Question

Find three consecutive odd integers whose sum is 111. Are there three consecutive odd integers whose sum is 1111? Explain.

Answer 1

Anyone have any ideas on this and if there is a pattern for: 11111, 111111, etc.?

Answer 2

There's no trick to it. You just have to solve the equation.

Answer 3

\[x + (x+2) + (x+4) = 1111 \implies 3x + 6 = 1111\]
But this will only have integer solutions when the sum - 6 is evenly divisible by 3.

Answer 4

Which, in the case of 1111, it isn't.

Answer 5

Thank you very much! Not sure I understand but definitely appreciate your assistance.

Answer 6

Which part don't you understand?

Answer 7

Imagine that x is the smallest of the odd numbers. Then the next consecutive odd number will be x+2 and the next one after that is x+4.

Answer 8

If the sum of all 3 is 1111 then you have that x + (x+2) + (x+4) = 1111.
Which in turn means that x + x + x + 2 + 4 = 1111.
Combining our 'like terms' we have 3x + 6 = 1111.

Answer 9

So for 1111 we have that
\[3x + 6 = 1111\]\[\implies 3x = 1111 - 6\]\[\implies 3x = 1105\]\[\implies x = \frac{1105}{3} = 368.\bar{33}\ \not \in Z\]
Therefore 1111 has no solution. And you can also come up with a general formula for any number to test by replacing the 1111 with whatever other number you choose.

Answer 10

Excellent! Thank you so much. I understand it totally now!