcuts. Anyone have any ideas on how to solve this in under 2 mins?
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by logitech » Thu Dec 11, 2008 11:24 pm
0 - 9999

5 = 5+0 0005 = 4!/3! = 4
5= 4+1 0041 = 4!/2! = 12
5= 3+2 0032 = 4!/2! = 12
5= 3+1+1 0311 = 4!/2! = 12
5=2+2+1 0221 = 4!/2! = 12

so IMO, it is 52

If OA is 56, I might be missing 4 numbers somewhere and I am sure Cramya can find out where they are hiding.
LGTCH
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by dreamzz2010 » Fri Dec 12, 2008 2:02 am
logitech wrote:0 - 9999

5 = 5+0 0005 = 4!/3! = 4
5= 4+1 0041 = 4!/2! = 12
5= 3+2 0032 = 4!/2! = 12
5= 3+1+1 0311 = 4!/2! = 12
5=2+2+1 0221 = 4!/2! = 12

so IMO, it is 52

If OA is 56, I might be missing 4 numbers somewhere and I am sure Cramya can find out where they are hiding.
Hey Logitech,
yr method seems nice and logical enough. I guess 52 is the right ans
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by Stuart@KaplanGMAT » Fri Dec 12, 2008 3:17 am
logitech wrote:0 - 9999

5 = 5+0 0005 = 4!/3! = 4
5= 4+1 0041 = 4!/2! = 12
5= 3+2 0032 = 4!/2! = 12
5= 3+1+1 0311 = 4!/2! = 12
5=2+2+1 0221 = 4!/2! = 12

so IMO, it is 52

If OA is 56, I might be missing 4 numbers somewhere and I am sure Cramya can find out where they are hiding.
5 = 1+1+1+2 1112 = 4!/3! = 4

There's your missing numbers!

Great solution.
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by dmateer25 » Fri Dec 12, 2008 4:25 am
logitech wrote:0 - 9999

5 = 5+0 0005 = 4!/3! = 4
5= 4+1 0041 = 4!/2! = 12
5= 3+2 0032 = 4!/2! = 12
5= 3+1+1 0311 = 4!/2! = 12
5=2+2+1 0221 = 4!/2! = 12

so IMO, it is 52

If OA is 56, I might be missing 4 numbers somewhere and I am sure Cramya can find out where they are hiding.
Great method logitech!

Also, thanks Stuart for the additional 4 numbers!
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by cramya » Fri Dec 12, 2008 6:03 am
If OA is 56, I might be missing 4 numbers somewhere and I am sure Cramya can find out where they are hiding.
Excellent thinking Logitech!!
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by logitech » Fri Dec 12, 2008 6:06 am
cramya wrote:
If OA is 56, I might be missing 4 numbers somewhere and I am sure Cramya can find out where they are hiding.
Excellent thinking Logitech!!
5 = 1+1+1+2!! Oh I am getting old ;-)
LGTCH
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by earth@work » Fri Dec 12, 2008 10:00 am
my answer was a blunder... thanks logitech, excellent explanation.
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by adilka » Fri Dec 12, 2008 10:44 am
awesome question. awesome answer, logitech and Stuart. Thank you.
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by orel » Sat Dec 13, 2008 9:09 pm
logitech wrote:0 - 9999

5 = 5+0 0005 = 4!/3! = 4
5= 4+1 0041 = 4!/2! = 12
5= 3+2 0032 = 4!/2! = 12
5= 3+1+1 0311 = 4!/2! = 12
5=2+2+1 0221 = 4!/2! = 12

so IMO, it is 52

If OA is 56, I might be missing 4 numbers somewhere and I am sure Cramya can find out where they are hiding.
Logitech! solution is amazing! thanks!
I got a question here:
How do you know that you have to divide 4! by 3! or 2! ?
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by logitech » Sun Dec 14, 2008 2:25 am
Feruza Matyakubova wrote:
logitech wrote:0 - 9999

5 = 5+0 0005 = 4!/3! = 4
5= 4+1 0041 = 4!/2! = 12
5= 3+2 0032 = 4!/2! = 12
5= 3+1+1 0311 = 4!/2! = 12
5=2+2+1 0221 = 4!/2! = 12

so IMO, it is 52

If OA is 56, I might be missing 4 numbers somewhere and I am sure Cramya can find out where they are hiding.
Logitech! solution is amazing! thanks!
I got a question here:
How do you know that you have to divide 4! by 3! or 2! ?
0005 = 4!/3! = 4

lets take this example

0005 can be sorted as 4! ways but since we have three 0's we need to find the unique ways so we should divide it by 3!

so 4!/3! = 4

5000
0500
0050
0005

4 :)
LGTCH
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by Stuart@KaplanGMAT » Sun Dec 14, 2008 11:29 am
Feruza Matyakubova wrote:
logitech wrote:0 - 9999

5 = 5+0 0005 = 4!/3! = 4
5= 4+1 0041 = 4!/2! = 12
5= 3+2 0032 = 4!/2! = 12
5= 3+1+1 0311 = 4!/2! = 12
5=2+2+1 0221 = 4!/2! = 12

so IMO, it is 52

If OA is 56, I might be missing 4 numbers somewhere and I am sure Cramya can find out where they are hiding.
Logitech! solution is amazing! thanks!
I got a question here:
How do you know that you have to divide 4! by 3! or 2! ?
Here's the general rule:

To find the number of permutations of n objects of which r are duplicates, we calculate n!/r!.

For example:

"How many different ways can the letters of the word DESERT be arranged?"

6 letters total, 2 "E"s, so 6!/2!

"How many different ways can the letters of the word DESSERT be arranged?"

7 letters total, 2 "E"s and 2 "S"s, so 7!/2!2!

"How many different ways can the letters of the word DESSERTS be arranged?"

8 letters total, 2 "E"s and 3 "S"s, so 8!/2!3!
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