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Let xyz be a three digit number such that x,y,z form sides of an equilateral triangle. How many such numbers are there?

The answer is 9 using p n c

9*1*1 = 9

Now, if x,y,z form sides of an isoceles triangle. How many such numbers are there?

My try :

the numbers we can use are 1,2,3,4,5,6,7,8,9 and not 0.

9*1*9 + 9*9*1 + 1*9*9 - .......

I am not getting how to subtract the repeating numbers. Can someone guide?

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    Keep in mind that any two sides must add up to strictly more than the third; otherwise, the sides do not form a triangle. (For instance, $(2, 2, 4)$ is not an isosceles triangle.) – 2017-01-26
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    Your concern about repeated patterns, as they come 3 by 3, is solved by dividing by three the number of ways... – 2017-01-26
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    @BrianTung so which method should I apply? – 2017-01-26

2 Answers 2

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Suppose the isosceles triangle had equal sides of length $n $. Let the other side length be $k $. In a triangle, we know that the sum of two sides $>$ than the third side.

We thus have $2n>k $ and $n+k >n $. The second condition is always true as $k \geq 1$. For the value of $k $ to satisfy the first condition, the permissible values of $k $ are $k \in [1,2n-1] \in \mathbb N $. For $n \in [1,9] \in \mathbb N $, can you thus find all permissible $k $?

The answer is $\boxed {165} $ if we also include the equilateral triangles as they are isosceles. If we exclude them , there are $\boxed {165-9=156} $ cases. Hope it helps.

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    Please don't mind.I guess your answer is somewhat wrong. I think you did not pondered that 223, 232 and 322 are individual numbers. – 2017-01-26
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    @Resorcinol Thank you for correcting the typo. The three factor comes due to the combinations that a particular case can have, that is $(3!)/(2!) =3$. Once again thank you. – 2017-01-26
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    Again the answer is somewhere wrong but I am not able to point it out. I am sure about it because the answer you got doesn't fit into the options I have – 2017-01-26
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    @Resorcinol Again sorry. I forgot to put the boundary on $k$ for $n>4$. Now it is all fine. Thank you for correcting once again. (+1). – 2017-01-26
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Let the equal side length be n and k be the length of the other side

For $n = 1, k = 1 \qquad (1)$

For $n = 2, k = 1,2,3 \qquad (3)$

For $n = 3, k = 1,2,3,4,5 \qquad (5)$

For $n = 4, k = 1,2,3,4,5,6,7 \qquad (7)$

For $n>4, k = 1,2,3,4,5,6,7,8,9 \quad (5\times 9=45)$

Total $ = 1+3+5+7+45 = 61$

Considering combination, total number of such numbers become $61 × (3!)/(2!) = 61×3 = 183$

Now eliminating the repeated numbers, we get $183 - (2×9) = 165$. Considering the Equilateral triangles are also isoceles triangles.

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    Looks like you understand it all. As a matter of preference, I would split the $61$ into $52$ non-equilateral and $9$ equilateral cases, then multiply only the non-equilateral cases by $3$ for digit order, – 2017-01-26