[bonanova]

Here's a little party trick.

A friend silently chooses three consecutive numbers [say 49, 50, 51] none greater than 60.

He then announces a multiple of 3 [say 54] that is less than 100.

Finally he multiplies their sum [204] by 67 [13668] and announces the last two digits [68].

You pause for a moment to think, then you tell him his starting number [49] and the remaining digits [1, 3, and 6].

Summarizing: he tells you 54 and 68; you tell him 49, 1, 3, and 6.

Wow, he gasps. How did you do that?

A real magician never tells. But you can tell us, in a spoiler.

[EventHorizon]

Here's a little party trick.

A friend silently chooses three consecutive numbers [say 49, 50, 51] none greater than 60.

He then announces a multiple of 3 [say 54] that is less than 100.

Finally he multiplies their sum [204] by 67 [13668] and announces the last two digits [68].

You pause for a moment to think, then you tell him his starting number [49] and the remaining digits [1, 3, and 6].

Summarizing: he tells you 54 and 68; you tell him 49, 1, 3, and 6.

Wow, he gasps. How did you do that?

A real magician never tells. But you can tell us, in a spoiler.

The sum of the three consecutive numbers will be 3x+3, where x is the first number.

He chooses a multiple of 3.....lets call it 3y.

3y=54, so y = 18

The sum is 3x+3+3y. Multiplying by 67 results in 67 (3x+3+3y) = 201 (x+1+y). Since x<=58 and y <=33, x+1+y < 100. Multiplying by 201 leaves the last two digits alone. He announced 3y, so you know y. This lets you determine x.

x = 68 - (y+1) = 68 - 19 = 49

Once you know x, you can determine 201(x+1+y), and list off the remaining digits.

201(x+1+y)=201(68) = 13668

Or more simply (and even without knowing x or y), the remaining digits are 2 times the last two digits 136=2*68

[bonanova]

Yeah.

[Guest]

I don't think that trick would go well at the parties I go to. Good one though.