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

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\text{(E) } 15</math>
 
\text{(E) } 15</math>
  
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Solution  
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== Solution ==
 
Because the only digits we are using are 1s and zeroes (they say that in question) we essentially perform the operation in binary and convert back to base 10 to get the answer.
 
Because the only digits we are using are 1s and zeroes (they say that in question) we essentially perform the operation in binary and convert back to base 10 to get the answer.
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Notice then that <math>R(n)_2=(2^n-1)_{10}</math>. It follows that <math>R(24)=2^{24}-1</math> and <math>R(4)=2^4-1</math>
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Notice then that <math>R(n)_2=(2^n-1)_{10}</math>. It follows that <math>R(24)=2^{24}-1</math> and <math>R(4)=2^4-1</math>
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Notice to compute <math>\frac{2^{24}-1}{2^4-1}</math> we take advantage of the fact that
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Notice to compute <math>\frac{2^{24}-1}{2^4-1}</math> we take advantage of the fact that
 
<math>x^6-1=(x-1)(x^5+x^4+x^3+x^2+x+1)</math>  
 
<math>x^6-1=(x-1)(x^5+x^4+x^3+x^2+x+1)</math>  
 
Our quotient then is just
 
Our quotient then is just
 
<math>2^{20}+2^{16}+2^{12}+2^{8}+2^4+1</math>
 
<math>2^{20}+2^{16}+2^{12}+2^{8}+2^4+1</math>
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Notice then this just a <math>21</math> digit in binary with <math>5</math> <math>1</math>s which occupy the <math>2^a</math> slots for the <math>6</math> <math>a</math> we have.
+
Notice then this just a <math>21</math> digit in binary with <math>5</math> <math>1</math>s which occupy the <math>2^a</math> slots for the <math>6</math> <math>a</math> we have.
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Our answer then is just <math>21-6=\boxed{15}</math>
+
Our answer then is just <math>21-6=\boxed{15}</math>
 
<math>\fbox{E}</math>
 
<math>\fbox{E}</math>
  

Revision as of 10:30, 24 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

Because the only digits we are using are 1s and zeroes (they say that in question) we essentially perform the operation in binary and convert back to base 10 to get the answer. Notice then that $R(n)_2=(2^n-1)_{10}$. It follows that $R(24)=2^{24}-1$ and $R(4)=2^4-1$ Notice to compute $\frac{2^{24}-1}{2^4-1}$ we take advantage of the fact that $x^6-1=(x-1)(x^5+x^4+x^3+x^2+x+1)$ Our quotient then is just $2^{20}+2^{16}+2^{12}+2^{8}+2^4+1$ Notice then this just a $21$ digit in binary with $5$ $1$s which occupy the $2^a$ slots for the $6$ $a$ we have. Our answer then is just $21-6=\boxed{15}$ $\fbox{E}$

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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