What is the Sum of Digits in the Power of Three Raised to 10,000?

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The discussion focuses on calculating the sum of digits for the number \( X = 3^{10000} \). Given \( \log_{10} 3 = 0.4771 \), it is established that \( X \) has 4772 digits. The sum of the digits of \( X \) is denoted as \( A \), which is constrained to \( A < 4772 \times 9 \) or \( A < 50000 \). The subsequent sums \( B \) and \( C \) are derived from \( A \), with \( B < 45 \) and \( C \) ultimately determined to be 9, as it cannot be zero.

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Albert1
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$ X=3^{10000}$
Given $log_{10} 3=0.4771$
All the digits sum of X =A
All the digits sum of A =B
All the digits sum of B =C
please find C
 
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Albert said:
$ X=3^{10000}$
Given $log_{10} 3=0.4771$
All the digits sum of X =A
All the digits sum of A =B
All the digits sum of B =C
please find C

this is 4771 digits so A < 4771 * 9 or A < 50000

B < 41 so C <= 12 ( if b were 39)

X mod 9 = C mod 9

ax x mod 9 = 0 so C= 0 or 9 as it cannot be 18

as C cannot be zero it is 9
 
kaliprasad said:
this is 4771 digits so A < 4771 * 9 or A < 50000

B < 41 so C <= 12 ( if b were 39)

X mod 9 = C mod 9

ax x mod 9 = 0 so C= 0 or 9 as it cannot be 18

as C cannot be zero it is 9

the answer is correct ,but X is not 4771 digits
 
Albert said:
the answer is correct ,but X is not 4771 digits

Thanks
it is my mistake x is 4772 digits
 
A < 4771 * 9 or A < 50000

B < 41 so C <= 12

here the range of A,B,and C should be edited

(the range of A,B,and C are not correct)
 
Albert said:
A < 4771 * 9 or A < 50000

B < 41 so C <= 12

here the range of A,B,and C should be edited

(the range of A,B,and C are not correct)

so I replace it with correct solution

A < 4772 * 9 or < 47720 or < 50000

I have taken for A to be the worst case. hence A and B range is OK
 
A is 5 digits number
$B<5\times 9=45$
B is 2 digits number
$C<2\times 9=18$
I think your range is OK (more rigid )
 
Last edited:

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