7
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$n$ is an arbitrary real number. By only using multiplication, you are asked to find $n^{23}$ with the least amount of multiplication operation.

Note: You can only use $n$ or the results you can with multiplications as examplified below.

For example if this question is asked for $n^4$, the answer would be $2$:

1.

$n\times n=n^2$

2.

$n^2\times n^2=n^4$

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  • 2
    $\begingroup$ You could also make this an addition problem - how do you sum to 23 using only "1", "+", and the result of previous sums? $\endgroup$ – mbeckish Aug 25 '17 at 20:18
10
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1.

$n \times n = n^2$

2.

$n\times n^2 = n^3$

3.

$n^3\times n^2=n^5$

4.

$n^5\times n^5=n^{10}$

5.

$n^{10}\times n^{10}=n^{20}$

6.

$n^{20}\times n^{3}=n^{23}$

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  • 1
    $\begingroup$ Welcome to Puzzling I was looking for this! $\endgroup$ – Oray Aug 24 '17 at 12:36
0
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Well, it is

7 times.

  1.  $a^{2\phantom0} = a^\phantom{{00}} \times a$
  2.  $a^{3\phantom0} = a^{2\phantom0} \times a$
  3.  $a^{5\phantom0} = a^{3\phantom0} \times a^2$
  4.  $a^{6\phantom0} = a^{5\phantom0} \times a$
  5.  $a^{11} = a^{6\phantom0} \times a^5$
  6.  $a^{22} = a^{11} \times a^{11}$
  7.  $a^{23} = a^{22} \times a$

as for $a^{22}$ it took 6 multiplications and for $a^{23}$ it is one more.

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