# Source of probably-the-most-simplest-math-trick

I am just 18, completely naive and new to the world of maths. I have made a few math tricks based on this trick (or tricks of this format - with the exact calculations and steps involved):

TRICK:                                 SOLUTION:
STEP 1: Take a secret number.          Let it be denoted by ‘x’.
STEP 2: Multiply by 5.                 => 5x
STEP 3: Add 25.                        => 5x +25
STEP 4: Divide by 5.                   => (5x+25)/5 = x+5
STEP 5: Subtract your secret number.   => x+5-x= 5


I am thinking of writing a paper on what I have made, but even after extensive searching on net, Google, Google Scholar and also websites like jstor.org, I am not able to find any material related to this exact math trick.

Can someone help me out here? Is it just that this trick is age old, and may be comes under "Creative Commons" (if that makes sense)?

I was in 8th when I figured it out, I know that it can be done by 6th grader, but it's not about this trick, it's about the 20 odd tricks made by me (based on this trick), some of which involving trigonometry and log too. I am writing a paper and I think it would be great if I find a reference to quote for this trick.

• I am not sure I see the "trick" except algebra but what sort of paper are you talking about?Where do you intend to publish it and who is your target audience? Bill Dubuque has talked about the machinery involved though in 'your" "trick". – user31029 Jun 22 '12 at 14:34
• @SabyasachiMukherjee I was in 8th when I figured it out, I know that it can be done by 6th grader, but it's not about this trick, it's about the 20 odd tricks made by me (based on this trick), some of which involving trigonometry and log too. I am writing a paper and I think it would be great if I find a reference to quote for this trick. – user33984 Jun 22 '12 at 14:42
• @SabyasachiMukherjee As I just naive, I would just send it to all the journals I find, I do not have any target audience and I am not sure whether it would ever be accepted. But what's wrong in trying just for the fun of it? – user33984 Jun 22 '12 at 14:44
• Try this: take any three digit number. Multiply by 7, add 11, multiply by 13. Give me the result. I then multiply by 11 in my head take the number of thousands less 1 - check the number of units less 1, and deduct another 1 from the thousands if necessary - and I have your original number. Can you see how that would work? – Mark Bennet Jun 22 '12 at 15:25
• @MarkBennet From how I understand the instructions, the trick doesn't works, it would have worked if you had said number of thousands less 1, number of 10thousands and number of 100thousands - then you would have had my original number. – user33984 Jun 22 '12 at 16:02

The "trick" is just linearity: $\rm\ f^{-1}(f(x)\!+\!f(n))\, =\, x\!+\!n.\:$ Above $\rm\ n=5,\ f(x) = 5x,\ f^{-1}(x) = x/5.$