The two consecutive positive integers has a product of 272? What are the 4 integers?

3 Answers
May 12, 2018

(-17,-16) and (16,17)

Explanation:

Let a be the smaller of the two integers and let a+1 be the bigger of the two integers:
(a)(a+1) = 272, easiest way to solve this is to take the square root of 272 and round down:
sqrt(272) = \pm16...
16*17 = 272
Thus, the integers are -17,-16 and 16,17

May 12, 2018

16 17

Explanation:

If we multiply two consecutive numbers, n and n+1
we get n^2+n. That is we square a number and add one more on.

16^2=256

256+16=272

So our two numbers are 16 and 17

May 12, 2018

16 and 17

Explanation:

color(blue)("A sort of cheat way")

The two number are very close to each other so lets 'fudge' it

sqrt(272) = 16.49... so the first number is close to 16

Test 16xx17=272 color(red)(larr"First guess gets the prize!")
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("The systematic way")

Let the first value be n then the next value is n+1

The product is n(n+1)=272

n^2+n-272=0

Compare to: ax^2+bx+c=0color(white)("ddd") -> color(white)("ddd") x=(-b+-sqrt(b^2-4ac))/(2a)
In this case x->n; color(white)("d")a=1; color(white)("d")b=1 and c=-272

n=(-1+-sqrt(1-4(1)(-272)))/(2(1))

n=-1/2+-sqrt(1089)/2

n=-1/2+-33/2 The negative is not logical so discard it

n=-1/2+33/2 = 16

The first number is 16 the second is 17