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# a*b*c*d = 770 a,b,c,d are positive integers a<b<c<d

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Director
Joined: 13 Mar 2007
Posts: 542

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Schools: MIT Sloan
a*b*c*d = 770 a,b,c,d are positive integers a<b<c<d [#permalink]

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20 Jun 2007, 20:24
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

a*b*c*d = 770

a,b,c,d are positive integers

a<b<c<d

What is c-b if a=1

a) 3
b) 4
c) 5
d) 7
e) 10

I think more than 1 choice works for the above .. wat do u guys say ?

Kudos [?]: 90 [0], given: 0

Senior Manager
Joined: 04 Jun 2007
Posts: 345

Kudos [?]: 33 [0], given: 0

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20 Jun 2007, 22:12
a*b*c*d = 770

a,b,c,d are positive integers

a<b<c<d

What is c-b if a=1

a) 3
b) 4
c) 5
d) 7
e) 10

I think more than 1 choice works for the above .. wat do u guys say ?

I am getting both 3 and 5 !!

Kudos [?]: 33 [0], given: 0

Director
Joined: 30 Nov 2006
Posts: 591

Kudos [?]: 314 [0], given: 0

Location: Kuwait

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20 Jun 2007, 22:37
bcd = 770 , given that a<b<c<d> c-b = 3
if d is 77, then c = 5, and b = 2 --> c-b = 3
if d is 22, then c= 7, and b = 5 --> c-b = 2
if d is 14, then c= 11, and b = 5 --> c-b = 6

I see the only choice is A. How did you get c-b = 5 ?

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Manager
Joined: 22 May 2006
Posts: 179

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20 Jun 2007, 22:57
me too, I got only 3, hence A.

Kudos [?]: 14 [0], given: 0

Senior Manager
Joined: 04 Jun 2007
Posts: 345

Kudos [?]: 33 [0], given: 0

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20 Jun 2007, 23:03
Mishari wrote:
bcd = 770 , given that a<b<c<d> c-b = 3
if d is 77, then c = 5, and b = 2 --> c-b = 3
if d is 22, then c= 7, and b = 5 --> c-b = 2
if d is 14, then c= 11, and b = 5 --> c-b = 6

I see the only choice is A. How did you get c-b = 5 ?

If d=55, c=7 and b=2, then c-b=5.

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Manager
Joined: 22 May 2007
Posts: 110

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21 Jun 2007, 01:44
my strategy: split in factors

770= 1*2*5*7*11

5 factors, but it has to be 4 (a,b,c,d) and a is 1

so we get 1 * 7 * 10 * 11 (multiply the smallest and re-sort the order)

c-b = 10-7 = 3

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Manager
Joined: 03 Mar 2007
Posts: 163

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21 Jun 2007, 06:46
ronron wrote:
my strategy: split in factors

770= 1*2*5*7*11

5 factors, but it has to be 4 (a,b,c,d) and a is 1

so we get 1 * 7 * 10 * 11 (multiply the smallest and re-sort the order)

c-b = 10-7 = 3

There are numerous 3 digit products that = 770.

I also used prime factorization 770 and got 2, 5, 7, 11.

so if:

b = 2 c = 5 d = 77 c - a = 5 - 2 = 3

However,

if we use 7 then:

b = 2 c = 7 d = 55 c - a = 5

I also get A and C.

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Senior Manager
Joined: 14 Jun 2007
Posts: 397

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22 Jun 2007, 21:23
a*b*c*d = 770

a,b,c,d are positive integers

a<b<c<d

What is c-b if a=1

a) 3
b) 4
c) 5
d) 7
e) 10

I think more than 1 choice works for the above .. wat do u guys say ?

having worked through a few challenges i am very skeptical of some of the challenge questions; for \$80 it is the most expensive lot of practice tests i have paid for and it is obvious they were not written by professional test writers - the explanations to some of the questions are just abysmal. this is a prime example; you can get two answers. i actually went shoonya's route on this one and started plugging in #s from the middle answer choice... and got 5 but i see 3 works too.

Kudos [?]: 15 [0], given: 0

Re: Challenge#23 q25   [#permalink] 22 Jun 2007, 21:23
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