Just a repeated digital problem

Number Theory Level pending

The digital root (also repeated digital sum) of a non-negative integer is a single digit value obtained by an iterative process of summing digits, on each iteration using the result from the previous iteration to compute a digit sum. (Source: Wikipedia)

What is the digital root of $\large 47^{81! - 3} ?$

The answer is 8.

1 solution

Efren Medallo
May 9, 2015

The digital root of 47 is 2 (4 + 7 = 11, then 1+1=2). Thus, the digital roots of the powers of 47 follow the same pattern as those of the powers of two. That is,

$\large dr (a^{n}) = dr [ dr(a) ^{n}]$

The digital roots of the powers of 2 form a sequence of 1, 2, 4, 8, 7, and 5. Since

$\large 81!-3 \equiv 3 (mod 6)$

then

$\large dr (47^{81!-3} )= dr(2^{81!-3}) = \boxed 8$

Just a repeated digital problem

The answer is 8.

1 solution

0 pending reports