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Difference between revisions of "1997 AHSME Problems/Problem 30"

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==Problem==
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For positive integers <math>n</math>, denote <math>D(n)</math> by the number of pairs of different adjacent digits in the binary (base two) representation of <math>n</math>. For example, <math> D(3) = D(11_{2}) = 0 </math>, <math> D(21) = D(10101_{2}) = 4 </math>, and <math> D(97) = D(1100001_{2}) = 2 </math>. For how many positive integers less than or equal <math>97</math> to does <math>D(n) = 2</math>?
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<math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 </math>
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==Solution 1==
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If <math>D(n)</math> is even, then the binary expansion of <math>n</math> will both begin and end with a <math>1</math>, because all positive binary numbers begin with a <math>1</math>, and if you switch digits twice, you will have a <math>1</math> at the end.  Thus, we are only concerned with the <math>49</math> odd numbers between <math>1</math> and <math>98</math> inclusive.
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All of these odd numbers will have an even <math>D(n)</math>.  <math>D(n) = 0</math> will be given by the numbers <math>1, 11, 111, 1111, 11111, 111111</math>, which is a total of <math>6</math> numbers.
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We skip <math>D(n) = 2</math> for now, and move to <math>D(n) = 4</math>, which is easier to count.  The smallest <math>D(n) = 4</math> happens when <math>n = 10101</math>.  To get another number such that <math>D(n) = 4</math>, we may extend any of the five blocks of zeros or ones by one digit.  This will form <math>110101, 100101, 101101, 101001, 101011</math>, all of which are odd numbers that have <math>D(n) = 4</math>.  To find seven digit numbers that have <math>D(n) = 4</math>, we can again extend any block by one, so long as it remains less than <math>1100001</math> or under.  There are five cases.
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1)  Extending <math>110101</math> is impossible without going over <math>1100001</math>.
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2)  Extending <math>100101</math> by putting a <math>1</math> at the beginning will go over <math>1100001</math>, but the other four extensions work, giving <math>1000101, 1001101, 1001001, 1001011</math>.
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3)  Extending <math>101101</math> by putting a <math>1</math> at the beginning will go over <math>1100001</math>, but the other four extensions give <math>1001101, 1011101, 1011001, 1011011</math>.  However, <math>1001101</math> already appeared in #2, giving only three new numbers.
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4)  Extending <math>101001</math> at the first group is impossible.  The other four extensions are <math>1001001, 1011001, 1010001, 1010011</math>, but the first two are repeats.  Thus, there are only two new numbers.
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5)  Extending <math>101011</math> at the first group is impossible.  The other four extensions give <math>1001011, 1011011, 1010011, 1010111</math>, but only the last number is new.
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Thus, there is <math>1</math> five digit number, <math>5</math> six digit numbers, and <math>4 + 3 + 2 + 1 = 10</math> seven digit numbers under <math>1100001</math> for which <math>D(n) = 4</math>.  That gives a total of <math>16</math> numbers.
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There smallest number for which <math>D(n) = 6</math> is <math>1010101</math>, which is under <math>98</math>.  Further extensions, as well as cases where <math>D(n) > 6</math>, are not possible.
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Thus, we know that there are <math>6</math> odd numbers that have <math>D(n) = 0</math>, and <math>16</math> odd numbers that have <math>D(n) = 4</math>, and <math>1</math> number that has <math>D(n) = 6</math>.  The remaining odd numbers must have <math>D(n) = 2</math>.  This means there are <math>49 - 6 - 16 - 1 = 26</math> numbers that have <math>D(n) = 2</math>, which is option <math>\boxed{C}</math>
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==Solution 2==
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Each pair of different adjacent digits corresponds to a change from one a chain of 1s to a chain of 0s or vice versa. For example, in the number <math>111001</math> there are 2 such pairs and also 2 changes (three 1s change to two 0s, then change back to 1). Thus, in a number with 2 changes, there will be three "blocks" of consecutive digits. Since the number always starts with a 1, each valid number will have the form <cmath>1\dots0\dots1\dots</cmath>
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The total number of digits cannot exceed <math>7</math>. If the number has 6 digits in binary or less: this is equivalent to solving <math>a+b+c \le 6</math>, where <math>a, b, c</math> represent how many 1s, 0s, and 1s are in the three blocks where <math>a, b, c</math> are positive integers. Using stars and bars, we find that there are <math>\binom{6}{3}=20</math> such binary numbers.
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If the number has <math>7</math> digits: we see that there are <math>5</math> numbers for which the second digit is 0, and only 1 valid number for which the second digit is 1 (97 is 1100001 in binary). Thus, we have a total of <math>20+5+1=\fbox{(C) 26}</math> numbers.
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== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=29|after=Last Question}}
 
