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17 Apr 2019, 23:36
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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32% (02:27) correct
68%
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wrong
based on 110
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Brian writes down four integers w > x > y > z whose sum is 44. The pairwise positive differences of these numbers are 1, 3, 4, 5, 6 and 9. What is the sum of the possible values of w?
(A) 16
(B) 31
(C) 48
(D) 62
(E) 93
(A) 16
(B) 31
(C) 48
(D) 62
(E) 93
22 Sep 2025, 06:17
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Bunuel
w - z = 9 since they must have the largest difference
Set:
w - x = a
w - y = b
Since a < b, you can try plugging in the remaining values:
| a | 1 | 1 | 1 | 1 | 3 | 3 | 3 | 4 | 4 | 5 |
| b | 3 | 4 | 5 | 6 | 4 | 5 | 6 | 5 | 6 | 6 |
w + x + y + z = 44
<=> w + (w - a) + (w - b) + (w - 9) = 44
<=> 4w - a - b - 9 = 44
<=> 4w = 53 + a + b
(53 + a + b) must be a multiple of 4
There are only 3 suitable pairs:
a = 1 ; b = 6 -> w = 15
a = 3 ; b = 4 -> w = 15
a = 5 ; b = 6 -> w = 16
15 + 16 = 31
(B) is the answer
18 Apr 2019, 00:38
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Let w-x=a, x-y=b and y-z=c
a+b+c should be equal to 9
and only 3 pairwise differences which leads to 9 are 1,3,5
Now a,b,c can take any of these 3 values
w+x+y+z=44
z+a+b+c+z+b+c+z+c+z=44
4z+3c+2b+a=44....(1)
There are 6 cases possible for (a,b,c)= (1,3,5), (1,5,3), (3,1,5),(3,5,1), (5,1,3) or(5,3,1)
By putting all combinations in equation (1), we get only 2 different values for w=15 and 16
Sum of all possible integer values of w= 15+16=31
a+b+c should be equal to 9
and only 3 pairwise differences which leads to 9 are 1,3,5
Now a,b,c can take any of these 3 values
w+x+y+z=44
z+a+b+c+z+b+c+z+c+z=44
4z+3c+2b+a=44....(1)
There are 6 cases possible for (a,b,c)= (1,3,5), (1,5,3), (3,1,5),(3,5,1), (5,1,3) or(5,3,1)
By putting all combinations in equation (1), we get only 2 different values for w=15 and 16
Sum of all possible integer values of w= 15+16=31
