# What is the value of $w+z$ if $1<w<x<y<z$

I am having solving the following problem:

If the product of the integer $w,x,y,z$ is 770. and if $1<w<x<y<z$ what is the value of $w+z$ ? (ans=$13$)

Any suggestions on how I could solve this problem ?

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## 4 Answers

The number 770 is the product of the prime numbers 2,5,7,11 , and 1<2<5<7<11 Thus, the answer is 2+11 = 13 . Hope this helps

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What do you mean by the word "u"??? – The Chaz 2.0 Sep 17 '12 at 11:24

Hint: $770$ is the product of four distinct primes. Why must $w$, $x$, $y$, $z$ each be one of those primes?

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I would proceed as follows. As you know that $$w \cdot x\cdot y\cdot z = 770$$ and $1 < w$, we can start with supposing that $w=2$. After dividing both sides of original equation by $w$, we are left with $$x \cdot y \cdot z = 385.$$ Now, 3 does not divide 385, so it cannot be the next integer I try for $x$. However, we can find an integer not much larger than 3. Continue in this manner until you have found all of the integers.

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Find the prime factorization of the number. That is always a great place to start when you have a problem involving a product of integer. Now here you are lucky, you find $4$ prime numbers to the power of one, so you know your answer is unique.

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