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Find $\displaystyle\binom{80}{40}\bmod 2000$.

So far, I've found that $\displaystyle\binom{80}{40}$ is divisible by $2^2$ and $5^1$, so the answer isn't $0$. Usually, with smaller numbers, I would split the $\bmod 2000$ into $2^4$ and $5^3$, find the answer through brute force for each of those, then use the Chinese Remainder theorem. But $80$ and $40$ are too big for this.

Jyrki Lahtonen
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Nathan Wang
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2 Answers2

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Since $${80 \choose 40} = \frac{80\cdot 79 \cdot 78 \cdot 77 \cdot ...\cdot 41}{40 \cdot 39 \cdot 38 \cdot 37 \cdot ... \cdot 1}$$ we can cancel the terms in the denominator with terms in the numerator leaving a factor of $2$, so $2^{40}$ is a factor of the answer. It only remains to see how many factors of 5 remain. $75, 65, 55$, and $45$ are all factors, so at least $5^4$ is a factor. As a result, the expression is definitely a multiple of $2000$, so modulo $2000$ the answer is $0$.

EDIT: A few minutes later I realized my mistake, and returned to find several people pointing it out: only the numbers down to 21 in the denominator cancel. That still leaves $$\frac{2 \cdot 79 \cdot 2 \cdot 77 \cdot 2 \cdot 75 \cdot...\cdot 2 \cdot 41}{ 20 \cdot 19 \cdot 18 \cdot...\cdot 3 \cdot 2 \cdot 1}$$ which gives $20$ factors of $2$ in the numerator and $10 + 5 + 2 + 1 = 18$ factors of $2$ in the denominator, so only $2^2$ is a factor. For the $5's$, we have $5$ factors in the numerator and $4$ in the denominator, so it is a multiple of $5$ but not $5^2$.

So now we have $$\frac{79 \cdot 77 \cdot 3 \cdot 73 \cdot...\cdot 43 \cdot 41}{ 19 \cdot 9 \cdot 17 \cdot 3 \cdot 7...\cdot 6 \cdot 3}$$. Continuing to cancel factors, we find $3^{8}$ in the numerator and $3^8$ in the denominator, and so forth. We end up computing ${80 \choose 40}$ and then can find the answer by inspection.

RobertTheTutor
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    2^40 does not divide it, you can verify with a calculator. Same with 5^4. – Nathan Wang Mar 06 '21 at 02:51
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    @RobertTheTutor that is incorrect. The answer is not $0$. – an4s Mar 06 '21 at 02:52
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    But there are $40$ factors downstairs, and only $20$ even factors upstairs. I think that your method is the preferred one, but I’m unconvinced that your count is correct. – Lubin Mar 06 '21 at 02:53
  • Notice that the numerator now has $40$ terms (of which only $20$ are even, as @Lubin points out), not $80$. And therefore, you cannot have $2^{40}$ in the numerator after performing the division. – an4s Mar 06 '21 at 02:55
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    OP is correct about the first part: the number is a multiple of $2^2$ and $5^1$, but is not a multiple of $2^3$ or $5^2$. – aschepler Mar 06 '21 at 02:57
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    Kummer’s Theorem clearly shows your computations are incorrect. The highest power of $2$ that divides $\binom{80}{40}$ is the number of carries when adding $80-40=40$ and $40$ in base $2$. In base $2$, $40=10100$, so you get exactly two carries. So $\binom{80}{40}$ is a multiple of $2^2$ but not of $2^3$. – Arturo Magidin Mar 06 '21 at 03:09
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    My apologies everyone, I caught my mistake but you caught it faster. Thank you. – RobertTheTutor Mar 06 '21 at 03:18
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I built Pascal's triangle in a spreadsheet program, but always reducing mod 2000. There's never a need to keep more than the result mod $2000$. The result I get is $1620$.

  • Cell A2: 0
  • Cell A3: =A2+1 and carry down to Cell A82 with value $80$.
  • Cell B1: 0
  • Cell C1: =B1+1 and carry rightward to Cell AP1 with value $40$.

Now these are the indices of Pascal's triangle. For $\binom{n}{k}$, $n$ is in column A and $k$ is in row 1.

  • Cell B2: 1 and carry all the way down.
  • Cell C3: =MOD(B2+C2,2000), carry all the way rightward, and then all the way down, filling out the table.

Now Cell AP82 is $\binom{80}{40}$ mod $2000$. It gives $1620$.

To do this by hand, at first it may seem like you would calculate about $80\cdot40=3200$ cell entries mod $2000$. But actually you would only need to compute about $\frac14$ of those entries, as represented in blue here:

enter image description here

The red is all $0$. The orange is not needed to reach the $\binom{80}{40}$ cell. And the green can be written using symmetry from the blue. So this would take about $800$ additions mod $2000$ plus some other less signiicant labor.

But we an do better, because there is the identity $$\binom{80}{40}=\sum_{n=0}^{40}\binom{40}{n}^2=\binom{40}{20}+2\sum_{n=0}^{19}\binom{40}{n}^2$$

So actually if we can get Pascal's triangle down to the first half of the 40th row, then we can square mod $2000$ about 20 times, double, sum the results mod $20$ and have our answer.

A similar analysis reveals that you would need to calculate about $400$ of the entries of Pascal's triangle mod $2000$ for this approach.

2'5 9'2
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