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D
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Difficulty:
95%
(hard)
Question Stats:
34% (02:38) correct
66%
(02:43)
wrong
based on 65
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Four positive integers a, b, c, and d have a product of 8! and satisfy:
ab + a + b = 524
bc + b + c = 146
cd + c + d = 104
What is a - d?
(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
Are You Up For the Challenge: 700 Level Questions
ab + a + b = 524
bc + b + c = 146
cd + c + d = 104
What is a - d?
(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
Are You Up For the Challenge: 700 Level Questions
22 Jun 2022, 06:38
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Bunuel
kntombat
If someone sees a way to get there in a reasonable amount of time, I'm all ears!!
ab + a + b +1 = 525
bc + b + c +1 = 147
cd + c + d +1 = 105
(a+1)(b+1) = 525 = 3*5*5*7
(b+1)(c+1) = 147 = 3*7*7
(c+1)(d+1) = 105 = 3*5*7
Let's look at the middle equation. There are only a few ways to multiply two numbers to get to 147: 1*147, 3*49, 7*21.
Both of the 7s from the middle equation can't come from (b+1) since there's only one 7 in the first equation. And both of the 7s from the middle equation can't come from (c+1) since there's only one 7 in the third equation. That means (b+1) and (c+1) are both multiples of 7. Looking back at the three ways to multiply two numbers to get to 147, only one of them has both numbers divisible by 7: 7*21.
So, either
Case 1: (b+1)=7 and (c+1)=21
or
Case 2: (b+1)=21 and (c+1)=7.
Case 1 would mean that (a+1)=75, (b+1)=7, (c+1)=21, and (d+1)=5. That means a=74, b=6, c=20, and d=4. Those four numbers won't multiply to 8! because there's no 7. Case 1 is not possible.
Case 2 would mean that (a+1)=25, (b+1)=21, (c+1)=7, and (d+1)=15. That means a=24, b=20, c=6, and d=14. d is the 2 and 7 in the 8!, so c must be the 1 and 6, b must be the 4 and 5, and a must be the 3 and 8. Okay, that works!
a-d = 24-14 = 10
Answer choice D
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