How many times does the digit $0$ occur in the list of numbers from $1$ to $10^{100}$, inclusive?

  • 5
    $\begingroup$ And list all the numbers in which it occurs... ;-) $\endgroup$
    – Stiv
    Dec 6, 2023 at 13:37
  • $\begingroup$ hehe good luck with that :) $\endgroup$ Dec 6, 2023 at 14:31
  • 1
    $\begingroup$ Are we considering leading zeros as well? $\endgroup$ Dec 6, 2023 at 15:11

2 Answers 2


Observe that the numbers of occurrences of zeroes from $0$ to $10^n-1$ follow this sequence 0, 1, 10, 190, 2890, 38890, ... (oeis)

those numbers can be obtained for any $n > 2$ as
$10^{n-1} + \sum_{i=2}^{n-1} (10^{n-1}-10^{n-i})$
Where each summation member is the number of occurrences of the zeroes in the $i$-th position starting from right and $10^{n-1}$ is the count of zeroes in the first position.

ie. for $n=4$ we count all zeroes between 0 and 9999
$10^3+(10^3-10^2) + (10^3-10) = 2890$

plugging in $n=100$ we obtain:
$ 10^{99} + \sum_{i=2}^{99} (10^{99}-10^{100-i})$

Now we have to move the range from $[0,10^n-1]$ to $[1,10^n]$.

To achieve this, we have to subtract the first zero count and add 100 zeroes from the googol itself to get the final answer:
$ 10^{99} + \sum_{i=2}^{99} (10^{99}-10^{100-i})$ - 1 + 100 = 98888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888989


To easily get the n-th entry (ignoring the first "$0$" term) of the OEIS serie above, for $n >= 3$ you can use this simple trick:

- write $n - 2$
- append $8$, $n-3$ times
- append $90$

for $n=100$
- $98$
- $8888.....8888$ (97 eights)
- $90$

  • $\begingroup$ Nice! Can you write the answer in closed form, just for clarity? Thanks. $\endgroup$ Dec 6, 2023 at 14:51
  • $\begingroup$ It is easier to count all zeros (proper and leading) first: N x 10^(N-1) and then subtract leading zeros: 1+10+100+1000+... (N terms). (Add back 1 if you want to count 0 as a legit 0 even though it is leading.) $\endgroup$ Dec 6, 2023 at 19:36
  • 1
    $\begingroup$ Great work and thanks for finding the sequence. There is actually a closed form formula in the OEIS, which you can use here. $\endgroup$ Dec 7, 2023 at 11:26

Consider the last $100$ digits of each of these numbers, padded with leading zeroes as necessary.

Now consider reversing each of these strings. Since we're counting all $10^{100}$ $100$-digit strings, this is a bijection, and it has the effect of turning leading zeroes into trailing zeroes, with the exception of the all-zeroes string we got from $10^{100}$. Therefore, the trailing zeroes that we wish to count are half of all the zeroes, plus the other $50$ from the all-zeroes string.
Each position in the string has a $0$ in exactly one-tenth of the strings each, and there are $100$ positions, so the total zeroes are $\frac{\frac{10^{100}}{10}100}{2}+50 = 5×10^{100}+50$.

  • $\begingroup$ I got a different answer which matches the other answer posted $\endgroup$ Dec 6, 2023 at 14:21
  • 3
    $\begingroup$ It's a good try, but you forgot about palindromes. $\endgroup$ Dec 6, 2023 at 14:53
  • 5
    $\begingroup$ What about zeros that are neither trailing nor leading? $\endgroup$ Dec 6, 2023 at 19:05

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