Difference between revisions of "1985 AHSME Problems/Problem 29"
(→Solution) |
Sevenoptimus (talk | contribs) (Improved solutions, formatting, and LaTeX) |
||
(2 intermediate revisions by one other user not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | 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>? | + | 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 the integer <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> | ||
==Solution== | ==Solution== | ||
− | + | By the formula for the sum of a geometric series, <cmath>\begin{align*}a &= 8 \cdot 10^0 + 8 \cdot 10^1 + \dotsb + 8 \cdot 10^{1984} \ &= \frac{8\left(10^{1985}-1\right)}{10-1} \ &= \frac{8\left(10^{1985}-1\right)}{9},\end{align*}</cmath> and similarly <cmath>b = \frac{5\left(10^{1985}-1\right)}{9},</cmath> so <cmath>\begin{align*}9ab &= 9\cdot\frac{8\left(10^{1985}-1\right)}{9}\cdot\frac{5\left(10^{1985}-1\right)}{9} \ &= \frac{40\left(10^{1985}-1\right)^2}{9} \ &= \frac{40\left(10^{3970}-2 \cdot 10^{1985}+1\right)}{9} \ &= \frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9}.\end{align*}</cmath> | |
− | + | We now compute the decimal expansion of this expression. Firstly, <math>10^{3971} = 100 \dotsb 0</math>, with <math>1</math> one and <math>3971</math> zeroes, and <math>2 \cdot 10^{1986} = 200 \dotsb 0</math>, with <math>1</math> two and <math>1986</math> zeroes. Subtracting therefore gives <cmath>10^{3971}-2 \cdot 10^{1986} = 99 \dotsb 9800 \dotsb 0,</cmath> where there are <math>3971-1986-1 = 1984</math> nines followed by <math>1</math> eight and then <math>1986</math> zeroes. Adding <math>10 </math> transforms this to <math>99 \dotsb 9800 \dotsb 010</math>, now with <math>1984</math> nines followed by <math>1</math> eight, <math>1984</math> zeroes, <math>1</math> one, and a final zero. | |
− | + | Using long division, and noting that <math>80 = 8 \cdot 9 + 8</math> and <math>81 = 9 \cdot 9</math>, it follows that <cmath>\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 11 \dotsb 1088 \dotsb 890,</cmath> with <math>1984</math> ones, <math>1</math> zero, then <math>1984</math> eights, <math>1</math> nine, and a final zero. Lastly, using long multiplication and noting that <math>9 \cdot 4 = 36</math>, <math>8 \cdot 4 = 32</math>, and <math>8 \cdot 4 + 3 = 35</math>, we obtain <cmath>\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 44 \dotsb 4355 \dotsb 560,</cmath> where there are <math>1984</math> fours, <math>1</math> three, <math>1984</math> fives, <math>1</math> six, and a final zero, so the sum of the digits is <cmath>\begin{align*}1984 \cdot 4 + 3 + 1984 \cdot 5 + 6 + 0 &= 1984 \cdot 9 + 9 \ &= 1985 \cdot 9 \ &= \boxed{\text{(C)} \ 17865}.\end{align*}</cmath> | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
==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}} | {{MAA Notice}} |
Latest revision as of 02:25, 20 March 2024
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 so
We now compute the decimal expansion of this expression. Firstly, , with one and zeroes, and , with two and zeroes. Subtracting therefore gives 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 with ones, zero, then eights, nine, and a final zero. Lastly, using long multiplication and noting that , , and , we obtain 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 | |
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.