Difference between revisions of "1985 AHSME Problems/Problem 29"
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==Problem== | ==Problem== | ||
− | In their base <math> 10 </math> | + | In their base <math> 10 </math> representations, the integer <math> a </math> consists of a sequence of <math> 1985 </math> eights and the integer <math> b </math> consists of a sequence of <math> 1985 </math> fives. What is the sum of the digits of the base <math> 10 </math> representation of <math> 9ab </math>? |
<math> \mathrm{(A)\ } 15880 \qquad \mathrm{(B) \ }17856 \qquad \mathrm{(C) \ } 17865 \qquad \mathrm{(D) \ } 17874 \qquad \mathrm{(E) \ }19851 </math> | <math> \mathrm{(A)\ } 15880 \qquad \mathrm{(B) \ }17856 \qquad \mathrm{(C) \ } 17865 \qquad \mathrm{(D) \ } 17874 \qquad \mathrm{(E) \ }19851 </math> | ||
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− | Finally, we have to multiply by <math> 4 </math>. Doing this, we find that the pattern continues, and the final outcome is <math> 4444\cdots443555\cdots5560 </math>, where there are <math> 1984 </math> | + | Finally, we have to multiply by <math> 4 </math>. Doing this, we find that the pattern continues, and the final outcome is <math> 4444\cdots443555\cdots5560 </math>, where there are <math> 1984 </math> fours, <math> 1 </math> three, <math> 1984 </math> fives, <math> 1 </math> six, and a final zero. Adding this up, the sum of the digits is <math> 1984(4)+3+1984(5)+6+0=17865, \boxed{\text{C}} </math>. |
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=28|num-a=30}} | {{AHSME box|year=1985|num-b=28|num-a=30}} | ||
+ | {{MAA Notice}} |
Latest revision as of 18:15, 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 fours, 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.