MHB Can You Solve the 4-Digit Number Puzzle?

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The challenge presented is to find a four-digit number represented as $\overline{abcd}$ that equals the sum of each digit raised to the power of itself, expressed as $a^a + b^b + c^c + d^d$. The solution provided is $3435$, which satisfies the equation since $3^3 + 4^4 + 3^3 + 5^5 = 3435$. It is noted that digits must be limited to 5 or less, as $6^6$ exceeds four digits. The calculations for the powers of 3, 4, and 5 are also provided to support the solution. This puzzle highlights the unique relationship between digits and their exponential values.
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I challenge users to find a four digit number $\overline{abcd}$ that is equal to $a^a+b^b+c^c+d^d$. (Drunk)
 
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I felt it was better to begin a new topic for this, especially since the other problem had not been solved yet. :D
 
eddybob123 said:
I challenge users to find a four digit number $\overline{abcd}$ that is equal to $a^a+b^b+c^c+d^d$. (Drunk)
Ans :$3435=3^3+4^4+3^3+5^5$
for $6^6=46656>9999$
$\therefore a,b,c,d \leq5$
$3^3=27$
$4^4=256$
$5^5=3125$
the next procedure is easy
 
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