alpenglow's blog

By alpenglow, history, 3 weeks ago, In English

Trying to solve this problem. I got the last non-zero digit by working with $$$2$$$-s and $$$5$$$-s, but here's an easier solution by using $$$10$$$ directly.

Can anyone explain why this works? How taking modulo $$$10^9$$$ does not make it wrong?

Code

Thanks.

 
 
 
 
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3 weeks ago, # |
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we're interested in the last digit so even taking modulo $$$10$$$ is right

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    3 weeks ago, # ^ |
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    But it gets wrong answer.

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    3 weeks ago, # ^ |
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    Taking modulo 10 actually does not work.

    For example, take 125*8, and you will get 1000, with a last non-zero digit of 1. But what if it was actually 125*18, which is equal to 2250, with a last non-zero digit of 5?

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3 weeks ago, # |
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I think it's modulo 1e9 because the zeros will never go past that far on a single iteration, for example instead it could be 1000 if N went up to 99 because the most the zeros will go up at once is twice because of 25 (4*25 = 100).

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    3 weeks ago, # ^ |
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    boocoo What I don't understand is when we do modulo, we lose some information about the number.

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      3 weeks ago, # ^ |
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      It is true that by doing modulo, we lose some information about the actual number. However, we are more interested in the least significant digit (rightmost non zero digit). Taking modulo removes part of the most significant digits (leftmost digits are lost).

      Here, the only purpose of having modulo 1e9 is to keep the answer in such a range that multiplying a big number like N ( <= 2e7 according to the question ), the product still fits in a long long variable.

      By repeatedly dividing ans by 10 inside the while loop, we are losing information about the least significant digits, only if these digits are 0 (ans % 10 == 0). In the same way, we don't need to keep track of unnecessary most significant digits. Doing so, we are focussed on the requirement, i.e., the last (rightmost) non zero digit.

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        3 weeks ago, # ^ |
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        I want to know exactly the reason for that, you haven't told how is it enough.

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          3 weeks ago, # ^ |
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          Let's say at some point you have ans = 123 (after removing the right hand side 0s).

          • Now you multiply ans with some big number 49834344.
          • ans becomes 6,129,624,312.
          • The next number you would be multiplying is 49834343
          • ans would become 305,465,800,425,347,016.
          • On going one step further, you will easily move beyond the range of long long.

          Instead, when you do modulo 1e9 at each step, it will look something like this

          • After multiplying 49834344, ans = 129,624,312 ( modulo 1e9 )
          • Multiply this again 49834343, ans = 6,459,742,425,347,016 = 425,347,016 ( modulo 1e9 ).

          Notice that doing the modulus operation doesn't change the last digits, It only gets rid of the digits to the left. With modulo and without modulo, in both cases the last digits are the same since the modulus is done with a power of 10.

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            3 weeks ago, # ^ |
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            But why it does not work on $$$10^5$$$?

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              3 weeks ago, # ^ |
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              Because you have N <= 2e7. If you take any modulo lesser than that, it will cause the multiplying value to be truncated. The nearest power of 10 above that value of N is 1e8. That might work.

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                3 weeks ago, # ^ |
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                Ok, thanks. But I'm still little bit confused.

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3 weeks ago, # |
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Maybe you need to calculate factorial[n] * inverse(factorial[m])?

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    3 weeks ago, # ^ |
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    Modulo 10 inverses should not work, it is composite.

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      3 weeks ago, # ^ |
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      oh yes, didn't notice that

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      3 weeks ago, # ^ |
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      what is the range of n and m?

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        3 weeks ago, # ^ |
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        both at most 20000000

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          3 weeks ago, # ^ |
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          I think the reason they choose the mod depends on the range of n and m, since the maximum is 20000000, the product after each steps always has no more than 8 trailing zeroes, so we need a number that larger than 10^8. Besides if we choose 10^10, the product each steps can be larger than 2^64-1, which leads to error. That's my opinion :)

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            3 weeks ago, # ^ |
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            But I first remove those 0-s and then take the mod, not before.

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              3 weeks ago, # ^ |
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              Take a minor example, 3125 mod 10^3 = 125, 3125 mod 10^2 = 25, multiply both by 4 and we get 500 and 100, after moving the zeroes we get two different solutions, so we need a large number to decrease this probability As I mentioned in the last comment, 10^9 is good enough since 10^10 is risky :)

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3 weeks ago, # |
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Any better explanations are still welcome.

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3 weeks ago, # |
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It doesn't work for $$$5^{10}=9765625$$$ as far as I understand

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    3 weeks ago, # ^ |
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    What do you mean does not work? There are two values in the problem $$$n$$$ and $$$m$$$.

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      3 weeks ago, # ^ |
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      I meant n=5^10, m=n so you are calculating factorial, basically

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        3 weeks ago, # ^ |
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        It seems like it's correct.

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          3 weeks ago, # ^ |
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          actually I bugged in my reimplementation, it's $$$n=m=5^{10}+1$$$, it prints 8 instead of 4

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            3 weeks ago, # ^ |
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            Ok, the interesting thing is how did u find it.

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              3 weeks ago, # ^ |
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              well, it's clear that the problem is with 2's and 5's , because without them (I mean if 10 wouldn't have divisors) everything would work, but not with 10's (because zeros will be handler OK automatically). So, tried nearby $$$5^{10}$$$

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      3 weeks ago, # ^ |
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      $$$n=5^{10}+1$$$

      $$$m=11$$$

      The correct answer is 8, but the program outputs 6.

      There is actually a smaller counterexample:

      $$$n=5^9+13$$$

      $$$m=14$$$

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        3 weeks ago, # ^ |
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        Thanks for the correct counterexamples. I also checked on my another code (with 2-s and 5-s) and the correct answer is 8 indeed.

        But the question is, how did you find it?

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          3 weeks ago, # ^ |
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          I compared the correct answer(using $$$mod=10^{18}$$$ and __int128) and the answer given by the code written in the blog on values of n close to a power of 5 and small values of m.

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3 weeks ago, # |
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I was convinced this was true and was trying to prove it, but got stuck in some part of the proof. I'll leave my attempt with a mark in the error.

False proof
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    3 weeks ago, # ^ |
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    The solution only works for $$$n<5^9$$$, so proving that it works for $$$n<10^9$$$ is impossible.