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# How many positive integers less than 10,000 are there in whi

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How many positive integers less than 10,000 are there in whi [#permalink]

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11 Jan 2010, 11:34
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55% (hard)

Question Stats:

61% (03:18) correct 39% (02:29) wrong based on 17 sessions

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How many positive integers less than 10,000 are there in which the sum of the digits equals 5?

A) 31
B) 51
C) 56
D) 62
E) 93
[Reveal] Spoiler: OA
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11 Jan 2010, 12:15
Since the sum of the 4 digits have to equal to 5, anyting #s >=6000 is eliminated.

5000s range - 1
5000 is the only possibly number in the 5000's range.

4000s range - 3
Since 4+1=5, there are only 3 spots to put the 1s, namely 4100, 4010, 4001.

3000s range - 6
We can have 3+2+0+0 (3 combinations) or 3+1+1+0 (3 combinations).

2000s range - 7
We can have 2+2+1+0 (6 combinations because of 3! ways of arranging the last 3 digits using 2,1,0) or 2+1+1+1 (1 combination).

1000s range - 15
We can have 1+1+1+2 (3 combinations), 1+2+2+0 (3 combinations), 1+4+0+0 (3 combinations), and 1+3+1+0 (6 combinations because of 3! ways of arranging the last 3 digits of 3,1,0).

The # of numbers is therefore 1+3+6+7+15=32. Notice that you get 3 combinations whenever 2 of the last 3 digits are the same.

Then use the same logic on 3 digits and 2 digits and 1 digit numbers.

Is there a fast way to do this?
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11 Jan 2010, 15:46
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Expert's post
zaarathelab wrote:
How many positive integers less than 10,000 are there in which the sum of the digits equals 5?

A) 31
B) 51
C) 56
D) 62
E) 93

Consider this: we have 5 $$d$$'s and 3 separators $$|$$, like: $$ddddd|||$$. How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$d$$'s and 3 $$|$$'s are identical, so $$\frac{8!}{5!3!}=56$$.

With these permutations we'll get combinations like: $$|dd|d|dd$$ this would be 3 digit number 212 OR $$|||ddddd$$ this would be single digit number 5 (smallest number less than 10,000 in which sum of digits equals 5) OR $$ddddd|||$$ this would be 4 digit number 5,000 (largest number less than 10,000 in which sum of digits equals 5)...

Basically this arrangements will give us all numbers less than 10,000 in which sum of the digits (sum of 5 d's=5) equals 5.

Hence the answer is $$\frac{8!}{5!3!}=56$$.

This can be done with direct formula as well:

The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is $${n+r-1}_C_{r-1}$$.

In our case we'll get: $${n+r-1}_C_{r-1}={5+4-1}_C_{4-1}={8}C3=\frac{8!}{5!3!}=56$$
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11 Jan 2010, 17:17
What's the concept behind the separators?
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11 Jan 2010, 22:58
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Expert's post
mrblack wrote:
What's the concept behind the separators?

Attachment:

pTNfS-2e270de4ca223ec2741fa10b386c7bfe.jpg [ 63.83 KiB | Viewed 2432 times ]

Hope it's clear.
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11 Jan 2010, 23:08
Bunuel wrote:
zaarathelab wrote:
How many positive integers less than 10,000 are there in which the sum of the digits equals 5?

A) 31
B) 51
C) 56
D) 62
E) 93

Consider this: we have 5 $$d$$'s and 3 separators $$|$$, like: $$ddddd|||$$. How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$d$$'s and 3 $$|$$'s are identical, so $$\frac{8!}{5!3!}=56$$.

With these permutations we'll get combinations like: $$|dd|d|dd$$ this would be 3 digit number 212 OR $$|||ddddd$$ this would be single digit number 5 (smallest number less than 10,000 in which sum of digits equals 5) OR $$ddddd|||$$ this would be 4 digit number 5,000 (largest number less than 10,000 in which sum of digits equals 5)...

Basically this arrangements will give us all numbers less than 10,000 in which sum of the digits (sum of 5 d's=5) equals 5.

Hence the answer is $$\frac{8!}{5!3!}=56$$.

This can be done with direct formula as well:

The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is $${n+r-1}_C_{r-1}$$.

In our case we'll get: $${n+r-1}_C_{r-1}={5+4-1}_C_{4-1}={8}C3=\frac{8!}{5!3!}=56$$

That was brilliant. Wow.
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12 Jan 2010, 07:04
Gr8 explanations guys

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09 Mar 2010, 22:41
How many positive integers less than 10,000 are there in which the sum of the digits equals 5?

A) 31
B) 51
C) 56
D) 62
E) 93
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10 Mar 2010, 03:23
Grt explanation Bunuel.
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10 Mar 2010, 22:31
there is a shortcut. For the problem, 4 digits are equally important in 0000-9999 set and it is impossible to build a number using only one digit (like 11111) So, answer has to be divisible by 4. Only 56 works.

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Re: Integers   [#permalink] 10 Mar 2010, 22:31
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