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Difference between revisions of "1985 AHSME Problems/Problem 29"

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==Problem==
 
==Problem==
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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>?
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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>
  
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==Solution==
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==Solution 1==
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Notice that <math> a=8\cdot10^0+8\cdot10^1+\cdots+8\cdot10^{1984}=\frac{8\cdot10^{1985}-8}{9} </math> by the formula for a geometric series.
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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>
  
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Similarly, <math> b=\frac{5\cdot10^{1985}-5}{9} </math>.
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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.
  
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Thus, <math> 9ab=9\left(\frac{8\cdot10^{1985}-8}{9}\right)\left(\frac{5\cdot10^{1985}-5}{9}\right)=\frac{40(10^{1985}-1)^2}{9} </math>.
+
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>
  
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==Solution 2==
  
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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{40(10^{3971}-2\cdot10^{1986}+10)}{9} </math>.
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Factoring out the 8 and the 5, we get <cmath>9ab = 10\cdot4\cdot9\cdot111 \dotsb 1 \cdot 111\dotsb1,</cmath> where there are <math>1985</math> ones in both numbers.
  
  
  
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We now find this in decimal form. <math> 10^{3971}=10000\cdots00 </math>, where there is <math> 1 </math> one and <math> 3971 </math> zeroes.  
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This is equal to <cmath>10 \cdot 4 \cdot (1000\dotsb 0 - 1) \cdot 111\dotsb 1</cmath> where there is an initial <math>1</math> followed by <math>1985</math> zeros and the second number has <math>1985</math> ones.
  
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<math> 2\cdot10^{1986}=2000\cdots00 </math>, where there is <math> 1 </math> two and <math> 1986 </math> zeroes.
 
  
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We subtract to find that <math> 10^{3971}-2\cdot10^{1986}=9999\cdots98000\cdots00 </math>, where there are <math> 1984 </math> nines, <math> 1 </math> eight, and <math> 1986 </math> zeroes.
 
  
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We now add <math> 10 </math> to get <math> 999\cdots998000\cdots010 </math>, where there are <math> 1984 </math> nines, <math> 1 </math> eight, <math> 1984 </math> zeroes, <math> 1 </math> one, and a final zero.
+
Multiplying then subtracting gives us <cmath>4 \cdot 10 \cdot 111\dotsb 10888 \dotsb 89,</cmath> where there are <math>1984</math> ones and <math>1984</math> eights.
  
  
  
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Next, we begin to divide by <math> 9 </math>. We get this to be <math> 111\cdots110888\cdots890 </math>, where there are <math> 1984 </math> ones, <math> 1 </math> zero, <math> 1984 </math> eights, <math> 1 </math> nine, and a final zero.
+
Multiplying one last time, then adding the <math>0</math> at the end gives us <cmath>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>1984(4+5)+6+3 = 1985 \cdot 9 = \boxed{ \text{ C) 
−
 
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}17865}</cmath>
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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> ones, <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}}
 
{{MAA Notice}}

Latest revision as of 11:56, 26 March 2025

Problem

In their base $10$ representations, the integer $a$ consists of a sequence of $1985$ eights and the integer $b$ consists of a sequence of $1985$ fives. What is the sum of the digits of the base $10$ representation of the integer $9ab$?

$\mathrm{(A)\ } 15880 \qquad \mathrm{(B) \ }17856 \qquad \mathrm{(C) \ } 17865 \qquad \mathrm{(D) \ } 17874 \qquad \mathrm{(E) \ }19851$

Solution 1

By the formula for the sum of a geometric series, \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*} and similarly \[b = \frac{5\left(10^{1985}-1\right)}{9},\] so \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*}

We now compute the decimal expansion of this expression. Firstly, $10^{3971} = 100 \dotsb 0$, with $1$ one and $3971$ zeroes, and $2 \cdot 10^{1986} = 200 \dotsb 0$, with $1$ two and $1986$ zeroes. Subtracting therefore gives \[10^{3971}-2 \cdot 10^{1986} = 99 \dotsb 9800 \dotsb 0,\] where there are $3971-1986-1 = 1984$ nines followed by $1$ eight and then $1986$ zeroes. Adding $10$ transforms this to $99 \dotsb 9800 \dotsb 010$, now with $1984$ nines followed by $1$ eight, $1984$ zeroes, $1$ one, and a final zero.

Using long division, and noting that $80 = 8 \cdot 9 + 8$ and $81 = 9 \cdot 9$, it follows that \[\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 11 \dotsb 1088 \dotsb 890,\] with $1984$ ones, $1$ zero, then $1984$ eights, $1$ nine, and a final zero. Lastly, using long multiplication and noting that $9 \cdot 4 = 36$, $8 \cdot 4 = 32$, and $8 \cdot 4 + 3 = 35$, we obtain \[\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 44 \dotsb 4355 \dotsb 560,\] where there are $1984$ fours, $1$ three, $1984$ fives, $1$ six, and a final zero, so the sum of the digits is \begin{align*}1984 \cdot 4 + 3 + 1984 \cdot 5 + 6 + 0 &= 1984 \cdot 9 + 9 \\ &= 1985 \cdot 9 \\ &= \boxed{\text{(C)} \ 17865}.\end{align*}

Solution 2

Factoring out the 8 and the 5, we get \[9ab = 10\cdot4\cdot9\cdot111 \dotsb 1 \cdot 111\dotsb1,\] where there are $1985$ ones in both numbers.


This is equal to \[10 \cdot 4 \cdot (1000\dotsb 0 - 1) \cdot 111\dotsb 1\] where there is an initial $1$ followed by $1985$ zeros and the second number has $1985$ ones.


Multiplying then subtracting gives us \[4 \cdot 10 \cdot 111\dotsb 10888 \dotsb 89,\] where there are $1984$ ones and $1984$ eights.


Multiplying one last time, then adding the $0$ at the end gives us \[44 \dotsb 4355 \dotsb 560,\] where there are $1984$ fours, $1$ three, $1984$ fives, $1$ six, and a final zero, so the sum of the digits is \[1984(4+5)+6+3 = 1985 \cdot 9 = \boxed{ \text{ C) }17865}\]

See Also

1985 AHSME (Problems • Answer Key • Resources)
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Problem 28 		Followed by
Problem 30
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