This one is well past the 800-level, but there are a lot of clever people on this site who will, I imagine, accept the challenge.
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
Answer: C
Try to resist examining each case and counting all of the possibilities. To discourage this strategy I could have asked "How many positive integers less than 1,000,000 are there in which the sum of the digits equals 8?"
Very tricky counting problem
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Brent,
Thanks a bunch for providing the OA. By the way, welcome to this forum.
Please refer to
https://www.beatthegmat.com/need-help-wi ... 25206.html
Good luck with ur preps!
Regards,
Cramya
Thanks a bunch for providing the OA. By the way, welcome to this forum.
Please refer to
https://www.beatthegmat.com/need-help-wi ... 25206.html
Good luck with ur preps!
Regards,
Cramya
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Yes - I guess some former students have posted them here already.
You can find a downloadable set of advanced math questions at www.ReadyForGMAT.com
You can find a downloadable set of advanced math questions at www.ReadyForGMAT.com
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cool!Brent Hanneson wrote:Yes - I guess some former students have posted them here already.
You can find a downloadable set of advanced math questions at www.ReadyForGMAT.com
https://www.readyforgmat.com/math/docume ... rsion2.pdf
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
thanks for the link...cool resource.Brent Hanneson wrote:Yes - I guess some former students have posted them here already.
You can find a downloadable set of advanced math questions at www.ReadyForGMAT.com
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Hi Brent -- Got this link from your latest post..Brent@GMATPrepNow wrote:This one is well past the 800-level, but there are a lot of clever people on this site who will, I imagine, accept the challenge.
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
Answer: C
Try to resist examining each case and counting all of the possibilities. To discourage this strategy I could have asked "How many positive integers less than 1,000,000 are there in which the sum of the digits equals 8?"
This is how I did it (though I used bit of a counting here)
------------------------------------------------
We can make numbers from ---0,1,2,3,4,5 ONLY
------------------------------------------------
1 digit number
only 1 = 5
TOTAL = 1 way
-----------------------------------------------
2 digit numbers
possible sets (2,3) = 2*1 = 2 ways
if (4,1) then 2*1 = 2 ways
if (5,0) then = 1 way
TOTAL = 5 ways
------------------------------------------
3 digit numbers
possible sets (5,0,0) then =1 way
if (4,0,1)
2*2*1 = 4 ways
if one is 3 then 2 sets possible
(3,0,2) = 2*2*1 = 4 ways
(3,1,1) = 3*2*1/2 = 3 ways
if one is 2 then set can be (2,2,1) = 3*2*1/2 = 3 ways ... NOTE- we have already used a set of (2,3,0)
TOTAL = 1+4+4+3+3 = 15 ways
---------------------------------------------------
4 digits
possible sets (1,2,2,0) (1,1,1,2) (1,3,1,0) (3,2,0,0) (4,1,0,0) (5000)
_ _ _ _
3*3*2*1 /2 = 9 ways
4*3*2*1/6 = 4 ways
3*3*2*1/2 = 9 ways
2*3*2*1/2 = 6 ways
2*3*2*!/2 = 6 ways
1 way
respectively
TOTAL = 18+16+1 = 35 ways
--------------------------------------------------------
GRAND TOTAL = 35+21 = 56
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That looks great, saketk!saketk wrote:Hi Brent -- Got this link from your latest post..Brent@GMATPrepNow wrote:This one is well past the 800-level, but there are a lot of clever people on this site who will, I imagine, accept the challenge.
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
Answer: C
Try to resist examining each case and counting all of the possibilities. To discourage this strategy I could have asked "How many positive integers less than 1,000,000 are there in which the sum of the digits equals 8?"
This is how I did it (though I used bit of a counting here)
------------------------------------------------
We can make numbers from ---0,1,2,3,4,5 ONLY
------------------------------------------------
1 digit number
only 1 = 5
TOTAL = 1 way
-----------------------------------------------
2 digit numbers
possible sets (2,3) = 2*1 = 2 ways
if (4,1) then 2*1 = 2 ways
if (5,0) then = 1 way
TOTAL = 5 ways
------------------------------------------
3 digit numbers
possible sets (5,0,0) then =1 way
if (4,0,1)
2*2*1 = 4 ways
if one is 3 then 2 sets possible
(3,0,2) = 2*2*1 = 4 ways
(3,1,1) = 3*2*1/2 = 3 ways
if one is 2 then set can be (2,2,1) = 3*2*1/2 = 3 ways ... NOTE- we have already used a set of (2,3,0)
TOTAL = 1+4+4+3+3 = 15 ways
---------------------------------------------------
4 digits
possible sets (1,2,2,0) (1,1,1,2) (1,3,1,0) (3,2,0,0) (4,1,0,0) (5000)
_ _ _ _
3*3*2*1 /2 = 9 ways
4*3*2*1/6 = 4 ways
3*3*2*1/2 = 9 ways
2*3*2*1/2 = 6 ways
2*3*2*!/2 = 6 ways
1 way
respectively
TOTAL = 18+16+1 = 35 ways
--------------------------------------------------------
GRAND TOTAL = 35+21 = 56
Lot of solid counting techniques here.
Cheers,
Brent
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Can you explain for me the way you count each set? I do not understand. For example (1,2,2,0) = 3 * 3 * 2 *1/2 (I don't understand where "3", "3", "2", "1/2" from?)saketk wrote: 4 digits
possible sets (1,2,2,0) (1,1,1,2) (1,3,1,0) (3,2,0,0) (4,1,0,0) (5000)
_ _ _ _
3*3*2*1 /2 = 9 ways
4*3*2*1/6 = 4 ways
3*3*2*1/2 = 9 ways
2*3*2*1/2 = 6 ways
2*3*2*!/2 = 6 ways
1 way
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using the inserting stick and seperator technique
(5+3)!/5!.3! = 56...C is the answer
is my approach to answer this question right?
(5+3)!/5!.3! = 56...C is the answer
is my approach to answer this question right?
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Perfect!!!sana.noor wrote:using the inserting stick and seperator technique
(5+3)!/5!.3! = 56...C is the answer
is my approach to answer this question right?
Cheers,
Brent