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I was reading about Friedman numbers a few days ago. They are numbers where you can place mathematical symbols around the digits to create an expression whose value is equal to the original number.

For example, 127 is a Friedman number because you can write:

-1 + 2^7 = 127.

16384 is another: 16^3 x 8 / √4 = 16384.

That one was easy because I knew 16384 is a power of 2. Other numbers on the list of puzzles I found were either easy because I knew something about the number (343 for example I know is a power of 7), fairly easy because I can factor them and get clues, or rather difficult because the numbers are unfamiliar.

That got me thinking about two-digit Friendman numbers. I thought they'd be a lot easier to find. Turns out they're pretty hard to find!

I have found six so far. I wonder if hubski can find more. I won't post any of my 6 right away. Maybe you'll find it as fun as me?

As a clue, I offer that the *factorial* and the *double factorial* funtions are useful. The factorial function is well-known. 4! for example = 4 x 3 x 2 x 1 = 24.

The double factorial is not well known but comes in very handy in this puzzle. Double factorial is like a "skip factorial", it is the product of every other integer equal to the number you're operating on down to 1 or 2. So for an even number n, n!! is the product of all the even numbers less than or equal to n. And for n odd, n!! is the product of all the odd numbers less than or equal to n. Examples are maybe better than notation, so for example:

8!! = 8 x 6 x 4 x 2 = 384

and 7!! = 7 x 5 x 3 x 1 = 105

I'd love to see what you find. I have two solutions for 48, and one of them is pretty crazy!

Here are my six solutions. flagamuffin found one of them - kudos!

15 = (1 x 5)!!

24 = (2+√4)!

24 = (√ ( 2 x 4!! ) )! (I'm pretty proud of that one!)

36 = 3 x 6!

46 = -√4 + 6!!

48 = ( (4! / 8)! ) !! (another tricky one!)

Four of my solutions make use of double factorial. 4!! = 8 and 6!! = 48.

I've worked on this on and off for a few days while I've been traveling -- a great airplane puzzle. And I've worked out a lot of strategies and smart ways to think about solutions. Key is not starting with a number and trying to make a solution, but to look at what operations are available and which ones will get near to the result you want.

By the way, if you don't like the double factorial (I hear sometimes from people that think it's cheating), it is a real function and I solved a problem some years ago dealing with probability that made use of this function, although I didn't know the function at the time.

It's not cheating, but I still don't like the double factorial. I had an inkling a lot of the other solutions made use of it, and since it's not something those of us who finished formal math for good by the age of 19 internalized, it's much harder to use in solutions that require intuitive leaps.

Nonetheless, best use of the quiz tag in a long time!!

I like your double exclamation point at the end of your comment!!

I'll argue it's not harder to use in solutions -- I gave an explanation in the puzzle to open up the possibility, and it's a pretty short list of operands that are useful:

4!! = 8

5!! = 15

6!! = 48

7!! = 105

after that, the numbers get too big to be useful. I didn't find a use for 5!! or 7!!, but certainly there are more solutions I haven't found!

I'm just too much of a damn stick in the mud, mike.

In sixth grade, we were given a homework assignment:

Create equations equal to every number from 1 to 50 using nothing but 4.

1 = 4/4

2 = 4/(4 plus 4)

3 = (4 plus 4 plus 4)/4

Not that bad, until you get to some of the weird ones, like 37. It took me about 28 hours to get all but seven of them, and then my family worked on it for another five or six hours. We still didn't get all of them.

Turns out nobody did. The teacher didn't even check the homework before assigning it. Since it was virtually impossible, she determined that it would be unfair to grade it. For those of us who actually tried, good job! For me, who got all but two... well, also good job. What do you want, extra credit or something?

I stopped doing homework for about four years.

Hope that bitch is dead.

If you were allowed to render 1 as 4/4, then...

That sort of annoying puzzle is simultaneously what I wish my math classes had been and what I am glad they weren't.

Seems like several commentators hate this kind of puzzle. I like them. Someone showed me this variation last week:

Set in symbols one the left side of the equals sign to make equations which equal six:

0 0 0 = 6

1 1 1 = 6

2 2 2 = 6

3 3 3 = 6

4 4 4 = 6

5 5 5 = 6

6 6 6 = 6

7 7 7 = 6

8 8 8 = 6

9 9 9 = 6

I found the 8s most difficult.

I found you can have as many exclamation points as you wish. This opens up lots of possibilities, and maybe is cause for headaches. A triple factorial multiplies every third factor, etc. Something like 9!!!!!!! multiplies every 7th number in the sequence down, so it is 9 x 2 = 18.

8!!!, for example is 8 x 5 x 2 = 80. Thus 10 more solutions are 8!!! ** 0 = 80, 8!!! ** 1 = 81, 8!!! + 2 = 82, etc.

Here's a similar idea I think:

**123789^2 + 561945^2 + 642864^2 = 242868^2 + 323787^2 + 761943^2**

Take the first, third and fifth digits of each number.

We have: 138^2 + 514^2 + 626^2 = 226^2 + 338^2 + 714^2

Take the second, fourth and sixth digits of each number.

We have: 279^2 + 695^2 + 484^2 = 488^2 + 277^2 + 693^2

Drop the middle two digits of each number.

We have: 1289^2 + 5645^2 + 6464^2 = 2468^2 + 3287^2 + 7643^2

Drop the middle two digits again.

We have: 19^2 + 55^2 + 64^2 = 28^2 + 37^2 + 73^2

We can drop the second and third digits.

We have: 1789^2 + 5945^2 + 6864^2 = 2868^2 + 3787^2 + 7943^2

and drop the second and third digits again.

We have: 19^2 + 55^2 + 64^2 = 28^2 + 37^2 + 73^2

Drop the leftmost digit of each number repeatedly ("beheading"):

23789^2 + 61945^2 + 42864^2 = 42868^2 + 23787^2 + 61943^2

3789^2 + 1945^2 + 2864^2 = 2868^2 + 3787^2 + 1943^2

789^2 + 945^2 + 864^2 = 868^2 + 787^2 + 943^2

89^2 + 45^2 + 64^2 = 68^2 + 87^2 + 43^2

9^2 + 5^2 + 4^2 = 8^2 + 7^2 + 3^2

Drop the rightmost digit of each number repeatedly ("curtailing"):

12378^2 + 56194^2 + 64286^2 = 24286^2 + 32378^2 + 76194^2

1237^2 + 5619^2 + 6428^2 = 2428^2 + 3237^2 + 6719^2

123^2 + 561^2 + 642^2 = 242^2 + 323^2 + 761^2

12^2 + 56^2 + 64^2 = 24^2 + 32^2 + 76^2

1^2 + 5^2 + 6^2 = 2^2 + 3^2 + 7^2

Drop the exponents from all of the preceding equations.

The equalities still hold!