2021 AIME I Problems/Problem 3

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 $\binom{10}{2}=55$ 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 $\binom{10}{2}-5$ 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: We start with the case 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 case.

Case 2: When our answer is in the form $2^n-2^i$ , where $i$ is an integer such that $1\le n\le 4$ . Using the same logic as Case 1, we have $8, 7, 6,$ and $5$ for for $i=1, i=2, i=3,$ and $i=4$ , respectively.

Case 3: When our answer is in the form $2^n-2^5$ . We notice that $2^5=32$ , and $2^{10}-2^5=992$ which is less than $1000$ , so the greatest result in this case is actually $2^{10}-2^5$ , and the lowest is $2^6-2^5$ . Thus, we have $5$ possibilities.

Case 4: When our answer is in the form of $2^n-2^j$ , where $j$ is an integer such that $6\le j\le 9$ . Using the same logic as Case 3, we have $4, 3, 2,$ and $1$ for these four subcases.

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 $9+8+7+6+5+4+3+2+1+5=\boxed{50}$ . ~jehu26

Video Solution by Punxsutawney Phil

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

2021 AIME I (Problems • Answer Key • Resources)
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2021 AIME I Problems/Problem 3

Contents

Problem

Solution 1

Solution 2 (Casework)

Video Solution by Punxsutawney Phil

See also