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

### GMAT/MBA Expert

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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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You can find a downloadable set of advanced math questions at www.ReadyForGMAT.com

- logitech
- Legendary Member
**Posts:**2134**Joined:**20 Oct 2008**Thanked**: 237 times**Followed by:**25 members**GMAT Score:**730

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

---------------------

"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

### GMAT/MBA Expert

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Hi Brent -- Got this link from your latest post..[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?"

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..[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?"

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

- tuanquang269
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**Posts:**296**Joined:**29 May 2011**Location:**Vietnam**Thanked**: 10 times**Followed by:**5 members

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

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?

Work hard in Silence, Let Success make the noise.

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If you found my Post really helpful, then don't forget to click the Thank/follow me button.

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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