# 2 - 3 - 4 - 5 Puzzler - alternate answer

**URL:** <https://community.cartalk.com/t/2-3-4-5-puzzler-alternate-answer/31336>\
**Category:** The Show\
**Created:** [June 2, 2009, 5:31pm UTC](https://community.cartalk.com/t/2-3-4-5-puzzler-alternate-answer/31336 "2009-06-02T17:31:47Z")\
**Posts on this page:** 11\
**Page:** 1

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**Author:** ![W\_Jeff\_S](https://avatars.discourse-cdn.com/v4/letter/w/439d5e/32.png) [@W\_Jeff\_S](https://community.cartalk.com/u/W_Jeff_S)\
**Post date:** [June 2, 2009, 5:31pm UTC](https://community.cartalk.com/t/2-3-4-5-puzzler-alternate-answer/31336/1 "2009-06-02T17:31:47Z")

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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

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**Author:** ![oldschool](https://avatars.discourse-cdn.com/v4/letter/o/e19adc/32.png) [@oldschool](https://community.cartalk.com/u/oldschool)\
**Post date:** [June 2, 2009, 9:11pm UTC](https://community.cartalk.com/t/2-3-4-5-puzzler-alternate-answer/31336/2 "2009-06-02T21:11:43Z")

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

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**Author:** ![texases](https://sea1.discourse-cdn.com/cartalk/user_avatar/community.cartalk.com/texases/32/862_2.png) [@texases](https://community.cartalk.com/u/texases)\
**Post date:** [June 2, 2009, 9:14pm UTC](https://community.cartalk.com/t/2-3-4-5-puzzler-alternate-answer/31336/3 "2009-06-02T21:14:00Z")

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

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**Author:** ![george4](https://avatars.discourse-cdn.com/v4/letter/g/ac8455/32.png) [@george4](https://community.cartalk.com/u/george4)\
**Post date:** [June 3, 2009, 1:46am UTC](https://community.cartalk.com/t/2-3-4-5-puzzler-alternate-answer/31336/4 "2009-06-03T01:46:14Z")

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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.

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**Author:** ![george4](https://avatars.discourse-cdn.com/v4/letter/g/ac8455/32.png) [@george4](https://community.cartalk.com/u/george4)\
**Post date:** [June 4, 2009, 12:48pm UTC](https://community.cartalk.com/t/2-3-4-5-puzzler-alternate-answer/31336/5 "2009-06-04T12:48:53Z")

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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.

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**Author:** ![tbg2222](https://avatars.discourse-cdn.com/v4/letter/t/f0a364/32.png) [@tbg2222](https://community.cartalk.com/u/tbg2222)\
**Post date:** [June 6, 2009, 4:50pm UTC](https://community.cartalk.com/t/2-3-4-5-puzzler-alternate-answer/31336/6 "2009-06-06T16:50:19Z")

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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.

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**Author:** ![leoshmu](https://avatars.discourse-cdn.com/v4/letter/l/90ced4/32.png) [@leoshmu](https://community.cartalk.com/u/leoshmu)\
**Post date:** [June 6, 2009, 11:09pm UTC](https://community.cartalk.com/t/2-3-4-5-puzzler-alternate-answer/31336/7 "2009-06-06T23:09:21Z")

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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!

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**Author:** ![george4](https://avatars.discourse-cdn.com/v4/letter/g/ac8455/32.png) [@george4](https://community.cartalk.com/u/george4)\
**Post date:** [June 7, 2009, 11:41pm UTC](https://community.cartalk.com/t/2-3-4-5-puzzler-alternate-answer/31336/8 "2009-06-07T23:41:53Z")

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> > 3!_5-2_sqrt(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.

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**Author:** ![nhstagehand](https://avatars.discourse-cdn.com/v4/letter/n/a88e57/32.png) [@nhstagehand](https://community.cartalk.com/u/nhstagehand)\
**Post date:** [July 3, 2009, 3:37am UTC](https://community.cartalk.com/t/2-3-4-5-puzzler-alternate-answer/31336/9 "2009-07-03T03:37:40Z")

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Here are some others…

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**Author:** ![nhstagehand](https://avatars.discourse-cdn.com/v4/letter/n/a88e57/32.png) [@nhstagehand](https://community.cartalk.com/u/nhstagehand)\
**Post date:** [July 3, 2009, 3:45am UTC](https://community.cartalk.com/t/2-3-4-5-puzzler-alternate-answer/31336/10 "2009-07-03T03:45:53Z")

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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
```

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**Author:** ![nhstagehand](https://avatars.discourse-cdn.com/v4/letter/n/a88e57/32.png) [@nhstagehand](https://community.cartalk.com/u/nhstagehand)\
**Post date:** [July 3, 2009, 4:01am UTC](https://community.cartalk.com/t/2-3-4-5-puzzler-alternate-answer/31336/11 "2009-07-03T04:01:41Z")

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