In last week’s show it was said that the answer given (4-3+5^2=26) was the only possible answer. I came up with (3/2+5)x4=26. Am I math illiterate or are there two answers?

- Jeff

Cant someone find the time to post the question not just the answer?

You found another. Oldschool, the question was to arrange the numbers 2,3,4,and 5 so that the answer is 26.

There are many correct answers, not just one. For example:

26 = 5^2 + 4 - 3 << per tom and ray

26 = (3/2 + 5)*4 << per W Jeff S

26 = 24 + 5 - 3 << assuming “24” is an allowed use of “2” and "4"

26 = 25 + 4 - 3 << same issue

26 = 4! + 2

<< assuming “not using each number more than once”

<< allows one or more of the digits to be omitted

26 = 4! + 5 - (3 2)

The 4! is 4 factorial.

The (3 2) is the binomial coefficient (n m) with n=3 and m=2.

I suspect there are other solutions as well.

Here are a few more solutions:

det |4 -2| = 26

|3 5|

det | 5 3| = 26

|-2 4|

<< The determinant of each 2x2 matrix is equal to 26.

<< The matrix members can be re-arranged several ways

<< with the determinant = 26.

Re{(5 - 3i)*(4 + 2i)} = 26
Re{(4 - 2i)*(5 + 3i)} = 26

<< This is similar to the matrix determinant.

<< The real and imaginary values of the two complex numbers

<< can be re-arranged several ways with the real part

<< of the complex product = 26.

and along the lines of the determinant solutions, there is the 2-D vector inner product (5,2)*(4,3) and its permutations.

If you can omit a number, then 3! + 4*5 works.

i got some more too, and they use up all the numbers and don’t create numbers by putting one in the tens place and another in the ones place:

Simple:

(4/2)^5-3!

3!*5-sqrt(2^4)

3!*5-sqrt(4^2)

3!*5-2*sqrt(4)

Involving the binomial coefficient or choose function

5*4+(3 choose 2)!

Involving binomial coefficent and modular arithmetic (n mod k)= the remainder when n is divided by k

4! + sqrt(2*(5 mod 3))

4! + sqrt(2^(5 mod 3))

I love the one below with complex arithemtic, but that requires introducing i, which is not an operation and so I think doesn’t count!

3!

5-2sqrt(4)

uses two * operations. I made this same mistake several times.

e.g. 5!/4 - 3! + 2 = 26, but this uses two factorial operations.

I know that using i may be questionable. My argument for using i is that it is no different from using + or - to locate a real number on the real number line. Complex numbers are typically represented as points in an x-y coordinate system. The +i and -i designation for the imaginary part shows where the y-component is located in the complex plane, and the + and - designation for the real part shows where the x-component is located in the complex plane.

It’s interesting to see all these alternates. As I say above, I suspect there are many more.

Here are some others…

In examples 2-4, the last term is the absolute value of the square root of 4.

```
If X > or = 0, then |X| = X
If X < 0, then |X| = -X
```

I re-read the question, and only example 4 meets the criteria.