Find the number of integers in the set {1,2,3, ...., 2009} whose sum of the digits is 11.
Thanks.
Help with Math Olympiad question pls.
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turquoise  BrownBelt
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turquoise  BrownBelt
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by verykiasu2010 » Mon May 31, 2010 11:45 pm
turquoise wrote:No one's able to solve this question?
17
 verykiasu2010
 KiasuGrandMaster
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Re: Help with Math Olympiad question pls.
by mjl » Tue Jun 01, 2010 9:24 am
turquoise wrote:Find the number of integers in the set {1,2,3, ...., 2009} whose sum of the digits is 11.
Thanks.
I think there is no simple way, you have to use the brute force method.
29...92 (8)
119..191 (9)
209..290 (10)
308..380 (9)
407..470 (8)
:
:
902..920 (3)
1019..1091 (9)
1109..1190 (10)
1208..1280 (9)
1307..1370 (8)
:
:
1901,1910 (2)
2009 (1)
total=8+9+(3+4+...+10)+9+(1+2+..+10)
=26+ ((3+10)/2) x8 + ((1+10)/2)x10
=133
Last edited by mjl on Tue Jun 01, 2010 9:33 am, edited 1 time in total.

mjl  BlueBelt
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by Vanilla Cake » Tue Jun 01, 2010 9:30 am
verykiasu2010 wrote:17turquoise wrote:No one's able to solve this question?
Hi verykiasu2010,
This is a 1mark question (Q24) from Singapore Mathematical Olympiad (SMO) 2009  Junior Section held on Tuesday, 2 June 2009 from 0930  1200 hrs.
The answer is 133 and the solution is also published. As the solution has subscript and superscript characters, it's not possible for me to type out the given solution for your review.The method used to solve is using combinatorics to find out the 3 sets of integers that satisfies the property.From 0001~0999  69 number of solutions, 1001 ~ 1999  63 number of solutions and 2001 ~ 2009  1 solution. 69+63+1=133.
Submitted by VC's mum

Vanilla Cake  BlackBelt
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by verykiasu2010 » Tue Jun 01, 2010 11:16 am
Vanilla Cake wrote:verykiasu2010 wrote:17turquoise wrote:No one's able to solve this question?
Hi verykiasu2010,
This is a 1mark question (Q24) from Singapore Mathematical Olympiad (SMO) 2009  Junior Section held on Tuesday, 2 June 2009 from 0930  1200 hrs.
The answer is 133 and the solution is also published. As the solution has subscript and superscript characters, it's not possible for me to type out the given solution for your review.The method used to solve is using combinatorics to find out the 3 sets of integers that satisfies the property.From 0001~0999  69 number of solutions, 1001 ~ 1999  63 number of solutions and 2001 ~ 2009  1 solution. 69+63+1=133.
Submitted by VC's mum
Thank you very much !
 verykiasu2010
 KiasuGrandMaster
 Posts: 11696
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