If the product of the integers w, x, y and z is 770, and if 1 < w < x

How can we be sure of these nos..It can be others also..How can one come with these nos

it can be any other number because these are prime factors of 770

770 = 77*10 = 7*11 * 2*5

hope this helps

prime factors of 770: 2, 5, 7, 11
Therefore:
w+z=2+11=13

Answer: B

Hi All,

We're told that the product of the integers W, X, Y and Z is 770, and that 1 < W < X < Y < Z. We're asked for the value of W + Z.

One of the great aspects of the 'math' behind factoring is that you can do the factoring in any order and you'll still get to the same result...

With 770, a few Number Properties stand out, so your first 'factor' would likely either be 2 (since 770 is EVEN), 7 (since 7 divides into 700 and 70) or 10 (since 770 ends in a 0). Let's go for the biggest option first....10

770 = (77)(10)

From here, you can further break down each of the two 'pieces'....

77 = (7)(11)
10 = (2)(5)

770 = (2)(5)(7)(11)

Based on the given inequality 'string', we know that W is the smallest value and Z is the largest value. 2 + 11 = 13

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Since 770 = 2 * 5 * 7 * 11, then w = 2, x = 5, y = 7 and z = 11. Therefore, the sum of w and z is 2 + 11 = 13.

Answer: B

1. Prime factorization of \(770\) yields - \(11 * 7 * 5 * 2\)
2. We are given that \(1 < w < x < y < z\) i.e. \(z = 11\), \(y = 7\), \(x = 5\) and \(w = 2\)
3. \(w + z = 2 + 11 = 13\)

Ans. B

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