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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this is little ambiguous to me, source writes it a 900 level question, how funny![/spoiler]
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the sum of the digits equals 5
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sanju09
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Sanjeev K Saxena
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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.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]
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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diebeatsthegmat
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we willl write down the number by this waysanju09 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]
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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Maciek
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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 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
"There is no greater wealth in a nation than that of being made up of learned citizens." Pope John Paul II
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if you have any questions, send me a private message!
should you find this post useful, please click on "thanks" button
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sanju09
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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 offMaciek 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
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
-
sanju09
- GMAT Instructor
- Posts: 3650
- Joined: Wed Jan 21, 2009 4:27 am
- Location: India
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Sheer waste of ink I thinkdiebeatsthegmat wrote:we willl write down the number by this waysanju09 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]
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
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
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Scott@TargetTestPrep
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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.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]
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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