MHB Can You Solve the 4-Digit Number Puzzle?

  • Thread starter Thread starter eddybob123
  • Start date Start date
AI Thread Summary
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.
eddybob123
Messages
177
Reaction score
0
I challenge users to find a four digit number $\overline{abcd}$ that is equal to $a^a+b^b+c^c+d^d$. (Drunk)
 
Mathematics news on Phys.org
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
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top