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Difference between revisions of "2021 AIME I Problems/Problem 3"

m (→‎Solution 3 (Bash): a -> one)
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==Solution 1==
 
==Solution 1==
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We want to find the number of positive integers <math>n<1000</math> which can be written in the form <math>n = 2^a - 2^b</math> for some non-negative integers <math>a > b \ge 0</math> (note that if <math>a=b</math>, then <math>2^a-2^b = 0</math>). We first observe <math>a</math> must be at most 10; if <math>a \ge 11</math>, then <math>2^a - 2^b \ge 2^{10} > 1000</math>. As <math>2^{10} = 1024 \approx 1000</math>, we can first choose two different numbers <math>a > b</math> from the set <math>\{0,1,2,\ldots,10\}</math> in <math>\binom{10}{2}=55</math> ways. This includes <math>(a,b) = (10,0)</math>, <math>(10,1)</math>, <math>(10,2)</math>, <math>(10,3)</math>, <math>(10,4)</math> which are invalid as <math>2^a - 2^b > 1000</math> in this case. For all other choices <math>a</math> and <math>b</math>, the value of <math>2^a - 2^b</math> is less than 1000.
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We want to find the number of positive integers <math>n<1000</math> which can be written in the form <math>n = 2^a - 2^b</math> for some non-negative integers <math>a > b \ge 0</math> (note that if <math>a=b</math>, then <math>2^a-2^b = 0</math>). We first observe <math>a</math> must be at most 10; if <math>a \ge 11</math>, then <math>2^a - 2^b \ge 2^{10} > 1000</math>. As <math>2^{10} = 1024 \approx 1000</math>, we can first choose two different numbers <math>a > b</math> from the set <math>\{0,1,2,\ldots,10\}</math> in <math>\binom{11}{2}=55</math> ways. This includes <math>(a,b) = (10,0)</math>, <math>(10,1)</math>, <math>(10,2)</math>, <math>(10,3)</math>, <math>(10,4)</math> which are invalid as <math>2^a - 2^b > 1000</math> in this case. For all other choices <math>a</math> and <math>b</math>, the value of <math>2^a - 2^b</math> is less than 1000.
  
 
We claim that for all other choices of <math>a</math> and <math>b</math>, the values of <math>2^a - 2^b</math> are pairwise distinct. More specifically, if <math>(a_1,b_1) \neq (a_2,b_2)</math> where <math>10 \ge a_1 > b_1 \ge 0</math> and <math>10 \ge a_2 > b_2 \ge 0</math>, we must show that <math>2^{a_1}-2^{b_1} \neq 2^{a_2} - 2^{b_2}</math>. Suppose otherwise for sake of contradiction; rearranging yields <math>2^{a_1}+2^{b_2} = 2^{a_2}+2^{b_1}</math>. We use the fact that every positive integer has a unique binary representation:
 
We claim that for all other choices of <math>a</math> and <math>b</math>, the values of <math>2^a - 2^b</math> are pairwise distinct. More specifically, if <math>(a_1,b_1) \neq (a_2,b_2)</math> where <math>10 \ge a_1 > b_1 \ge 0</math> and <math>10 \ge a_2 > b_2 \ge 0</math>, we must show that <math>2^{a_1}-2^{b_1} \neq 2^{a_2} - 2^{b_2}</math>. Suppose otherwise for sake of contradiction; rearranging yields <math>2^{a_1}+2^{b_2} = 2^{a_2}+2^{b_1}</math>. We use the fact that every positive integer has a unique binary representation:
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If <math>a_1 = b_2</math> then <math>2^{a_1}+2^{b_2} = 2 \times 2^{a_1}</math>, and it follows that <math>a_1=a_2=b_1=b_2</math>, also contradicting the assumption <math>(a_1,b_1) \neq (a_2,b_2)</math>. Hence we obtain contradiction.
 
If <math>a_1 = b_2</math> then <math>2^{a_1}+2^{b_2} = 2 \times 2^{a_1}</math>, and it follows that <math>a_1=a_2=b_1=b_2</math>, also contradicting the assumption <math>(a_1,b_1) \neq (a_2,b_2)</math>. Hence we obtain contradiction.
  
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Then there are <math>\binom{10}{2}-5</math> choices for <math>(a,b)</math> for which <math>2^a - 2^b</math> is a positive integer less than 1000; by the above claim, each choice of <math>(a,b)</math> results in a different positive integer <math>n</math>. Then there are <math>55-5 = \boxed{050}</math> integers which can be expressed as a difference of two powers of 2.
+
Then there are <math>\binom{11}{2}-5</math> choices for <math>(a,b)</math> for which <math>2^a - 2^b</math> is a positive integer less than 1000; by the above claim, each choice of <math>(a,b)</math> results in a different positive integer <math>n</math>. Then there are <math>55-5 = \boxed{050}</math> integers which can be expressed as a difference of two powers of 2.
  
 
==Solution 2 (Casework)==
 
==Solution 2 (Casework)==

Revision as of 19:46, 15 March 2021

Problem

Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$

Solution 1

We want to find the number of positive integers $n<1000$ which can be written in the form $n = 2^a - 2^b$ for some non-negative integers $a > b \ge 0$ (note that if $a=b$, then $2^a-2^b = 0$). We first observe $a$ must be at most 10; if $a \ge 11$, then $2^a - 2^b \ge 2^{10} > 1000$. As $2^{10} = 1024 \approx 1000$, we can first choose two different numbers $a > b$ from the set $\{0,1,2,\ldots,10\}$ in $\binom{11}{2}=55$ ways. This includes $(a,b) = (10,0)$, $(10,1)$, $(10,2)$, $(10,3)$, $(10,4)$ which are invalid as $2^a - 2^b > 1000$ in this case. For all other choices $a$ and $b$, the value of $2^a - 2^b$ is less than 1000.

