Can anyone show me how to solve this crazy math riddle?

Question

You have 4 consecutive integers. When you divide each by my age(an integer), then add all 4 remainders you get 40. You have 4 other consecutive integers. When you divided each by my brothers age(a different integer), and added all 4 remainders, you again get 40. What is the sum of our ages?

Answer 1

You don't have enough information to calculate this!

Proof:

let x, x+1, x+2 and x+3 are the first four consecutive integets

let y, y-1, y-2 and y-3 are the other four consecutive integets

let a = your age, let b = your brothers age

(x + x+1 + x+2 + x+3)/a = 40

4x + 6 = 40a (1.)

(y + y-1 + y-2 + y-3)/b = 40

4y - 6 = 40b (2.)

from (1.) and (2.) => 40a + 40b = 4x + 6 + 4y - 6

40(a + b) = 4(x+y)

a + b = 4(x+y)/40

a + b = (x+y)/10

Therefore the sum of your ages equals the sum of the smallest integer from the first four consecutive numbers with the biggest integer from the second four consecutive numbers, divided by 10.

Bottomline: you can express the sum however you want but you can never get an exact actual number based on the information you gave us. Too many unknowns.

Answer 2

I don't know///// are you g*y???

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Answer 3

For the first part of the question, your age could be forty, divide forty by four, and thats ten. add all four tens together to get 40. I don't feel like doing the other half.

Answer 4

is that an assignment? well it sounds like one..............