Difference between revisions of "1985 AHSME Problems/Problem 29"
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| − | We can multiply out <math> (10^{1985}-1)^2 </math> to get <math> 9ab=\frac{40(10^{3970}-2\cdot10^{1985}+1)}{9}=\frac{ | + | We can multiply out <math> (10^{1985}-1)^2 </math> to get <math> 9ab=\frac{40(10^{3970}-2\cdot10^{1985}+1)}{9}=\frac{4(10^{3971}-2\cdot10^{1986}+10)}{9} </math>. |
Revision as of 17:10, 15 May 2021
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 
?
Solution
Notice that 
by the formula for a geometric series.
Similarly, 
.
Thus, 
.
We can multiply out 
to get 
.
We now find this in decimal form. 
, where there is 
one and 
zeroes.
, where there is two and 
zeroes.
We subtract to find that 
, where there are 
nines, eight, and zeroes.
We now add to get 
, where there are nines, eight, zeroes, one, and a final zero.
Next, we begin to divide by 
. We get this to be 
, where there are ones, zero, eights, nine, and a final zero.
Finally, we have to multiply by 
. Doing this, we find that the pattern continues, and the final outcome is 
, where there are ones, three, fives, six, and a final zero. Adding this up, the sum of the digits is 
.
See Also
| 1985 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 28 |
Followed by Problem 30 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
