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the sum of the digits equals 5

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by sanju09 » Wed Sep 15, 2010 9:45 pm
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


[spoiler]Source: https://readyforgmat.com

this is little ambiguous to me, source writes it a 900 level question, how funny![/spoiler]
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by MariaS » Thu Sep 16, 2010 5:27 am
sanju09 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


[spoiler]Source: https://readyforgmat.com

this is little ambiguous to me, source writes it a 900 level question, how funny![/spoiler]
We are looking for 4 digit numbers, or less, that their digits sum up to five. The biggest maximum number that suites this description is 5000.

the different possible combinations are: 5=4+1=3+2=3+1+1=2+2+1=2+1+1+1

5--> 5000, 500, 50, 5 (total of 4 options)
4+1-->4001,4010,4100,1004,1040,1400, 401,410,104,140,41,14 (total of 12 options)
3+2--> also 12 options
3+1+1-->3110,3101,3011,1130,1103,1301,1310,1031,1013,113,131,311 (total of 12 options)
2+2+1--> also 12 options
2+1+1+1--> 2111, 1211,1121,1112 (total of 4 options)
---------------------------------------------------------------------------
56 numbers total
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by diebeatsthegmat » Thu Sep 16, 2010 5:33 am
sanju09 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


[spoiler]Source: https://readyforgmat.com

this is little ambiguous to me, source writes it a 900 level question, how funny![/spoiler]
we willl write down the number by this way
5000 ( ABBB)=4!/3!1!=4
1004=4!/1!2!1!=12
2003=4!/2!1!1!=12
1103=4!/2!1!1!=12
1112=4!/1!3!=4
2201=4!/1!2!1!=12
thus total=4x12+4+4=56
hope it helps
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by Maciek » Thu Sep 16, 2010 9:23 am
Hi!

This question is also included in "300+ GMAT Math questions with Best Solutions".

The explanation is following:

Here, we have 4 digits (positive integer less than 10000 means that 9999 is the biggest and we can pretend that "5" is "0005") that must sum to 5.
Since we have 4 digits, we'll have 3 partitions. We're summing to 5, so we have 5 "donuts".
O O O O O
Since we can use 0, we can have multiple partitions in the same spot. For example, we could have:
|||OOOOO (which translates to 0005)
we could have:
||O|OOOO (which translates to 0014, or 14).
So, we view this as a permutation question: we have 8 total objects, 3 of which are identical to each other (the partitions) and 5 of which are identical to each other (the donuts). Using the permutation formula for which some objects are identical:
Total permutations = n!/r!s! = 8!/3!5! = 8*7*6/3*2*1 = 8*7 = 56... Choose (C).

Source: 300+ GMAT Math questions with Best Solutions
https://www.beatthegmat.com/download-gma ... 59366.html

Hope it helps!
Best,
Maciek
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by sanju09 » Thu Sep 16, 2010 8:19 pm
Maciek wrote:Hi!

This question is also included in "300+ GMAT Math questions with Best Solutions".

The explanation is following:

Here, we have 4 digits (positive integer less than 10000 means that 9999 is the biggest and we can pretend that "5" is "0005") that must sum to 5.
Since we have 4 digits, we'll have 3 partitions. We're summing to 5, so we have 5 "donuts".
O O O O O
Since we can use 0, we can have multiple partitions in the same spot. For example, we could have:
|||OOOOO (which translates to 0005)
we could have:
||O|OOOO (which translates to 0014, or 14).
So, we view this as a permutation question: we have 8 total objects, 3 of which are identical to each other (the partitions) and 5 of which are identical to each other (the donuts). Using the permutation formula for which some objects are identical:
Total permutations = n!/r!s! = 8!/3!5! = 8*7*6/3*2*1 = 8*7 = 56... Choose (C).

Source: 300+ GMAT Math questions with Best Solutions
https://www.beatthegmat.com/download-gma ... 59366.html

Hope it helps!
Best,
Maciek
This is a masterpiece, but what MariaS and diebeatsthegmat did, is certainly worth give a standing ovation. I only anticipate that their ideas must have clicked their minds within 30 seconds after reading the question. Hats off
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by sanju09 » Thu Sep 16, 2010 8:26 pm
diebeatsthegmat wrote:
sanju09 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


[spoiler]Source: https://readyforgmat.com

this is little ambiguous to me, source writes it a 900 level question, how funny![/spoiler]
we willl write down the number by this way
5000 ( ABBB)=4!/3!1!=4
1004=4!/1!2!1!=12
2003=4!/2!1!1!=12
1103=4!/2!1!1!=12
1112=4!/1!3!=4
2201=4!/1!2!1!=12
thus total=4x12+4+4=56
hope it helps
Sheer waste of ink I think
The mind is everything. What you think you become. -Lord Buddha



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Lucknow-226001

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by Scott@TargetTestPrep » Fri Jan 10, 2020 6:57 am
sanju09 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


[spoiler]Source: https://readyforgmat.com

this is little ambiguous to me, source writes it a 900 level question, how funny![/spoiler]
If the sum of digits of a number is equal to 5, then each digit must be less than or equal to 5. Furthermore, we can consider each number as a "4-digit" number. For example, 5 can be considered as 0005 and 104 can be considered as 0104. We can classify the numbers into cases according to the largest digit of the number.
1) If the largest digit is 5, then the other three digits must be all 0. The number of ways to arrange one 5 and three 0's is 4!/3! = 4.
2) If the largest digit is 4, then the other three digits must consist of one 1 and two 0's. The number of ways to arrange one 4, one 1 and two 0's is 4!/2! = 4 x 3 = 12.
3) If the largest digit is 3, then the other three digits must consist of one 2 and two 0's OR two 1's and one 0. The number of ways to arrange one 3, one 2 and two 0's is 4!/2! = 12, and the number of ways to arrange one 3, two 1's and one 0 is also 4!/2! = 12. So the total number of ways in this case is 12 + 12 = 24.
4) If the largest digit is 2, then the other three digits must consist of all 1 OR one 2, one 1 and one 0. The number of ways to arrange one 2 and three 1's is 4!/3! = 4, and the number of ways to arrange two 2's, one 1 and one 0 is also 4!/2! = 12. So the total number of ways in this case is 4 + 12 = 16.
Since the largest digit of the number can't be 1 or 0, the total number of ways in which the sum of the digits is equal to 5 is 4 + 12 + 24 + 16 = 56.
Answer: C

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