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# Four positive integers a, b, c, and d have a product of 8! and satisfy

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Joined: 02 Sep 2009
Posts: 58381
Four positive integers a, b, c, and d have a product of 8! and satisfy  [#permalink]

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18 Mar 2019, 03:02
00:00

Difficulty:

65% (hard)

Question Stats:

40% (02:55) correct 60% (02:43) wrong based on 10 sessions

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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

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Re: Four positive integers a, b, c, and d have a product of 8! and satisfy  [#permalink]

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18 Mar 2019, 03:22
1
we can rewrite the three equations as follows:

(a+1)(b+1) & = 525
(b+1)(c+1) & = 147
(c+1)(d+1) & = 105 Note that
(a + 1)(b + 1)
= ab + a + b + 1
= 524 + 1
= 525
= $$3 · 5^2 · 7$$,
and
(b + 1)(c + 1)
= bc + b + c + 1
= 146 + 1
= 147 = $$3 · 7^2$$
Since (a + 1)(b + 1) is a multiple of 25 and (b + 1)(c + 1) is not a multiple of
5,
it follows that a + 1 must be a multiple of 25.
Since a + 1 divides 525, a is
one of 24, 74, 174, or 524.
Among these only 24 is a divisor of 8!,
so a = 24.
This implies that b + 1 = 21, and b = 20.
From this it follows that c + 1 = 7
and c = 6. Finally,
(c + 1)(d + 1) = 105 = 3 · 5 · 7,
so d + 1 = 15 and d = 14.
Therefore, a − d = 24 − 14 = 10.
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Four positive integers a, b, c, and d have a product of 8! and satisfy  [#permalink]

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18 Mar 2019, 03:32
Bunuel wrote:
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

ab + a + b = 524
Adding 1 both sides
ab + a + b +1 = 524+1

i.e. (a+1)*(b+1) = 525
the possible combinations are (Not necessarily in order)
1*525 i.e. a = 0 and b = 524
3*175 i.e. a = 2 and b = 174
5*105 i.e. a = 4 and b = 104
7*75 i.e. a = 6 and b = 74
15*35 i.e. a = 14 and b = 34
21*25 i.e. a = 20 and b = 24

bc + b + c = 146
Adding 1 both sides
bc + b + c +1 = 146+1
i.e. (b+1)*(c+1) = 147

the possible combinations are (Not necessarily in order)
3*49 i.e. b = 2 and c = 48
7*21 i.e. b = 6 and c = 20

cd + c + d = 104
Adding 1 both sides
cd + c + d +1 = 104+1
i.e. (d+1)*(c+1) = 105
the possible combinations are (Not necessarily in order)
3*35 i.e. c = 2 and d = 34
5*21 i.e. c = 4 and d = 20
7*15 i.e. c = 6 and d = 14

i.e. c = 6, d = 14, b = 20 and a = 24

i.e. a - d = 24-14 = 10

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Four positive integers a, b, c, and d have a product of 8! and satisfy   [#permalink] 18 Mar 2019, 03:32
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# Four positive integers a, b, c, and d have a product of 8! and satisfy

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