The three sides of a triangle are all integers, not necessarily different. If two are 7 and 10, how many different lengths can the third one be?

Solution to June's Problem

What is the sum of the first 100 positive odd integers?

ANSWER: 10,000

Solution:

1. Thank you Carl Friedrich Gauss:

1 + 3 + 5 + 7 +...+ 199 = (1+199) + (3+197) + (5+195) + ... + (99 + 101)

= (200) + (200) + (200) + ... + (200) {50 times} = 10,000

2. The sum of N consecutive odd integers beginning with 1 equals N^2:

Shown below is 1 making a square 1x1 =1, 1 + 3 making a square 2x2 = 4,

1 + 3 + 5 = 3x3 = 9, then 1 + 3 + 5 + 7 = 4x4 = 16, and finally 1 + 3 + 5 + 7 + 9 = 5x5 = 25.

So 1 + 3 + 5 + ... + 199 = 100x100 = 10,000

3. Thank You, follow up question:

Summing the first 100 evens as well as odds (i.e. the first 200 integers) can be calculated as follows:

1 + 2 + 3 + ... + 200 = (1 + 200) + (2 + 199) + (3 + 197) + ... + (100 + 101) =

(201) + (201) + (201) + ... + (201) {100 times} = 20,100.

Since EVEN = ODD + 1, ODD + EVEN = ODD + (ODD + 1). The 20,100 can be considered as the (sum of the first 100 odds) + (sum of first 100 odds) + (100 x 1's) which will be 10000 + (10000 + 100). So the sum of the odds is 10,000, while the sum of the evens is (10000 + 100) = 10,100

Follow-up question:

Similar to #3 above, each even = the preceeding odd + 1. Since there are 100 of them, the total of the evens is 100 greater than the sum of the odds. 10,000 for the total of the odds means 10,100 is the sum of the evens.

ANSWER: 10,100