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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Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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