Throw any three dice. Double the number on the first die, add 5 then multiply by 5, add the number on the second die, then multiply by 10, add the number on the third die, then subtract 250.

The digits of your result will always be the numbers on your three dice.

Private Sub Form_Load()
For die1 = 1 To 6
For die2 = 1 To 6
For die3 = 1 To 6
formula = ((((die1 * 2) + 5) * 5) + die2) * 10 + die3 - 250
Debug.Print die1 & die2 & die3 & " = " & formula
Next
Next
Next
End Sub

Yep... Much easier to multiply the die1 by 100 and die2 by 10 and add all three die to get the same result :D

Reminds me of a trick my grandad use to play on me when I was little.

Think of a number between 1 and 20.

Add 16 to your number. (If you are playing this trick on someone vary this number)

Divide the sum by 2.

Take away half the number you originally thought of.

The number in your head is 8.


There is a relationship between the two bold numbers - the result is always half of what you asks them to add. You can add more maths from then on to make it harder for them to figure it out because by now you have the number

I am quoting my mentor, Max, here

"And your point is?"

x

Actually, I misread the question and I wrote a program hoping to find the answer to a teaser.

Once I've seen the result, I gave a summary and....

...just like LouiseB said, "And your point is?"

Yep... Much easier to multiply the die1 by 100 and die2 by 10 and add all three die to get the same result
Actually, I misread the question and I wrote a program hoping to find the answer to a teaser.

Just like LouiseB said, "And your point is?"
- John

The point is in the title ie. dicey - two meanings.

And, of course it's easier to do as you say, as all 3-digit numbers are of the form 100a + 10b + c.

However, as a parlour trick, and that's all it is, then my version is a little-bit more 'amazing'.

No big deal - but then we'd be getting into card-tricks - wouldn't we.

Nevertheless, sorry for the downer!