Cause we're young and we're reckless

Logic Level 3

$2^0 \, \square \, 2^1 \, \square \, 2^2 \, \square \, \ldots \, \square \, 2^n = 1991$

For positive integer $n$ , there are $2^n$ ways in which we can fill the squares with $+, -$ . To make the equation true, how many of these squares are addition signs?

The answer is 7.

4 solutions

Brian Charlesworth
Aug 20, 2015

Note first that the least possible value of $n$ is $10,$ since $\sum_{k=0}^{9} 2^{k} = 1023.$

Now since $\sum_{k=0}^{10} 2^{k} = 2047$ differs from $1991$ by $56,$ we need to switch the addition signs in front of those powers of $2$ that sum to $56 \div 2 = 28$ to subtraction signs in order to satisfy the given equation. Since $28 = 2^{2} + 2^{3} + 2^{4},$ we need to switch a total of $3$ addition signs, leaving us with $10 - 3 = 7$ addition signs.

Now for $n \gt 10,$ we will need to switch the same addition signs as above, as well as those from $2^{10}$ to $2^{n-1},$ with the addition sign before $2^{n}$ left intact, in order to satisfy the given equation. This changes the number of subtraction signs but leaves the number of addition signs at $7.$

So for any $n \ge 10,$ the number of addition signs is $\boxed{7}.$

How can u assume that this is the only possible case where u can arrive at the desired 1991 as the solution.

Harish Kumaar - 5 years, 9 months ago

We can only get from $2047$ to $1991$ by subtracting $56 = 2*28.$ All we have to work with are powers of $2,$ and when we switch the addition signs to subtraction signs we in effect subtract twice the chosen powers of $2.$ So we need to find those powers of $2$ that add to $28,$ and $2^{2} + 2^{3} + 2^{4}$ is the unique representation of $28$ in this format, i.e., it is the only possible case where we can arrive at $1991.$

If we have $n = 11,$ if we have all addition signs then we start at $4095,$ and hence have to subtract $2104$ to get to $1991.$ Due to the aforementioned "doubling" effect, this means we have to switch addition signs on powers of $2$ totaling to $1052,$ the unique representation of which in this format being $2^{2} + 2^{3} + 2^{4} + 2^{10}.$ Similarly for any other $n,$ there will be unique representation to achieve the desired sum of $1991,$ and in every case, the number of addition signs will be $7.$

Brian Charlesworth - 5 years, 9 months ago

Andrey Bessonov
Aug 21, 2015

1991 = 11111000111 (binary).

The number of pluses is equal to number of 1 in binary form.

Сouse all subsequence like 0001 we can make with 2^n-2^(n-1)-...-2^k

So, we have 8 pluses, but there no operation before 0 power. Answer = 8-1 = 7

Mohamed Hassan
Aug 23, 2015

1+2+4+8+16+32+64+128+256+512+1024=2047

2047-1991=56

56/2=28

4+8+16=28

1+2-4-8-16+32+64+128+256+512+1024

Jorge Fernández
Aug 20, 2015

1+2-4-8-16+32+64+128+256+512+1024

Cause we're young and we're reckless

The answer is 7.

4 solutions

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