positive integers less than 10,000

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krishnasty: Master | Next Rank: 500 Posts; Posts: 142; Joined: Mon Jan 10, 2011 8:03 am; Thanked: 19 times

by krishnasty » Fri Jul 22, 2011 10:38 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) 934

OA : C

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Source: Beat The GMAT — Problem Solving | Fri Jul 22, 2011 10:38 pm

cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members; Location: India; Thanked: 310 times; GMAT Score:750

by cans » Sat Jul 23, 2011 12:34 am

# of one digit numbers = 1 (only 5 satisfies)
# of 2 digit numbers = 5 (14,41,23,32,50)
# of 3 digit numbers = 15(122 (*3) ,131 (*3),410 (*4),230 (*4),500 (*1))
# of 4 digit numbers =35 (5000 (*1), 4100 (*6), 3110 (*9), 3200 (*6), 2210(*9), 2111(*4))
OA C

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Jul 23, 2011 1:36 am

I posted a solution here:

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Ozlemg: Master | Next Rank: 500 Posts; Posts: 407; Joined: Tue Jan 25, 2011 9:19 amFollowed by: 7 members; Thanked: 25 times

by Ozlemg » Sat Jul 23, 2011 3:51 am

cans wrote:# of one digit numbers = 1 (only 5 satisfies)
# of 2 digit numbers = 5 (14,41,23,32,50)
# of 3 digit numbers = 15(122 (*3) ,131 (*3),410 (*4),230 (*4),500 (*1))
# of 4 digit numbers =35 (5000 (*1), 4100 (*6), 3110 (*9), 3200 (*6), 2210(*9), 2111(*4))
OA C

Hi Cans

Thank you for your help!
I did not understand the bold numbers?

The more you suffer before the test, the less you will do so in the test!

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Ozlemg: Master | Next Rank: 500 Posts; Posts: 407; Joined: Tue Jan 25, 2011 9:19 amFollowed by: 7 members; Thanked: 25 times

by Ozlemg » Sat Jul 23, 2011 4:01 am

krishnasty 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) 934

OA : C

This can be done with direct formula as well:

The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is-->n+r+1Cr-1

8C3 is 56.

Hence, C

The more you suffer before the test, the less you will do so in the test!

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winniethepooh: Master | Next Rank: 500 Posts; Posts: 370; Joined: Sat Jun 11, 2011 8:50 pmFollowed by: 2 members; Location: Arlington, MA.; Thanked: 27 times

by winniethepooh » Sat Jul 23, 2011 4:12 am

That is the number of ways those digits can be written to give a sum of five:
122 can be arranged in 3!/2! ways. = 3 ways

410 can be arrangen in 6(3!/0!) ways in all = 6 ways - 2 ways in which 0 appears as the first digit as already considered earlier in two digits(14). Hence, 4 ways.

Similarly others.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sat Jul 23, 2011 8:19 am

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

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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