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Difference between revisions of "2025 AIME I Problems/Problem 15"
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==Problem==
 
==Problem==
  
−
Let <math>N</math> denote the number of ordered triples of positive integers <math>(a, b, c)</math> such that <math>a, b, c \leq 3^6</math> and <math>a^3 + b^3 + c^3</math> is a multiple of <math>3^7</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>.  
+
Let <math>N</math> denote the number of ordered triples of positive integers <math>(a, b, c)</math> such that <math>a, b, c \leq 3^6</math> and <math>a^3 + b^3 + c^3</math> is a multiple of <math>3^7</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>.
 +
 
 +
==Solution==
 +
 
 +
First, state the LTE Lemma for <math>p = 3, n = 3</math>, which we might use.
 +
 
 +
 
 +
    <math>\bullet</math> <math> \nu_3(n) = \begin{cases}
 +
        \max \{k : 3^k \mid n\} &n\neq 0\\
 +
        \infty &n=0
 +
    \end{cases}</math>
 +
 
 +
    <math>\bullet</math> If <math>3 \nmid x, 3\nmid y, 3\mid x+y</math>, then <math>\nu_3(x^3+y^3) = \nu_3(x+y) + \nu_3(3) = \nu_3(x+y) + 1</math>
 +
   
 +
    <math>\bullet</math> If <math>3 \nmid x, 3\nmid y, 3\mid x-y</math>, then <math>\nu_3(x^3-y^3) = \nu_3(x-y) + \nu_3(3) = \nu_3(x-y) + 1</math>
 +
   
 +
 
 +
Then we classify all cube numbers <math>\mod{3^7}</math>
 +
 
 +
 
 +
 
 +
    <math>\bullet</math> <math>A = \{a : 1\leq a \leq 3^6, 9\mid a\}</math>
 +
 
 +
    We can write <math>A = \{a: a=9k, 1\leq k \leq 3^4\}</math>, so <math>|A| = 3^4</math>.
 +
   
 +
        <math>\bullet</math> If <math>k \equiv 0 \mod{3}</math>, <math>k^3 \equiv 0 \mod{3}, a^3 \equiv 3^6k^3 \equiv 0 \mod{3^7}</math>, there are 27 roots.
 +
        <math>\bullet</math> If <math>k \equiv 1 \mod{3}</math>, <math>k^3 \equiv 1 \mod{3}, a^3 \equiv 3^6k^3 \equiv 3^6 \mod{3^7}</math>, there are 27 roots.
 +
        <math>\bullet</math> If <math>k \equiv 2 \mod{3}</math>, <math>k^3 \equiv 2 \mod{3}, a^3 \equiv 3^6k^3 \equiv 2\times 3^6 \mod{3^7}</math>, there are 27 roots.
 +
   
 +
   
 +
    <math>\bullet</math> <math>B = \{a : 1\leq a \leq 3^6, 3\mid a, 9 \nmid a\}</math>
 +
 
 +
    We can write <math>B = \{a: a=9k+3 \text{ or }a=9k+6, 0\leq k < 3^4\}</math>, so <math>|B| = 2 \times 3^4</math>.
 +
 
 +
    For <math>x,y \in \{a: a=9k+3, 0\leq k < 3^4\}</math>, let <math>x = 3a, y= 3b</math> and hence <math>3 \nmid a, 3\nmid b, 3\mid a-b</math>.
 +
   
 +
        <math>(3a)^3 - (3b)^3 \equiv 0 \mod{3^7}</math>
 +
        <math>\iff a^3 - b^3 \equiv 0 \mod{3^4}</math>
 +
        <math>\iff \nu(a^3-b^3) \geq 4</math>
 +
        <math>\iff \nu(a-b) \geq 3</math>
 +
        <math>\iff a - b\equiv 0 \mod{3^3}</math>
 +
        <math>\iff x - y\equiv 0 \mod{3^4}</math>
 +
   
