Finding the last 5 digits of this huge number.

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[waynemv]

Michael's solution, while more general and very instructional, was way overcomplicated and tedious for this particular problem.

The original sum can be split into ten sums, each with the same last five digits.
Any of those sums can, in turn, be split into ten sums each with the same last four digits.
Any of those sums can, in turn, be split into ten sums each with the same last three digits.
Any of those sums can, in turn, be split into ten sums each with the same last two digits.
Any of those sums can, in turn, be split into ten sums each with the same last digit.

Thus, our original sum must be equivalent (mod 10^5) to 10*10*10*10*10*something; that is, its last five digits must all be zeros.

[LuxiBelle]

These number theory questions always boogle my mind. It is a lot of tricks you need to know from experience

[jursamaj]

14:50 Once you produced N2, it's is clear that you could keep going, 3 more time reducing by a factor of 5 each. This will easily produce a result with a factor of 5^6, which of course will be 0 mod 5^5. No need to count the 1's.

[andreiherghelegiu]

For any number A < 1000 we have A^100 mod 10^5 = (K*1000+A)^100 mod 10^5 for any positive K, since in the binomial expansion we have A^100 + 100*(A^99)*(K*1000) + terms with 1000 raised to something bigger than or equal to 2. Therefore, the remainders repeat after each 1000 terms in the sum. Since all multiples of 1000 raised to 100 are equal to 0 mod 10^5, the initial sum reduces to 1000*(1^100+...+999^100) which ends in 3 zeros. We have to find the other last two digits, i.e. (1^100+...+999^100) mod 100. Using the same reasoning, the remainders repeat after 10 terms, we have 100*(1^100+...+9^100) mod 100 = 0. Therefore, the sum ends in 5 zeros.

[koenth]

I just went about as follows
It takes just a few terms of Newton's binomium to prove that (x+1000)^100 = x^100 (mod
From here, our result is 1000 N (mod where N= 0^100+1^100+2^200+..+999^100.

Similarly a few terms of the binomium show that (x+10)^100 = x^100 (mod 1000), so that N =100 M (mod 1000) where M=(0^100+1^100+..+9^100), but we need not even calculate it, since

our result is 1000 N = M = 0 (mod

[s]

Can't we just apply/derive gerneralized Faulhaber's formula and factor? That seems much easier if your end term (1, 000, 000) is divisible by your divisor (100, 000).

[nothayley]

You described the 5-case as "more grindy, " but honestly it seemed more interesting than the 2-case! The recursive reductions were satisfying and clear to see. My instinct would have been to continue, defining N_3, N_4, and N_5, until it was immediately clear what was happening.

As a side note, I really hate that as a community we've settled on x ≡ 0 (mod N) for our notation; I'd much rather see ≡_N (≡ subscript N), because the modulus is a property of the equivalence relationship, and sometimes it can be hard to see. Maybe the subscript notation would be harder to see in small text, though.

[kisupahn]

While Michael's approach is more general and has some insight in it, I think the answer for this particular problem can be found more easily:

A^100 = (10a+b)^100 = 10^100*c+d^100, where d∈[0, 9]
so the original problem is equivalent to finding the last five digit of
{1^100+2^100+...+0^100} *100, 000
and the answer is obvious

But what if the power was less or the sum range was different or something? well, the trick doesn't work easily..

[krisbrandenberger]

Hey, Michael! At 9:27 the bottom line needs to have 5 raised to the 6th power, not the 1st power.

[sugarfrosted]

that reduction in the mod 5^5 case was neat.

[HershO.]

For the case of 5, if one goes through the effort of going up to N_4 then it becomes immediately clear (without even using a pen) that the number is 0 (mod 5^5)

[orionspur]

is extremely suspicious. Surely there is a quick and clever trick.

[MarsAnonymous]

As interesting as finding the last 5 digits of a large number like this is, what about finding the last 5 digits of an infinite number like lim_{n \to \infty} 2^{(10^n)}?

[wesleydeng]

No need to use mod 2^5 and 5^5.
N so Its last 2 digits are 0s. And, (mod 1000) = 100*(1^100+2^100+...999^100) (mod 1000), so its next last 2 digits are 0s too. Finally, 1^100+2^100+...999^100 (mod 10) = 10*(1^100+2^100+...99^100) (mod 10). So, its 5th last digit is 0 as well.

[teeweezeven]

I have a strong feeling this generalizes, i.e. the sum of the first n digits raised to some power like 100 is divisible by n and n+1

[krisbrandenberger]

So in general, for any number N not equal to 0, if N is divisible by 10^k, the last k digits of N will all be zeroes.

[jamesfortune]

It's interesting that Euler didn't come up with the phi function to solve problems like that.

[abrahammekonnen]

That was hard for me to follow, but it was a cool problem, thank you.

[marc-andredesrosiers]

I didn't understand this one. Not sure how come.

[robert-skibelo]

As a Swiss citizen I always feel a little twinge of pride when one of Euler's results comes in handy. Surely the most useful Swiss of all time. Bet most of you guys didn't even know he was Swiss.😂