Let us say you think up a 5 digit number , say, X=15683.

Add up the digits: Y=1+5+6+8+3=23

Subtract this from the original number : Z=X-Y=15683-23=15660.

Let us say, from Z we take out the number 5. So we have 1660.

The Problem is to find the missing number.

Now look carefully, here is the trick for finding it out. The number you see is 1660. Add them up and you get 1+6+6+0=**13**.

Now you know the multiplication table of ‘9’.

It is 9×1=9

9×2=18

9×3=27

etc.

Now we have the remaining sum=**13**. Ask yourself the question What is the number in the 9’s table which is closest to **13**? The answer is 18, right? So subtract 13 from 18, which is 18-13=5. MAGIC: That is the missing number. You got it.(Remember, I took out 5 from 15660 to get 1660.) NOTE: You can’t hide a 9 or 0

Now, Here is the challenge for you. The number I give you is = 122, tell me which number is missing?

If you have any doubts about the solution , don’t hesitate to contact me. Have a nice mathemagic. :))

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PS: Proof of the trick.Let us say a 4 digit number X=1000a+100b+10c+d. Then Y=a+b+c+d

Z=X-Y=999a+99b+9c=0(mod 9).

Which means that the sum of the digits of Z will also be divisible by 9[1]. So, all you have to do to find the missing digit is to get a number which makes the sum divisible by 9.

The proof can easily be extended to n-digit numbers.

PPS: Proof of [1]

Let us say a number X=0(mod 9). Say, X=1000a+100b+10c+d=9k.

X=(999a+99b+9c) + (a+b+c+d)=9k. Obviously (a+b+c+d) is divisible by 9 to make the LHS divisible by 9. Hence, the result. Again, this proof can be easily extended to n-digit numbers.

(Those who attended my talk–Here are the slides )

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