 +
    For <math>k = 0, ..., 8</math>, each has 9 roots.
 +
   
 +
    Since <math>(9k+3)^3 \equiv 3^6k^3+3^6k^2+3^5k + 3^3 \equiv 3^5m + 3^3 \mod{3^7}</math>, <math>0\leq m \leq 8</math>. They are the corresponding classes.
 +
 
 +
    Same apply to <math>x,y \in \{a: a=9k+6, 0\leq k < 3^4\}</math>. For <math>k = 0, ..., 8</math>, each has 9 roots.
 +
 
 +
    Since <math>(9k+6)^3 \equiv 3^6k^3+2\times3^6k^2+4\times3^5k + 8\times3^3 \equiv 3^5m - 3^3 \mod{3^7}</math>, <math>0\leq m \leq 8</math>. They are the corresponding classes.
 +
 
 +
    <math>\bullet</math> <math>C = \{a : 1\leq a \leq 3^6, 3\nmid a\}</math>
 +
 
 +
    We write <math>C = \{a: a=3k+1 \text{ or }a=3k+2, 0\leq k < 3^5\}</math>, so <math>|C| = 2 \times 3^5</math>.
 +
 
 +
    For <math>a,b \in \{a: a=3k+1, 0\leq k < 3^5\}</math>, then <math>3 \nmid a, 3\nmid b, 3\mid a-b</math>.
 +
   
 +
        <math>a^3 - b^3 \equiv 0 \mod{3^7}</math>
 +
        <math>\iff \nu(a^3-b^3) \geq 7</math>
 +
        <math>\iff \nu(a-b) \geq 6</math>
 +
        <math>\iff a - b\equiv 0 \mod{3^6}</math>
 +
    For <math>k = 0, ..., 3^5-1</math>, 1 root each.
 +
   
 +
    <math>(3k+1)^3 \equiv 3^3k^3+3^3k^2+3^2k + 1 \equiv 3^2m + 1 \mod{3^7}</math>, <math>0\leq m < 3^5</math>. They are the corresponding classes.
 +
 
 +
    Same apply to <math>x,y \in \{a: a=3k+2, 0\leq k < 3^5\}</math>. For <math>k = 0, ..., 3^5-1</math>, 1 root each.
 +
 
 +
    <math>(3k+2)^3 \equiv 3^3k^3+2\times3^3k^2+4\times3^2k + 8 \equiv 3^2m - 1 \mod{3^7}</math>, <math>0\leq m < 3^5</math>. They are the corresponding classes.
 +
 
 +
 
 +
Summarized the results:
 +
 
 +
    <math>\bullet</math> <math>A</math>: If <math>x \equiv 0, 3^6, 2\times3^6 \mod{3^7}</math>, then <math>x</math> has 27 roots. <math>|A| = 3^4</math>.
 +
    <math>\bullet</math> <math>B</math>: If <math>x \equiv 3^5m \pm 3^3 \mod{3^7}</math>, then <math>x</math> has 9 roots. <math>|B| = 2\times3^4</math>.
 +
    <math>\bullet</math> <math>C</math>: If <math>x \equiv 3^2m \pm 1 \mod{3^7}</math>, then <math>x</math> has 1 root. <math>|C| = 2\times3^5</math>.
 +
    <math>\bullet</math> Otherwise, <math>x</math> has no roots.
 +
 
