Very tricky counting problem

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Very tricky counting problem

by [email protected] » Sat Dec 13, 2008 6:01 pm
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?"

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by cramya » Sat Dec 13, 2008 6:06 pm
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

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by dmateer25 » Sat Dec 13, 2008 6:13 pm
Brent, did you create these questions??

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by [email protected] » Sat Dec 13, 2008 8:13 pm
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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by logitech » Sat Dec 13, 2008 8:45 pm
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
cool!

https://www.readyforgmat.com/math/docume ... rsion2.pdf
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by krazzy4 » Sun Dec 14, 2008 12:37 am
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
thanks for the link...cool resource.

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by dmateer25 » Sun Dec 14, 2008 5:25 am
Brent,

Thanks for taking the time to help out!

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by adilka » Sun Dec 14, 2008 1:46 pm
Thanks for the link, Brent

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by [email protected] » Sun Dec 14, 2008 3:02 pm
I checked the link to the previous solutions to this question, but didn't see any shortcuts, so I'll post this solution.

Image


We can extend this solution and conclude that the number of integers less than 1,000,000 in which the sum of the digits equals 8 will be 13C5

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by mental » Sun Dec 14, 2008 11:09 pm
:shock: amazed

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by saketk » Sun Sep 18, 2011 3:27 am
[email protected] 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?"
Hi Brent -- Got this link from your latest post..

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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by [email protected] » Sun Sep 18, 2011 6:38 am
saketk wrote:
[email protected] 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?"
Hi Brent -- Got this link from your latest post..

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
That looks great, saketk!
Lot of solid counting techniques here.

Cheers,
Brent
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by tuanquang269 » Wed Sep 21, 2011 6:38 pm
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
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?)

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by sana.noor » Sun Jul 21, 2013 10:49 am
using the inserting stick and seperator technique

(5+3)!/5!.3! = 56...C is the answer

is my approach to answer this question right?
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by [email protected] » Sun Jul 21, 2013 10:51 am
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?
Perfect!!!

Cheers,
Brent
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