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Bunuel
If a, b, c are three consecutive integers such that c > b > a and abc = 3360. What is the value of b?

A. 13
B. 14
C. 15
D. 16
E. 17


3360 = 14*15*16

b would be 15 IMO C
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Bunuel
If a, b, c are three consecutive integers such that c > b > a and abc = 3360. What is the value of b?

A. 13
B. 14
C. 15
D. 16
E. 17
\(3360 = 14*15*16\)

So, \(c = 16\) , \(b = 15\) & \(a = 14\)

Thus, Answer must be (C) 15
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Bunuel
If a, b, c are three consecutive integers such that c > b > a and abc = 3360. What is the value of b?

A. 13
B. 14
C. 15
D. 16
E. 17
Given that \(a, b, c\) are three consecutive integers and \(abc = 3360\).
Since the integers are consecutive, their values are very close to each other. Therefore, the product \(a \times b \times c\) will be approximately equal to the cube of the middle term (\(b^3\)).

Method 1: Approximation (Fastest)
We need to find \(b\) such that \(b^3 \approx 3360\).

Let's look at perfect cubes we might know:
\(10^3 = 1000\) (Too small)
\(20^3 = 8000\) (Too big)

Let's check the midpoint, 15:
\(15^3 = 3375\)

The given value (3360) is extremely close to 3375.
This suggests the middle term \(b\) is 15.

Let's verify with the integers centered around 15: \(14, 15, 16\).
\(14 \times 15 \times 16\)
\(= 210 \times 16\)
\(= 3360\)
This matches exactly.

[hr]

Method 2: Prime Factorization
If the approximation isn't immediately obvious, break 3360 into prime factors to find three consecutive numbers.

\(3360 = 10 \times 336\)
\(= 2 \times 5 \times 3 \times 112\)
\(= 2 \times 5 \times 3 \times 2 \times 56\)
\(= 2^2 \times 3 \times 5 \times 7 \times 8\)
\(= 2^5 \times 3 \times 5 \times 7\)

Now, group these factors to form three consecutive integers:
1. We see a \(5\) and a \(3\). \(5 \times 3 = 15\).
2. We see a \(7\) and a \(2\). \(7 \times 2 = 14\).
3. The remaining factors are four \(2\)'s (\(2^4\)). \(16\).

The integers are 14, 15, and 16.
Since \(c > b > a\), \(b\) is the middle integer.
\(b = 15\).

Answer: C
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