This is an original paper written by Uttkarsh Kohli.
(27/03/2020)
Introduction
This paper is going to be about discovering and figuring out the working behind the magical number 2997.
2997 is a number that can be arrived at by a very simple trick.
Take any positive number say ‘x y z’ multiply each digit by 111 and take the sum.
(i.e, x*111 + y*111 +z*111).
[Note: Does not have to be a 3 digit number]
Repeating these steps of multiplying and adding with all positive numbers will guarantee arrival at 2997 in a maximum of 4 tries.
Why does this happen?
There is no simple answer other than thinking it is magic.
But I am here to find out the real reason behind this interesting trick.
Consider the number 111, the number that will be multiplied to each digit.
111 is divisible by 3, it can be confirmed by taking the sum of all its digits and checking if the sum is divisible by 3.
1+1+1 = 3 which is the first multiple of 3.
Consider a number (say xyz)
Multiply each digit by 111 and take the sum,
The sum is x*111 + y*111 +z*111
Which can also be written as 111(x+y+z)
Thus making the sum divisible by 3.
For example in the case of 999:
(The only 3 digit number that will end up at 2997 in the first series of steps.)
Step 1: Multiply each digit by 111
999 999 999.
Step 2: Take the sum
999+999+999 = 2997.
For the rest of the numbers,
The sum may not add up to 2997 after the first try.
Let the sum 111(x+y+z)
from the first try be a new arbitrary number N.
Now, this number N will be divisible by 3.
Multiplying each digit of the number with 111 and taking the sum again would make the sum divisible by 9.
Let us consider an example: 486
Step 1: Multiply each digit by 111
444 888 666.
Step 2: Take the sum
444+888+666 = 1998.
Repeat Step 1 and Step 2:
111+999+999+888 = 2997.
There you have it.
For the rest of the numbers (which are not originally divisible by 3)
It is necessary to repeat the steps to add another factor of 3.
The trick is that a factor of 3 is added each time it is multiplied by 111, once the sum (x*111 + y*111 +z*111)
becomes 111*(27),
The number 2997 is obtained.
And when the sum reaches 999(The largest 3 digit number) or another number that adds up to 27(The third multiple of 9)
The result of multiplying the digits by 111 and taking the sum will be 2997.
Doing this continuously would guarantee an end at 2997.
The beautiful part is when each digit of the number 2997 itself is multiplied by 111 and the sum of the digits is taken, it adds up to 2997.
Why 2997?
After adding factors of 3, once the sum reaches 27, multiplying the sum by 111 gives 2997 and this is where the series stops because 2997 is the only number that has the magical property of adding up to itself following the steps of multiplying each digit by 111 and summing them up.
And for this reason if the sum reaches any other number the series will continue.
[NOTE: The number does not have a significance in complex mathematics but it serves as a recreational and entertaining fact which shows that math continues to surprise us with the patterns which are just waiting to be discovered.]
To show examples and simplify working only 3 digit numbers have been used above as examples but it works for all positive integers as shown below.
Examples :
1) 8
888
=888
888+888+888
=2664
222+666+666+444
=1998
111+999+999+888
=2997.
2) 29
222+999
=1221
111+222+222+111
=666
666+666+666
=1998
111+999+999+888
=2997.
3) 739
777+333+999
=2109
222+111+0+999
=1332
111+333+333+222
=999
999+999+999
=2997.
4) 6307
666+333+0+777
=1776
111+777+777+666
=2331
222+333+333+111
=999
999+999+999
=2997.
5) 39658
333+999+666+555+888
=3441
333+444+444+111
=1332
111+333+333+222
=999
999+999+999
=2997.
PROOF
Let x (an arbitrary variable) be the number with say n(another arbitrary variable) digits.
Step 1:
Multiply all digits by 111
111*a1 111*a2 ….111*an
Step 2:
Take the sum
111(a1+a2+....an)
If (a1+a2+..an) is equal to 27
Sum: 2997.
Else
Repeat the 2 steps thereby adding a factor of 3 each time.
3*37(a1+a2+....an)
Or 37*(3a1+3a2+....3an)
Continue until the sum(a1+a2+....an) becomes 27
Sum: 2997.