Math Is Fun Forum

Hannah,
Mathematics is fun when you try to understand the subject.
Trying to memorize the tables and the formulae may be difficult for some.
Yet, it is easier to get a good score in Mathematics than any other subject.
For example, When you forget what 7x7 is,
try to remember what 7x6 is, then add 7 to the result.
You can get doubts clarified here.

Maybe, I should have waited for some more days
before giving the solution.

I am unable to find any serious flaw, yet,
I am not fully convinced.
I shall try with a much simpler number, viz. y76.
y76 is (y*10^2)+(7*10^1)+(6*10^0).
When this number is squared,
[This is of the form (a+b+c)²
which is equal to a² + b²  + c² + 2ab + 2bc + 2ac]
we get
10000y²  + 4900 + 36 + 14000y + 840 + 1200y
which is the same as
10000y² + 15200y + 5776
From this, it is clear, for any y belonging to Natural Numbers,
the last two digits are not affected. They continue to be 76.
But, does this conclusively prove there exists a value y such that
the last three digits of y76² are y76?

Mathsy, I am confused.

Mathsy, how did you get '2(y*(6*10^0))+a constant made up of the expansion of the other powers of 10, but that does not involve y'?

Both x! and x^y work, but for larger values, as you rightly pointed out, it becomes very slow.
However, a useful tool.

42576576769103890995893380022607743740081787109376

Mathsy, did you get this number too by exhaustion, just like you got the 30 digits????:)

I'll give my opinion tomorrow on Mathsy's proof, when I start the day afresh;
Regarding including n!, it would be of immense help as combinatorics works mostly on
factorials;
nPr= n!/(n-r)!
and
nCr = n!/(n-r)!*r!
nPr gives the number of Permutations
and
nCr gives the number of combinations.
There is an approximation for factorials of higher numbers;
it is called the James-Stirling formula
n! is approximately equal to
√(2*Pi*n) multiplied by (n/e)^n
This is a brilliant approximation as the value of n goes up.

The layout is fine. Would it be possible to add 2^n? Thats the most important of powers.
And also n! if possible.

And Mathsy, its difficult to prove or disprove that there would always be a number in the sequence. But I strongly feel, the sequence would go on and on.

Wow! Great job by you two!
I got to really thank both of you, Admin for putting the calculator on the site,
and Mathsy for making my job pretty easy!
I had 17 digits with me already, now you have given me 30!
I know, Mathsy, you have done a lot of homework....
Testing for all numbers 0-9 for twenty digits is a demanding job....
it requires squaring to be done 200 times! ( after 1787109376)

Regarding my ideas, my explanation is (not too sure),
2 and 5 are numbers that perfectly compliment each other in multiplication,
and I go back to my earlier post....
The last n digits of powers of 2 start repeating at every
2^(4 x 5^n-1). This is a hypothesis, it isn't proven yet.
It works till 2^2500, beyond that, not too sure! 2^2500 ends in 09376.

This is indeed, a great work by the Mathsisfun team!