The sum of two consecutive integers is 68, what is the smaller number?

1 Answer
Jul 18, 2016

color(red)("This question is wrong!")

Explanation:

color(blue)("Why this question is wrong")

Two consecutive number means that one of them is even and the other odd. Consequently their sum will be odd.

For the sum to be 68 the question has to be one of:
Two consecutive even numbers give an even number answer.
Two consecutive odd numbers give an even number answer.
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color(brown)("Alternative questions")

color(blue)("Solution for two consecutive even numbers sum to 68")

Let n be any number

Then 2n is even

So 2n+2 is the next even number

Thus 2n+(2n+2)=68

So 4n+2=68

Subtract 2 from both sides

4n=66

n=66/4=16.5 larr" seed value"

Thus the 1st even number is 2n->2xx16.5=33
Thus the next even number is 33+2=35

color(blue)(33+35 = 68)
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color(blue)("Solution for two consecutive odd numbers sum to 68")

Using the notation from the first solution

If 2n is even then 2n+1 is odd and the first number
The second odd number will be (2n+1)+2=2n+3

So (2n+1)+(2n+3)=68

=> 4n+4=68

=> 4n=64

Divide both sides by 4

=> n= 64/4=16larr" Seed value"

So the first odd number is 2n+1=2(16)+1=33
So the second odd number is 33+2=35

color(blue)(33+35=68)