Can You Solve the 4-Digit Number Puzzle?

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SUMMARY

The four-digit number puzzle challenges users to find a number $\overline{abcd}$ such that it equals the sum of its digits raised to the power of themselves, specifically $a^a + b^b + c^c + d^d$. The solution provided is $3435$, which satisfies the equation as $3^3 + 4^4 + 3^3 + 5^5 = 3435$. It is established that the digits must be less than or equal to 5, as demonstrated by the calculation of $6^6 = 46656$, which exceeds the four-digit limit.

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eddybob123
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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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