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805+ (Hard)|   Algebra|      
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Bunuel
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The number of solutions (x, y, z) to the equation x – y – z = 25, wher 02 Apr 2020, 10:44
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95% (hard)

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40% (02:56) correct 60% (02:53) wrong based on 72 sessions

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The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is

A. 105
B. 101
C. 100
D. 99
E. 87
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D
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nick1816
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The number of solutions (x, y, z) to the equation x – y – z = 25, wher 02 Apr 2020, 15:37
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x= 25+y+z

Since y and z are positive integers, minimum value x can take is 27.

27≤x≤40
2≤y+z≤15

Case 1 - when y=1, z can take 12 values
Case 2 - when y=2, z can take 12 values
Case 3 - when y=3, z can take 12 values
Case 4 - when y=4, z can take 11 values
Case 5 - when y=5, z can take 10 values
......

Case 12- when y=12, z can take 3 values

Total possible solutions = \(3+4+5+6+7+8+9+10+11+12+12+12= \frac{(13*12)}{2} -3 +24= 99\)

Bunuel
The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is

A. 105
B. 101
C. 100
D. 99
E. 87
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angysinghdhillon
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The number of solutions (x, y, z) to the equation x – y – z = 25, wher 19 Aug 2023, 10:51
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nick1816 can you please explain the solution in detail?
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HarshavardhanR
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The number of solutions (x, y, z) to the equation x – y – z = 25, wher 20 Oct 2024, 10:12
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1) x, y, and z are positive integers (>=1).
2) y and z <=12. So, y and z can each take any integer value from 1 to 12.
3) x<=40.

x - y - z = 25

So,

x = y + z + 25

Remember: x<=40. Hence, y+z cannot be >15.

At this point, I found it quicker to manually count ->

-> If y = 1, z can be any number from 1 to 12. 12 sets of (x,y,z) possible.
-> If y = 2, z can be any number from 1 to 12. 12 sets of (x,y,z) possible.
-> If y = 3, z can be any number from 1 to 12. 12 sets of (x,y,z) possible.
-> If y = 4, z can be any number from 1 to 11. z cannot be 12. 11 sets of (x,y,z) possible.
-> If y = 5, z can be any number from 1 to 10. z cannot be [11,12]. 10 sets of (x,y,z) possible.
-> If y = 6, z can be any number from 1 to 9. z cannot be [10,12]. 9 sets of (x,y,z) possible.
-> If y = 7, z can be any number from 1 to 8. z cannot be [9,12]. 8 sets of (x,y,z) possible.
-> If y = 8, z can be any number from 1 to 7. z cannot be [8,12]. 7 sets of (x,y,z) possible.
-> If y = 9, z can be any number from 1 to 6. z cannot be [7,12]. 6 sets of (x,y,z) possible.
-> If y = 10, z can be any number from 1 to 5. z cannot be [6,12]. 5 sets of (x,y,z) possible.
-> If y = 11, z can be any number from 1 to 4. z cannot be [5,12]. 4 sets of (x,y,z) possible.
-> If y = 12, z can be any number from 1 to 3. z cannot be [4,12]. 3 sets of (x,y,z) possible.

Total number of solutions possible -> (3 + 4 + 5 + ....11) + 12 + 12 + 12 = 7*9 + 36 = 63 + 36 = 99. Choice D.

Alter:

-> Given that y and z can each take 12 values (1 to 12), totally 12 x 12 = 144 sets are possible, if we did not have to worry about the constraint that x <=40.

-> From 144, we simply need to remove those sets of (x,y,z) where x>40 i.e., y+z>15.

From y = 1 till y = 3, there is no set where y+z>15.

For y =4, there is one case we need to reject (z=12).
For y = 5, there are 2 cases we need to reject (z = 12,11).
And so on.

So overall, number of cases to reject = 1 + 2 + 3 + .....9 = 45.

Final answer: 144 - 45 = 99. Choice D.

Hope this helps!
Harsha
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