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Difference between revisions of "1993 AHSME Problems/Problem 7"

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Therefore <math>\frac{R_{24}}{R_4} = \frac{ 10^{24}-1 }{10^4-1}</math>.   
 
Therefore <math>\frac{R_{24}}{R_4} = \frac{ 10^{24}-1 }{10^4-1}</math>.   
  
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But we can recognize this form as the sum of a geometric series <math>1+10^4 + (10^4)^2 + \dots + (10^4)^5 = 1+ 10^4 + 10^8 + \dots + 10^{20}</math>
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We can recognize this is also the formula for the sum of a geometric series <math>1+10^4 + (10^4)^2 + \dots + (10^4)^5 = 1+ 10^4 + 10^8 + \dots + 10^{20}</math>
  
 
Now the 1's place has a 1, but the 10's, 100's and 1,000's place have 0's.  The 10,000's place has a 1, but the <math>10^5</math>, <math>10^6</math> and <math>10^7</math> places have 0's.  Between successive 1's in the decimal expansion, there are three 0's, which gives <math>5\times 3=15</math> zeros altogether.
 
Now the 1's place has a 1, but the 10's, 100's and 1,000's place have 0's.  The 10,000's place has a 1, but the <math>10^5</math>, <math>10^6</math> and <math>10^7</math> places have 0's.  Between successive 1's in the decimal expansion, there are three 0's, which gives <math>5\times 3=15</math> zeros altogether.

Latest revision as of 21:28, 27 May 2021

Problem

The symbol $R_k$ stands for an integer whose base-ten representation is a sequence of $k$ ones. For example, $R_3=111,R_5=11111$, etc. When $R_{24}$ is divided by $R_4$, the quotient $Q=R_{24}/R_4$ is an integer whose base-ten representation is a sequence containing only ones and zeroes. The number of zeros in $Q$ is:

$\text{(A) } 10\quad \text{(B) } 11\quad \text{(C) } 12\quad \text{(D) } 13\quad \text{(E) } 15$

Solution

Note $R_n = \sum_{k=0}^{n-1} 10^k = \frac{10^n - 1}{10-1}$.

Therefore $\frac{R_{24}}{R_4} = \frac{ 10^{24}-1 }{10^4-1}$.

We can recognize this is also the formula for the sum of a geometric series $1+10^4 + (10^4)^2 + \dots + (10^4)^5 = 1+ 10^4 + 10^8 + \dots + 10^{20}$

Now the 1's place has a 1, but the 10's, 100's and 1,000's place have 0's. The 10,000's place has a 1, but the $10^5$, $10^6$ and $10^7$ places have 0's. Between successive 1's in the decimal expansion, there are three 0's, which gives $5\times 3=15$ zeros altogether.

The answer is $\fbox{E}$

See also

1993 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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

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