Question

Find sum of digits of 4444^4444.

Answer 1

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♦ IMO 1975 ♦
→ Modular Arithmetic
→ Logarithmic Estimation of Size 
→ Number Theory 

◘ Problem 4 : When 4444⁴⁴⁴⁴ is written in decimal notation, the sum of its digits is A. Let B be the sum of the digits of A . Find the sum of the digits of B. ( A and B are written in Decimal Notation )

► Solution : 
 • Note, the number 4444⁴⁴⁴⁴ is really big .

 • The number of digits in any number 'X' written in decimal notation is given as :     10ᵃ ≤ X ≤ 10ᵇ
   =>     a ≤ log X ≤ b  [  where 'a' and 'b' ∈ N₀  ]
   => Number of digits of 'X' is : ( a + 1 )

→ Number of digits of 4444⁴⁴⁴⁴ is given by :
    • 4444⁴⁴⁴⁴ < 10000⁴⁴⁴⁴ = 10¹⁷⁷⁷⁶
  => Number of digits in 'X' is lesser than 17776 ( approx. 16211 ) 
  =>  A < 16211 • 9
  => A < 145899 
  => Sum of digits of 'A' is maximum when A = 99999 
  => B ≤ 45

→ Sum of digits of 'B' is maximum at B = 39 , i.e., 12
 => B ≤ 12
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◙ Main Step is to figure out, " The remainder of the sum of digits of any natural number 'X' is the same as the remainder when 'X' is divided by 9 "

◙ Mathematically interpreted as :
  → If S( x ) defines the sum of digits of x :
      •  X ≡ S( x ) ≡ S( S( x )) ( mod 9 )

→ Proof : 
     • X =  A₀ x 10ᵃ¹ + A₁ x 10ᵃ² + ... + A x 10ᵃⁿ 
    But, 10 ≡ 1 ( mod 9 )
   => X ≡ A₀ x 1 + A₁ x 1 + ... + A = S ( X ) mod 9
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◘ Coming back to our problem : 
  → 4444⁴⁴⁴⁴ ≡ 4444³ ⁽ ¹⁴⁸¹ ⁾ x 4444 ≡ 4444 ≡ 7 ( mod 9 )

• Now, since S( B ) ≤ 12 and S( B ) ≡ 7 ( mod 9 )
  => S( B ) = 7
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→ Hence, the solution to this question was '7' 
→ Additionally, this question was quite troublesome back then, but thanks to modular arithmetic, it consumes less time and space 
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• A healthy discussion here : 
  → https://math.stackexchange.com/questions/169797 
          might be just helpful 
 
  → http://www.math.wayne.edu/~brcs/POTW/archives/W12/Prob10SOL.pdf
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^_^ Hope this helps

Answer 2

4444^4444 is

ans= 5.103632503725508048204E16210