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 April 2nd, 2013, 07:55 PM #1 Newbie   Joined: Apr 2013 Posts: 6 Thanks: 0 Geometry(Triangle)/Algebra(Probability) Problem A triangle will be called almost equilateral if the sum of the differences between all pairs of sides is 6 or less. Thus, a 3-4-5 triangle is almost equilateral, since 1 + 2 + 1 = 4. However, a 4-6-8 triangle is not, since 2 + 4 + 2 = 8. How many almost equilateral triangles are there with sides whose lengths are integers 100 or less, and such that no side length ends in the digits 4 or 6? (Note: a 3-4-5 triangle is considered the same as a a 3-5-4 triangle; the order of the sides does not matter.) Any and all help is much appreciated. Thank you!
 April 2nd, 2013, 07:57 PM #2 Newbie   Joined: Apr 2013 Posts: 6 Thanks: 0 Re: Geometry(Triangle)/Algebra(Probability) Problem I also have a variant on the same problem that I also must solve: How many almost equilateral triangles are there with sides whose lengths are integers 100 or less, and such that no side length ends in the digit 6? Once again, any and all help is much appreciated. Thank you and have a good day!
April 2nd, 2013, 08:34 PM   #3
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Re: Geometry(Triangle)/Algebra(Probability) Problem

Quote:
 Originally Posted by ItisBowtime A triangle will be called almost equilateral if the sum of the differences between all pairs of sides is 6 or less. Thus, a 3-4-5 triangle is almost equilateral, since 1 + 2 + 1 = 4. However, a 4-6-8 triangle is not, since 2 + 4 + 2 = 8. How many almost equilateral triangles are there with sides whose lengths are integers 100 or less, and such that no side length ends in the digits 4 or 6? (Note: a 3-4-5 triangle is considered the same as a a 3-5-4 triangle; the order of the sides does not matter.)
I get 870.
1st 4: 1-2-2, 1-3-3, 1-4-4, 2-2-3
last 4: 98-99-100, 98-100-100, 99-99-100, 99-100-100

Call the sides a-b-c
a=1: 3
a=2: 6
a=3: 8

a=4 to 97: 9

a=98: 5
a=99: 2
a=100: 0

3 + 6 + 8 + 94*9 + 5 + 2 + 0 = 870

b-a + c-a + c-b = 2c - 2a
2c - 2a = 6
c - a = 3
If c - a <= 3 then ok

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