Difference between revisions of "2024 SSMO Team Round Problems/Problem 1"
(Created page with "==Problem== How many ordered triples of positive integers <math>(a, b, c)</math> satisfy the equation <math>2(a^b)^c+1=513</math>? ==Solution==") |
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==Solution== | ==Solution== | ||
| + | We have | ||
| + | \begin{align*} | ||
| + | 2(a^b)^c+1 &= 513\\\implies | ||
| + | a^{bc}&=256\implies a = 2^{a_1} \mid a_1 \in \mathbb{Z}_{\ge 0}\\\implies | ||
| + | 2^{a_1bc} &= 2^{8}\\\implies | ||
| + | a_1bc &= 2^3\implies a_1 = 2^{a_2},b=2^{b_1}, c = 2^{c_1} \mid a_2,b_1,c_1 \in \mathbb{Z}_{\ge 0}\\\implies | ||
| + | 2^{a_2+b_1+c_1} &=2^3\\\implies | ||
| + | a_2+b_1+c_1&=3. | ||
| + | \end{align*} | ||
| + | From the Hockey Stick Identity, it follows that this equation has <math>\binom{5}{2} = \boxed{10}</math> solutions. | ||
| + | |||
| + | ~SMO_Team | ||
Latest revision as of 14:35, 10 September 2025
Problem
How many ordered triples of positive integers 
satisfy the equation 
?
Solution
We have
\begin{align*}
2(a^b)^c+1 &= 513\\\implies
a^{bc}&=256\implies a = 2^{a_1} \mid a_1 \in \mathbb{Z}_{\ge 0}\\\implies
2^{a_1bc} &= 2^{8}\\\implies
a_1bc &= 2^3\implies a_1 = 2^{a_2},b=2^{b_1}, c = 2^{c_1} \mid a_2,b_1,c_1 \in \mathbb{Z}_{\ge 0}\\\implies
2^{a_2+b_1+c_1} &=2^3\\\implies
a_2+b_1+c_1&=3.
\end{align*}
From the Hockey Stick Identity, it follows that this equation has 
solutions.
~SMO_Team
