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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x, y and z are non-negative integers such that x < y < z,

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

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Top Contributor
19 00:00

Difficulty:   85% (hard)

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

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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

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Senior Manager  D
Joined: 15 Feb 2018
Posts: 450
If x, y and z are non-negative integers such that x < y < z,  [#permalink]

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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  S
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]

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

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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
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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]

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