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1155 is divisible by 11 from the first glance, which lets us easily split it into 11*105 and then into 11*7*5*3.
As it turns out we don't have to think much about possible products of 4 numbers, because we pretty much got 4 numbers for 4 spots and after arranging them in the corresponding order its easy to figure out that the answer is 11 - 3 = 8 which matches answer D
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Bunuel
If the product of the integers a, b, c, and d is 1,155 and if a > b > c > d > 1, then what is the value of a – d?

(A) 2
(B) 8
(C) 10
(D) 11
(E) 14


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1155= 5 * 3* 11*7
a=11
d=3

a-d = 8

answer B
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1155
divided by 5 first, got 231. Then by 3 gives 77.
thus 1155 = 3.5.7.11
a-d = 11-3 = 8
Ans D
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Factories 1155 in prime factors, we get 1155=11.7.5.3
from the question stem we know that a=11,d=3 so a-d=8
Hence answer is B

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Bunuel
If the product of the integers a, b, c, and d is 1,155 and if a > b > c > d > 1, then what is the value of a – d?

(A) 2
(B) 8
(C) 10
(D) 11
(E) 14


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VERITAS PREP OFFICIAL SOLUTION:

Upon first glance, 1155 and four variables might look really messy. But take the first step – you know it’s divisible b y 11 and that you have to factor it. 1100 is 11*100 and 55 is 11*5, so you have 11*105. And 105 is much easier to divide out since it ends in a 5. That’s 21*5, which is 7*3*5. Once you’ve factored it down, it’s 11*7*5*3, which are all prime, so when 1 has to be less than any of these, that’s exactly a, b, c, and d. You need the biggest minus the smallest, and 11-3 is 8. What may have looked like a big, intimidating number was actually not so bad once you took the first step. It’s always darkest before the light goes on.
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Hello everybody!
I have a confusion, whether this factorization technique is the way to approach such type of question or it's appropriate with the information given in the question.
For instance i have a number 910 which is the product of abc where a>b>c and i have to find a-c.
If I take factorization technique there are four prime factors of 910, which is 2,5,7,13, so if i take 2,35(5x7),13 then answer is 33. however if i take, 10(2x5),7,13, answer is 6. Because the question ask me that a,b,c,d are integer and integers could b prime and non-prime as well.
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Ahtisham
Hello everybody!
I have a confusion, whether this factorization technique is the way to approach such type of question or it's appropriate with the information given in the question.
For instance i have a number 910 which is the product of abc where a>b>c and i have to find a-c.
If I take factorization technique there are four prime factors of 910, which is 2,5,7,13, so if i take 2,35(5x7),13 then answer is 33. however if i take, 10(2x5),7,13, answer is 6. Because the question ask me that a,b,c,d are integer and integers could b prime and non-prime as well.
I don't think that questions would be that confusing. Even if there are given just 3 integers (a,b,c) there will be given a>b>c>2.

Sent from my Redmi Note 3 using GMAT Club Forum mobile app
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Bunuel
If the product of the integers a, b, c, and d is 1,155 and if a > b > c > d > 1, then what is the value of a – d?

(A) 2
(B) 8
(C) 10
(D) 11
(E) 14


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\(1155 = 3*5*7*11\)

\(a > b > c > d > 1\)

So, \(a = 11\), \(b = 7\) , \(c = 5\) & \(d = 11\)

So, \(a - d = 11 - 3 => 8\)

Thus, answer will be (B) 8
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1155= 3*5*7*11

a>b>c>d>1
11>7>5>3

a-d=11-3=8
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Bunuel
If the product of the integers a, b, c, and d is 1,155 and if a > b > c > d > 1, then what is the value of a – d?

(A) 2
(B) 8
(C) 10
(D) 11
(E) 14

We can start by expressing 1,155 as the product of its prime factors.

1,155 = 5 x 231 = 5 x 77 x 3 = 5 x 7 x 11 x 3

This means that a = 11, b = 7, c = 5, and d = 3.

Thus, a - d = 11 - 3 = 8.

Answer: B
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