Question

The sum of five consecutive integers is 46 less than 9 times the least of the integers. Find the largest consecutive integer.

Question 6 options:

A)

17

B)

10

C)

18

D)

19

Answer 1

C) 18

Explanation:

Let x represent the largest of the five consecutive integers. Then (x -4) is the smallest, and the sum of the five of them is ...

sum = 5((x +(x -4))/2 = 5x -10

That sum is 46 less than 9 times the smallest.

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This relation can be written as ...

5x -10 = 9(x -4) -46

72 = 4x . . . . . . . . . . . . add 82-5x and simplify

18 = x . . . . . . . . . divide by 4

The largest of the consecutive integers is 18.

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Additional comment

Here, we have used the variable to represent what the problem is asking for--the largest of the integers. Often, in consecutive integer problems, it is convenient to let a variable represent the middle (or average) value of the consecutive integers. If we do that here, then sum of the 5 integers would be 5x, and the relation to find x would be ...

5x +46 = 9(x -2)

This resolves to 64=4x ⇒ x=16, so the largest integer is x+2=18. The number of steps is about the same, but the sum is easier to find, and there aren't as many constant terms to play with in solving for x.