ov25 wrote:A 3-digit integrer consists of digits 1, 2, 3, 4,5,6,7 and8 and the digits can be used more than once. If the sum of the digits is 16, how many such numbers are possible?
a) 32
b) 40
c) 42
D) 56
E) 64

Any quick methods to resolve such problems is appreciated

Sorry I could not find any short method. Short method always welcome

A B C be the number.

In order to get A+B+C = 16 we have to consider digit by digit

Let A = 1 then B + C = 15: 2 ways (78 or 87)
Let A = 2 then B + C = 14: 3 ways (68 or 86 or 77)
Let A = 3 then B + C = 13: 4 ways (76 or 67 or 85 or 58)
Let A = 4 then B + C = 12: 5 ways (66 or 75 or 57 or 84 or 48)
Let A = 5 then B + C = 11: 6 ways (65 or 56 or 47 or 74 or 83 or 38)
Let A = 6 then B + C = 10: 7 ways (55 or 64 or 46 or 73 or 37 or 28 or 82)
Let A = 7 then B + C = 09: 8 ways (54 or 45 or 36 or 36 or 27 or 27 or 81 or 18)
Let A = 8 then B + C = 08: 7 ways (44 or 35 or 53 or 17 or 71 or 26 or 62)

Total = 42

Hi there,

In terms of GMAT, I guess Shovan´s solution is the BEST one, because it is "sufficiently quick" and pretty safe (in terms of being hard to get confused).

Is there a "pure Combinatorics approach"? Yes.
Is it a bit too-sophisticated? Yes.

I will give to you "the idea" below and, if you really want me to solve it in details, please let me know.

Regards,
Fabio.

01. There is an easy "classical way" of finding the number of POSITIVE INTEGER solutions for the problem: x1 + x2 + x3 = 16. (The answer is C(16-1, 3-1))

02. There is a not-so-easy (and not that known) way of counting the number of positive integer solutions to the problem above with the restriction that xn > 8, where n = 1, 2, 3.

03. To find the solution to the problem originally posted, you "would do" the following:

number of elements in En = {(x1, x2, x3) from item 01 above, such that xn > 8} where n=1,2,3

Then you would use answer = number(item 01) - sum of numbers of (En) + sum of numbers of (Ek and Ej) - sum of numbers of (E1 and E2 and E3), where Ek and Ej are all two choices between indexes 1,2 and 3 (not counting orders).

(For the technical reader: I´m using Polya´s inclusion/exclusion principle, for sure.)

I know this seems "hard", that´s why I believe you should not bother, in terms of GMAT preparation...

Regards,
Fabio.

Let me suggest the same Shovan´s approach, but a bit more "spicy"... that could be as safe as his, but a bit quicker!

Start with "1" in the middle, then you must have a+c = 15, impossible to have a = c (why?) then you imagine a>c without loss of generality, and you find (very easy) only the following solution a = 8 and c = 7 therefore you DOUBLE to get 2 solutions.

Now "2" in the middle, you must have a+c = 14, get the a=c=7 (first solution) then you imagine a>c WLOG and you find (easy) only the following solutions: a = 8 and c = 6 therefore you DOUBLE this, SUM 1 and get 3 solutions.

Now "3" in the middle, you must have a+c = 13, no a=c solutions, then a>c WLOG implies a= 7 and c = 6 or a= 8 and c=5, therefore you DOUBLE to get 4 solutions...

Now "4", "5", "6", "7" and "8" in the middle. That´s it.

ov25 wrote:A 3-digit integrer consists of digits 1, 2, 3, 4,5,6,7 and8 and the digits can be used more than once. If the sum of the digits is 16, how many such numbers are possible?
a) 32
b) 40
c) 42
D) 56
E) 64

Any quick methods to resolve such problems is appreciated

consd 8:
8,7,1 (6)
8,6,2 (6)
8,5,3 (6)
8,4,4 (3)
consd 7:
7,7,2 (3)
7,6,3 (6)
7,5,4 (6)
consd 6:
6,6,4 (3)
6,5,5 (3)

C is answer.

this is quick method not slow and only method i guess O_o

24+15+6

sorry for that thing in end

HI OV 25 , Rahul has given a very beautiful approach this kind of problem in a different post, I am sure you will like it,
https://www.beatthegmat.com/12-t70715.html#319654

fskilnik wrote:
01. There is an easy "classical way" of finding the number of POSITIVE INTEGER solutions for the problem: x1 + x2 + x3 = 16. (The answer is C(16-1, 3-1))

HI ! Fabio,

This Problem is even more tough, Because 9 is not there other wise C(16-1, 3-1) Would have given us the answer straight,

Now the problem is we are not considering 9 so from the total of C ( 15, 2 ) = 105,
We must subtract those cases where 9 is there,

105 - 42 = 63 cases must be there where we can considering 9

But I am only able to figure out 18

1-) 9,3,4
2-) 9,6,1
3-) 9, 2 , 5

63 - 18 = 45 more numbers must be there,

Please share your views.

goyalsau wrote: Please share your views.

