GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 22 Jan 2020, 01:40

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# If x, y and z are non-negative integers such that x < y < z,

Author Message
TAGS:

### Hide Tags

GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4219
If x, y and z are non-negative integers such that x < y < z,  [#permalink]

### Show Tags

05 Nov 2018, 10:28
Top Contributor
19
00:00

Difficulty:

85% (hard)

Question Stats:

46% (02:24) correct 54% (02:13) wrong based on 136 sessions

### HideShow timer Statistics

If x, y and z are non-negative integers such that x < y < z, then the equation x + y + z = 11 has how many distinct solutions?

A) 5
B) 10
C) 11
D) 22
E) 78

_________________
Test confidently with gmatprepnow.com
Senior Manager
Joined: 15 Feb 2018
Posts: 450
If x, y and z are non-negative integers such that x < y < z,  [#permalink]

### Show Tags

Updated on: 05 Nov 2018, 22:11
Misinterpreted non-negative for positive.

Originally posted by philipssonicare on 05 Nov 2018, 21:23.
Last edited by philipssonicare on 05 Nov 2018, 22:11, edited 1 time in total.
Manager
Joined: 08 Sep 2008
Posts: 128
Location: India
Concentration: Operations, General Management
Schools: ISB '20
GPA: 3.8
WE: Operations (Transportation)
Re: If x, y and z are non-negative integers such that x < y < z,  [#permalink]

### Show Tags

05 Nov 2018, 22:04
3
X, Y, Z Is non negative integer, and x+y+z=11, and x<y<z
We can fix value of x and calculate distinct solutions
Case-1
X=0
Y can have 1,2,3,4,5 (Total 5 values) to reach x+y+z =11.
For each value of y z will have a fix value. So total 5 distinct solutions.
Case-2
x=1
Y can be 2,3,4, for each value, z will have a fix value ( 6,7, and 8)
Total 3 distinct solutions in this case.
Case -3
X=2
Y can be 3,4 for each value, z will have a fix value ( 5,6,)
Total 2 distinct solutions in this case.
Case -4
X=3
No value of y can satisfy x+y+z=11, and x<y<z.

Total
5+3+2=10
B.
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4219
If x, y and z are non-negative integers such that x < y < z,  [#permalink]

### Show Tags

07 Nov 2018, 07:43
1
Top Contributor
GMATPrepNow wrote:
If x, y and z are non-negative integers such that x < y < z, then the equation x + y + z = 11 has how many distinct solutions?

A) 5
B) 10
C) 11
D) 22
E) 78

I created this question to show that there can be times when the best (i.e., fastest) way to solve a counting question is by listing and counting

How do we know when it's not a bad idea to use listing and counting?
Here, the answer choices are reasonably small, so listing and counting shouldn't take long.
ASIDE: Yes, 78 (answer choice E) is pretty big. However, if you start listing possible outcomes and you eventually list more than 22 outcomes (answer choice D), you can stop because the answer must be E.

Okay, let's list possible outcomes in a systematic way.
We'll list outcomes as follows: x, y, z

Since x is the smallest value.
Let's list the outcomes in which x = 0. We get:
0, 1, 10
0, 2, 9
0, 3, 8
0, 4, 7
0, 5, 6

Now, the outcomes in which x = 1. We get:
1, 2, 8
1, 3, 7
1, 4, 6

Now, the outcomes in which x = 2. We get:
2, 3, 6
2, 4, 5

Now, the outcomes in which x = 3. We get:
3, 4...hmmm, this won't work.

So, we're done listing!
Count the outcomes to get a total of 10

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
VP
Joined: 24 Nov 2016
Posts: 1086
Location: United States
Re: If x, y and z are non-negative integers such that x < y < z,  [#permalink]

### Show Tags

11 Oct 2019, 07:48
1
GMATPrepNow wrote:
If x, y and z are non-negative integers such that x < y < z, then the equation x + y + z = 11 has how many distinct solutions?

A) 5
B) 10
C) 11
D) 22
E) 78

0≤x<y<z
x+y+z=11
x=0, y=1, z=11-1=10: average (10+1)=11/2=5.5 so y={1 to 5}={1,2,3,4,5}=5
x=1, y=2, z=11-3=8: average (8+2)=10/2=5 so y={2 to 4}={2,3,4}=3
x=2, y=3, z=11-5=6: average (6+3)=9/2=4.5 so y={3 to 4}={3,4}=2
x=3, y=4, z=11-7=4: invalid

total solutions: 5+3+2=10

Re: If x, y and z are non-negative integers such that x < y < z,   [#permalink] 11 Oct 2019, 07:48
Display posts from previous: Sort by