Question

“Think of a 3-digit number, add 7 to it, then double it, subtract 4 and then divide it
by 2. Now, subtract the original number from this. You will be left with 5. Can you
explain how this trick works?​

Answer 1

Hey Pretty Stranger!

okay.. so in my childhood period I've been fooled by this trick several times. After doing hard struggle of such a long Calculation, I used to get that answer and I was like "Dude, you're a magician, How did you do that? imma be your fan now" ;-; How innocent I was or we were xD

Well, lem'me show you today how it actually happens. There's all MATH behind it.

Let's construct a 3 digit number where

• a is the hundreds digit

• b is the tenth digit and

• c is the ones digit

So, the number would be 100a + 10b + c

Add 7 to it

$\sf \: 100a + 10b + c + 7$

Double it

$\sf \: 2(100a + 10b + c + 7)$

$\sf \: 200a + 20b + 2c + 14$

Subtract 4

$\sf \: 200a + 20b + 2c + 14 - 4$

$\sf \: 200a + 20b + 2c + 10$

Divide it by 2

$\sf \: \dfrac{200a + 20b + 2c + 10}{2}$

$\sf \: 100a + 10b + c + 5$

Subtract the original number from it

$\sf \: 100a + 10b + c +5 - (100a + 10b + c)$

$\sf \:100a + 10b + c + 5- 100a - 10b - c$

$\sf \: \huge \: \: 5$

We're left with 5, aren't we?

So, basically choose what'ver, you'll be left with 5.

yeah, I'm a grown up kid now xD I don't get fooled easily. Thanks to Math ;)