We claim that for all other choices of $a$ and $b$, the values of $2^a - 2^b$ are pairwise distinct. More specifically, if $(a_1,b_1) \neq (a_2,b_2)$ where $10 \ge a_1 > b_1 \ge 0$ and $10 \ge a_2 > b_2 \ge 0$, we must show that $2^{a_1}-2^{b_1} \neq 2^{a_2} - 2^{b_2}$. Suppose otherwise for sake of contradiction; rearranging yields $2^{a_1}+2^{b_2} = 2^{a_2}+2^{b_1}$. We use the fact that every positive integer has a unique binary representation:

If $a_1 \neq b_2$ then $\{a_1,b_2\} = \{a_2,b_1\}$; from here we can deduce either $a_1=a_2$ and $b_1=b_2$ (contradicting the assumption that $(a_1,b_1) \neq (a_2,b_2)$, or $a_1=b_1$ and $a_2=b_2$ (contradicting the assumption $a_1>b_1$ and $a_2>b_2$).

If $a_1 = b_2$ then $2^{a_1}+2^{b_2} = 2 \times 2^{a_1}$, and it follows that $a_1=a_2=b_1=b_2$, also contradicting the assumption $(a_1,b_1) \neq (a_2,b_2)$. Hence we obtain contradiction.

Then there are $\binom{11}{2}-5$ choices for $(a,b)$ for which $2^a - 2^b$ is a positive integer less than 1000; by the above claim, each choice of $(a,b)$ results in a different positive integer $n$. Then there are $55-5 = \boxed{050}$ integers which can be expressed as a difference of two powers of 2.

Solution 2 (Casework)

Case 1: When our answer is in the form $2^n-2^i$, where $i$ is an integer such that $0\le i\le 4$.

We start with the subcase where it is $2^n-2^0$, for some integer $n$ where $n>0$ (this is because the case where $n=0$ yields $2^0-2^0=0$, which doesn't work because it must be a positive integer.) Note that $2^{10}=1024$, and $2^9=512$. Our answer needs to be less than $1000$, so the maximum possible result (in this case) is $2^9-2^0$. Our lowest result is $2^1-2^0$. All the positive powers of two less than $1024$ work, so we have $9$ possibilities for this subcase. For subcases $i=1, i=2, i=3,$ and $i=4$, we have $8, 7, 6,$ and $5$ possibilities, respectively.

Case 2: When our answer is in the form of $2^n-2^j$, where $j$ is an integer such that $5\le j\le 9$.

We can start with the subcase where $j=5$. We notice that $2^5=32$, and $2^{10}-2^5=992$ which is less than $1000$, so the greatest result in this subcase is actually $2^{10}-2^5$, and the lowest is $2^6-2^5$. Thus, we have $5$ possibilities. For the other four subcases, we have $4, 3, 2,$ and $1$ possibilities, respectively.

Answer: We note that these are our only cases, as numbers in the form of $2^n-2^{10}$ and beyond are greater than $1000$.

Thus, our result is $9+8+7+6+5+5+4+3+2+1=(9+8+7+6+5+4+3+2+1)+5=\boxed{50}$. ~jehu26

Solution 3 (Bash)

We look for all positive integers of the form $2^a-2^b<1000,$ where $0\leq b<a.$ Performing casework on $a,$ we can enumerate all possibilities in the table below: \[\begin{array}{c|c} \boldsymbol{a} & \boldsymbol{b} \\ \hline & \\ [-2ex] 1 & 0 \\ 2 & 0,1 \\ 3 & 0,1,2 \\ 4 & 0,1,2,3 \\ 5 & 0,1,2,3,4 \\ 6 & 0,1,2,3,4,5 \\ 7 & 0,1,2,3,4,5,6 \\ 8 & 0,1,2,3,4,5,6,7 \\ 9 & 0,1,2,3,4,5,6,7,8 \\ 10 & \xcancel{0},\xcancel{1},\xcancel{2},\xcancel{3},\xcancel{4},5,6,7,8,9 \end{array}\] As indicated by the X-marks, the ordered pairs $(a,b)=(10,0),(10,1),(10,2),(10,3),(10,4)$ generate $2^a-2^b>1000,$ which are invalid.

Note that each of the remaining ordered pairs generates one unique desired positive integer. We prove as follows:

  1. The positive integers generated for each value of $a$ are clearly different.
  2. For all integers $1\leq k\leq9,$ the largest positive integer generated for $a=k$ is $1$ less than the smallest positive integer generated for $a=k+1.$

Together, we justified our claim in bold. Our answer is \[1+2+3+4+5+6+7+8+9+5=\boxed{050}.\]

~MRENTHUSIASM

Solution 4

First, you need to notice that it is impossible to have overlapping, making the problem easier.

Case 1 : $2^0$ There are $9$ ways here, from $2^1$ to $2^9$

It is easy to see here that this continues all the way down to one. However, when the case gets to $32$, there are 5 ways instead of 4 because $2^{10}-2^5$ is smaller than 1000.

Thus, $9+8+7+6+5+5+4+3+2+1 = 50$

So the answer is $\boxed{050}$

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=H17E9n2nIyY&t=569s

Video Solution

https://youtu.be/M3DsERqhiDk?t=749

See also

2021 AIME I (Problems • Answer Key • Resources)
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Problem 2 		Followed by
Problem 4
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All AIME Problems and Solutions

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