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the sum of all my digits (x + 8 + 2 + 7) is the middle number of three consecutive...

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scoogie | Student, Grade 11 | eNotes Newbie

Posted August 27, 2012 at 8:57 PM via web

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the sum of all my digits (x + 8 + 2 + 7) is the middle number of three consecutive integers such that the sum of the second and third is twenty more than the first

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rugbykats | High School Teacher | (Level 2) Adjunct Educator

Posted August 28, 2012 at 4:30 AM (Answer #1)

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First, solve the three consecutive integers part of the problem.

Let the three consecutive integers be y, y+1 and y+2. Then,

y+20=(y+1)+(y+2)

Adding: y+20=2y+3

Subtract y from each side: 20=y+3

Subtract 3 from each side: 17=y

(Check: 17+20=18+19. Each side equals 37, so we are correct.)

Now, solve the second part of the problem. We know the sum of the digits is equal to the middle integer, so

(x+8+2+7) = (y+1)

Plug in the known quantity for y+1: (x+8+2+7) = 18

Simplify: x+17=18

Subtract 17 from each side: x=1

(Check: 1+8+2+7=18. Each side equals 18, so we are correct.)

From there, the correct answer depends on how the question was asked. The apparent answer is 1827. However, your question just said the sum of my digits, which would mean any combination of those four digits would work. If that's the case, there are 24 correct answers.

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