# The product of two consecutive integers is 600. Find the integers.

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### 1 Answer

There are two ways to solve this, depending on how you want to think about it. For starters, we represent the problem with the following two equations:

xy = 600

y=x+1

And plugging the second into the first we get:

x(x+1) = 600

From there we can develop and solve a polynomial equation:

x^2 + x = 600

`x^2 + x - 600 = 0`

Which we then plug into quadratic formula:

`[-b +- sqrt(b^2 - 4ac) ] / (2a)`

for:

`[-1 +-sqrt(1-(4)(1)(-600))]/[(2)(1)]`

This is now arithmetic to solve for x and then y.

The second method involves making approximations. We know that the product of x and y is 600. Take two numbers that we can multiply to reach 600 - say, x is 20 and y is 30. These aren't consecutive integers, but we know now that the values of x and y are between 20 and 30. Before we start trying different numbers, thnk of the last digits... when multiplied together, the resulting final digit must be zero because the last digit is zero. So if we want to do the simplest math possible, we go through the consecutive integers between 20 and 30 - but only the last digits:

2*3 = 6 (nope, not 0)

3*4 = 12 (nope not 0)

4*5 = 20 (let's try this)

The two consecutive integers between 20 and 30 with last digits 4 and 5 are 24 and 25. Does 24*25 equal 600? It does.

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