1985 AHSME Problems/Problem 29
Problem
In their base 
representations, the integer 
consists of a sequence of 
eights and the integer 
consists of a sequence of fives. What is the sum of the digits of the base representation of the integer 
?
Solution
By the formula for the sum of a geometric series, 
and similarly ![\[b = \frac{5\left(10^{1985}-1\right)}{9},\]](http://latex.artofproblemsolving.com/e/f/3/ef3998cb14dbc7d6460692f9b0673ebb0c693eb8.png)
so 
We now compute the decimal expansion of this expression. Firstly, 
, with 
one and 
zeroes, and 
, with two and 
zeroes. Subtracting therefore gives ![\[10^{3971}-2 \cdot 10^{1986} = 99 \dotsb 9800 \dotsb 0,\]](http://latex.artofproblemsolving.com/7/a/7/7a7d5e2cb3aaeb2d06521dfeb0736a0fe6a87df1.png)
where there are 
nines followed by eight and then zeroes. Adding transforms this to 
, now with 
nines followed by eight, zeroes, one, and a final zero.
Using long division, and noting that 
and 
, it follows that ![\[\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 11 \dotsb 1088 \dotsb 890,\]](http://latex.artofproblemsolving.com/c/0/d/c0d1c79617d4c8ad909ee33fc3da8ed802fc8fab.png)
with ones, zero, then eights, nine, and a final zero. Lastly, using long multiplication and noting that 
, 
, and 
, we obtain ![\[\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 44 \dotsb 4355 \dotsb 560,\]](http://latex.artofproblemsolving.com/b/2/6/b263b4e0d6af2ab56e7202884dc0d559cfca0f77.png)
where there are fours, three, fives, six, and a final zero, so the sum of the digits is 
See Also
| 1985 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 28 |
Followed by Problem 30 | |
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