 +
 
 +
We do the final combinatorial problem.
 +
 
 +
 
 +
    <math>\bullet</math> Case: <math>A,A,A</math>: <math>|A| \times |A| \times 27 = \boxed{3 \times 3^{10}}</math>
 +
    <math>\bullet</math> Case <math>A,A,B</math>: No solution.
 +
    <math>\bullet</math> Case <math>A,A,C</math>: No solution.
 +
    <math>\bullet</math> Case: <math>A,B,B</math>: <math>3 \times |A| \times |B| \times 9 = \boxed{6 \times 3^{10}}</math>
 +
    <math>\bullet</math> Case <math>A,A,B</math>: No solution.
 +
    <math>\bullet</math> Case: <math>A,C,C</math>: <math>3 \times |A| \times |C| \times 1 = \boxed{2 \times 3^{10}}</math>
 +
    <math>\bullet</math> Case <math>B,B,C</math>: No solution.
 +
    <math>\bullet</math> Case <math>B,B,B</math>: No solution.
 +
    <math>\bullet</math> Case <math>B,C,C</math>: <math>3 \times |B| \times |C| \times 1 = \boxed{4 \times 3^{10}}</math>
 +
    <math>\bullet</math> Case <math>C,C,C</math>: No solution.   
 +
 
 +
Total is <math>(3+6+2+4)3^{10}=15\times 3^{10}</math>.
 +
 
 +
<math>3^5 = 243 \equiv 43 \mod{200},43^2=1600+240+9\equiv49\mod{200}</math>
 +
 
 +
Hence <math>15\times3^{10}\equiv15\times49\equiv735\mod{1000}</math>
 +
 
 +
Answer is <math>\boxed{735}</math>.
 +
 
 +
~Rakko12
  
 
==See also==
 
==See also==

Revision as of 23:39, 14 February 2025

Problem

Let $N$ denote the number of ordered triples of positive integers $(a, b, c)$ such that $a, b, c \leq 3^6$ and $a^3 + b^3 + c^3$ is a multiple of $3^7$. Find the remainder when $N$ is divided by $1000$.

Solution

First, state the LTE Lemma for $p = 3, n = 3$, which we might use. 


   $\bullet$ $\nu_3(n) = \begin{cases}         \max \{k : 3^k \mid n\} &n\neq 0\\         \infty &n=0     \end{cases}$

   $\bullet$ If $3 \nmid x, 3\nmid y, 3\mid x+y$, then $\nu_3(x^3+y^3) = \nu_3(x+y) + \nu_3(3) = \nu_3(x+y) + 1$
   
   $\bullet$ If $3 \nmid x, 3\nmid y, 3\mid x-y$, then $\nu_3(x^3-y^3) = \nu_3(x-y) + \nu_3(3) = \nu_3(x-y) + 1$

Then we classify all cube numbers $\mod{3^7}$ 


   $\bullet$ $A = \{a : 1\leq a \leq 3^6, 9\mid a\}$

   We can write $A = \{a: a=9k, 1\leq k \leq 3^4\}$, so $|A| = 3^4$.
   
       $\bullet$ If $k \equiv 0 \mod{3}$, $k^3 \equiv 0 \mod{3}, a^3 \equiv 3^6k^3 \equiv 0 \mod{3^7}$, there are 27 roots.
       $\bullet$ If $k \equiv 1 \mod{3}$, $k^3 \equiv 1 \mod{3}, a^3 \equiv 3^6k^3 \equiv 3^6 \mod{3^7}$, there are 27 roots.
       $\bullet$ If $k \equiv 2 \mod{3}$, $k^3 \equiv 2 \mod{3}, a^3 \equiv 3^6k^3 \equiv 2\times 3^6 \mod{3^7}$, there are 27 roots.
   
   
   $\bullet$ $B = \{a : 1\leq a \leq 3^6, 3\mid a, 9 \nmid a\}$

   We can write $B = \{a: a=9k+3 \text{ or }a=9k+6, 0\leq k < 3^4\}$, so $|B| = 2 \times 3^4$. 

   For $x,y \in \{a: a=9k+3, 0\leq k < 3^4\}$, let $x = 3a, y= 3b$ and hence $3 \nmid a, 3\nmid b, 3\mid a-b$.
   
       $(3a)^3 - (3b)^3 \equiv 0 \mod{3^7}$
       $\iff a^3 - b^3 \equiv 0 \mod{3^4}$
       $\iff \nu(a^3-b^3) \geq 4$
       $\iff \nu(a-b) \geq 3$
       $\iff a - b\equiv 0 \mod{3^3}$
       $\iff x - y\equiv 0 \mod{3^4}$
   
   For $k = 0, ..., 8$, each has 9 roots.
   
