ALGEBRA.

posted by .

The product of 2 positive integers is 1000. What is the smallest possible sum of these 2 integers?

  • ALGEBRA. -

    possibilies
    1x1000
    2x500
    4x250
    5x200
    8x125
    10 x 100
    20x50
    25x40
    40x25 --- repeating from here on

    so the smallest possible sum of those factors
    = 25 + 40 = 65

    OR

    by the "Just Think About It" theorem, the two number would have to be as close as possible to each other, that is, near the √1000
    which is appr 31.6
    the closest number to 31.6 which divides evenly into 1000 is 25, so the other must be 40

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Integers

    The larger of two positive integers is five more than twice the smaller integer. The product of the integers is 52. Find the integers.
  2. Math

    Find the smallest possible sum of six consecutive integers such that none of the integers is prime.
  3. smallest of 3 integers

    The sum of the reciprocals of three consecutive positive integers is equal to 47 divided by the product of the integers. What is the smallest of the three integers?
  4. math

    the sum of the squares of four consecutive positive integers is 734. what is the smallest of the integers?
  5. math - integers

    The sum of 25 consecutive integers is 1000. What is the smallest integer used as an addendum?
  6. math

    there are three consecutive positive integers such that the sum of the squares of the smallest two is 221. write and equation to find the three consecutive positive integers let x= the smallest integer
  7. math

    the sum of the reciprocals of 3 consecutive positive integers is 47 divided by the product of the integers. what is the smallest of the 3 integers
  8. math

    the sum of the reciprocals of three consecutive positive integers is equal to 47 divided by the product of the integers. what is the smallest of the three integers?
  9. Maths

    The number 1000 can be written in several ways as a sum of one or more consecutive positive integers, for instance, 1000=1000 (one summand) or 1000=198+199+200+201+202 (five summands). Find the largest possible number of summands in …
  10. algebra!!!! please help me!!!!

    The smallest possible positive value of 1−[(1/w)+(1/x)+(1/y)+(1/z)] where w, x, y, z are odd positive integers, has the form a/b, where a,b are coprime positive integers. Find a+b.

More Similar Questions