Hi, goyalsau.

I guess I could not make myself entirely clear... let me explain a bit more what I meant:

The problem asks for positive integer solutions with an EXTRA restriction, that is, the integers must be positive and LESS than 9.

The "classical" way of calculating the number of positive integer solutions that I used (and that Rahul explained in the post you suggested to ov25) is not enough because, as you said, that number would include values over 8. My explanation was exactly how to get this undesired solutions out.

Please take into account that my suggestion does not count on the fact that there are only a few cases to exclude in our particular case. If the number of digits were greater and/or the sum of them were also greater, not to mention the restrictions that could be "the first digit no less than 14", the "second digit no more than 7", etc, I presented the general case scenario, with the general classical mathematical approach.

I hope things make more sense now.

Regards,
Fabio.

fskilnik wrote:

goyalsau wrote: Please share your views.

Hi, goyalsau.

I guess I could not make myself entirely clear... let me explain a bit more what I meant:

The problem asks for positive integer solutions with an EXTRA restriction, that is, the integers must be positive and LESS than 9.

The "classical" way of calculating the number of positive integer solutions that I used (and that Rahul explained in the post you suggested to ov25) is not enough because, as you said, that number would include values over 8. My explanation was exactly how to get this undesired solutions out.

Please take into account that my suggestion does not count on the fact that there are only a few cases to exclude in our particular case. If the number of digits were greater and/or the sum of them were also greater, not to mention the restrictions that could be "the first digit no less than 14", the "second digit no more than 7", etc, I presented the general case scenario, with the general classical mathematical approach.

I hope things make more sense now.

Regards,
Fabio.

May be i am to dumb, that's why not able to figure out the solution of this problem.....

Don´t write such a thing, goyalsau. The approach I suggested is really conceptually harder, far above GMAT-like arguments. In other words, this is not "piece of cake" for ANYONE... myself certainly included.

Let me ask you 2 things:

01. Did you understand the idea/approach I explained and, important, why the several items would be really needed?

02. Did you understand how my "smart" suggestion to Shovan´s method would really be super-safe and super-quick?

Regards,
Fabio.

ov25 wrote:A 3-digit integrer consists of digits 1, 2, 3, 4,5,6,7 and8 and the digits can be used more than once. If the sum of the digits is 16, how many such numbers are possible?
a) 32
b) 40
c) 42
D) 56
E) 64

Any quick methods to resolve such problems is appreciated

Smallest digit 1 : others are 7 and 8 : 3!
Smallest digit 2 : others are 7 and 7 or 6 and 8 : 3 +3!
Smallest digit 3 : others are 6 and 7 or 5 and 8: 3! + 3!
Smallest digit 4 : others are 4 and 8 or 5 and 7 or 6 and 6 : 3 + 3! +3
Smallest digit 5: others are 5 and 6: 3

Total: 5(3!) + 4(3)= 42

kevincanspain wrote:Smallest digit 1 : others are 7 and 8 : 3!
Smallest digit 2 : others are 7 and 7 or 6 and 8 : 3 +3!
Smallest digit 3 : others are 6 and 7 or 5 and 8: 3! + 3!
Smallest digit 4 : others are 4 and 8 or 5 and 7 or 6 and 6 : 3 + 3! +3
Smallest digit 5: others are 5 and 6: 3

Total: 5(3!) + 4(3)= 42

Hi Kevin.

Yep, good method! (Although yours and mine would get a "tie", wouldn´t they?)

Regards,
Fabio.

P.S. (for all other readers): when Kevin mentions "smallest digit" and he uses the very same digit inside "that case" there is nothing wrong with it: in Math we accept smallest as "not greater than any other one", therefore "ties" (on the smaller elements) are perfectly allowed, "mathematically-semantically" speaking.

fskilnik wrote: Although yours and mine would get a "tie", wouldn´t they?

No, they wouldn´t. I myself want to be the first person to admit your idea is better.

Reason: [spoiler]I need to check all middle digits from "1" to "8" and Kevin does not need to bother with "6", "7" or "8".[/spoiler]

(I realized that when I was preparing this problem to be added to my online course.)

Congrats!
Fabio.