   Since $(9k+3)^3 \equiv 3^6k^3+3^6k^2+3^5k + 3^3 \equiv 3^5m + 3^3 \mod{3^7}$, $0\leq m \leq 8$. They are the corresponding classes.

   Same apply to $x,y \in \{a: a=9k+6, 0\leq k < 3^4\}$. For $k = 0, ..., 8$, each has 9 roots.

   Since $(9k+6)^3 \equiv 3^6k^3+2\times3^6k^2+4\times3^5k + 8\times3^3 \equiv 3^5m - 3^3 \mod{3^7}$, $0\leq m \leq 8$. They are the corresponding classes.

   $\bullet$ $C = \{a : 1\leq a \leq 3^6, 3\nmid a\}$

   We write $C = \{a: a=3k+1 \text{ or }a=3k+2, 0\leq k < 3^5\}$, so $|C| = 2 \times 3^5$.

   For $a,b \in \{a: a=3k+1, 0\leq k < 3^5\}$, then $3 \nmid a, 3\nmid b, 3\mid a-b$.
   
       $a^3 - b^3 \equiv 0 \mod{3^7}$
       $\iff \nu(a^3-b^3) \geq 7$
       $\iff \nu(a-b) \geq 6$
       $\iff a - b\equiv 0 \mod{3^6}$
   For $k = 0, ..., 3^5-1$, 1 root each.
   
   $(3k+1)^3 \equiv 3^3k^3+3^3k^2+3^2k + 1 \equiv 3^2m + 1 \mod{3^7}$, $0\leq m < 3^5$. They are the corresponding classes.

   Same apply to $x,y \in \{a: a=3k+2, 0\leq k < 3^5\}$. For $k = 0, ..., 3^5-1$, 1 root each.

   $(3k+2)^3 \equiv 3^3k^3+2\times3^3k^2+4\times3^2k + 8 \equiv 3^2m - 1 \mod{3^7}$, $0\leq m < 3^5$. They are the corresponding classes.


Summarized the results: 

   $\bullet$ $A$: If $x \equiv 0, 3^6, 2\times3^6 \mod{3^7}$, then $x$ has 27 roots. $|A| = 3^4$.
   $\bullet$ $B$: If $x \equiv 3^5m \pm 3^3 \mod{3^7}$, then $x$ has 9 roots. $|B| = 2\times3^4$.
   $\bullet$ $C$: If $x \equiv 3^2m \pm 1 \mod{3^7}$, then $x$ has 1 root. $|C| = 2\times3^5$.
   $\bullet$ Otherwise, $x$ has no roots.


We do the final combinatorial problem. 


   $\bullet$ Case: $A,A,A$: $|A| \times |A| \times 27 = \boxed{3 \times 3^{10}}$
   $\bullet$ Case $A,A,B$: No solution.
   $\bullet$ Case $A,A,C$: No solution.
   $\bullet$ Case: $A,B,B$: $3 \times |A| \times |B| \times 9 = \boxed{6 \times 3^{10}}$
   $\bullet$ Case $A,A,B$: No solution.
   $\bullet$ Case: $A,C,C$: $3 \times |A| \times |C| \times 1 = \boxed{2 \times 3^{10}}$
   $\bullet$ Case $B,B,C$: No solution.
   $\bullet$ Case $B,B,B$: No solution.
   $\bullet$ Case $B,C,C$: $3 \times |B| \times |C| \times 1 = \boxed{4 \times 3^{10}}$
   $\bullet$ Case $C,C,C$: No solution.

Total is $(3+6+2+4)3^{10}=15\times 3^{10}$.

$3^5 = 243 \equiv 43 \mod{200},43^2=1600+240+9\equiv49\mod{200}$

Hence $15\times3^{10}\equiv15\times49\equiv735\mod{1000}$

Answer is $\boxed{735}$.

~Rakko12

See also

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