{{AHSME box|year=1997|num-b=29|after=Last Question}}
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{{MAA Notice}}

Revision as of 21:05, 1 August 2016

Problem

For positive integers $n$, denote $D(n)$ by the number of pairs of different adjacent digits in the binary (base two) representation of $n$. For example, $D(3) = D(11_{2}) = 0$, $D(21) = D(10101_{2}) = 4$, and $D(97) = D(1100001_{2}) = 2$. For how many positive integers less than or equal $97$ to does $D(n) = 2$?

$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$

Solution 1

If $D(n)$ is even, then the binary expansion of $n$ will both begin and end with a $1$, because all positive binary numbers begin with a $1$, and if you switch digits twice, you will have a $1$ at the end. Thus, we are only concerned with the $49$ odd numbers between $1$ and $98$ inclusive.

All of these odd numbers will have an even $D(n)$. $D(n) = 0$ will be given by the numbers $1, 11, 111, 1111, 11111, 111111$, which is a total of $6$ numbers.

We skip $D(n) = 2$ for now, and move to $D(n) = 4$, which is easier to count. The smallest $D(n) = 4$ happens when $n = 10101$. To get another number such that $D(n) = 4$, we may extend any of the five blocks of zeros or ones by one digit. This will form $110101, 100101, 101101, 101001, 101011$, all of which are odd numbers that have $D(n) = 4$. To find seven digit numbers that have $D(n) = 4$, we can again extend any block by one, so long as it remains less than $1100001$ or under. There are five cases.

1) Extending $110101$ is impossible without going over $1100001$.

2) Extending $100101$ by putting a $1$ at the beginning will go over $1100001$, but the other four extensions work, giving $1000101, 1001101, 1001001, 1001011$.

3) Extending $101101$ by putting a $1$ at the beginning will go over $1100001$, but the other four extensions give $1001101, 1011101, 1011001, 1011011$. However, $1001101$ already appeared in #2, giving only three new numbers.

4) Extending $101001$ at the first group is impossible. The other four extensions are $1001001, 1011001, 1010001, 1010011$, but the first two are repeats. Thus, there are only two new numbers.

5) Extending $101011$ at the first group is impossible. The other four extensions give $1001011, 1011011, 1010011, 1010111$, but only the last number is new.

Thus, there is $1$ five digit number, $5$ six digit numbers, and $4 + 3 + 2 + 1 = 10$ seven digit numbers under $1100001$ for which $D(n) = 4$. That gives a total of $16$ numbers.

There smallest number for which $D(n) = 6$ is $1010101$, which is under $98$. Further extensions, as well as cases where $D(n) > 6$, are not possible.

Thus, we know that there are $6$ odd numbers that have $D(n) = 0$, and $16$ odd numbers that have $D(n) = 4$, and $1$ number that has $D(n) = 6$. The remaining odd numbers must have $D(n) = 2$. This means there are $49 - 6 - 16 - 1 = 26$ numbers that have $D(n) = 2$, which is option $\boxed{C}$

Solution 2

Each pair of different adjacent digits corresponds to a change from one a chain of 1s to a chain of 0s or vice versa. For example, in the number $111001$ there are 2 such pairs and also 2 changes (three 1s change to two 0s, then change back to 1). Thus, in a number with 2 changes, there will be three "blocks" of consecutive digits. Since the number always starts with a 1, each valid number will have the form \[1\dots0\dots1\dots\]

The total number of digits cannot exceed $7$. If the number has 6 digits in binary or less: this is equivalent to solving $a+b+c \le 6$, where $a, b, c$ represent how many 1s, 0s, and 1s are in the three blocks where $a, b, c$ are positive integers. Using stars and bars, we find that there are $\binom{6}{3}=20$ such binary numbers.

If the number has $7$ digits: we see that there are $5$ numbers for which the second digit is 0, and only 1 valid number for which the second digit is 1 (97 is 1100001 in binary). Thus, we have a total of $20+5+1=\fbox{(C) 26}$ numbers.

See also

1997 AHSME (Problems • Answer Key • Resources)
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