Find the missing digits in the expansion of $34!$

Question

If $34!=\overline{295232799cd96041408476186096435ab000000}$ then find the value of $a,b,c$ and $d.$

My Attempt: I can find that $b=0$ because it has seven five integers.

$\lfloor{\frac{34}{5}}\rfloor+\lfloor{\frac{34}{25}}\rfloor=6+1=7$

Also I think $a$ can be found using divisibility rule of $8$ that means $\overline{35a}$ is divisible by $8$ which gives us $a=2$. But I'm stuck in finding $c$ and $d$. Using the divisibility rule of $9$ I get an equation. And using divisibility rule of $11$ I get another equation but when I solve them I get two values for $c$ and $d$.

Note: calculator is not allowed.

Answer 1

$34! = 295232799cd96041408476186096435ab000000$

$\left \lfloor \dfrac{34}{5} \right \rfloor = 6$

$\left \lfloor \dfrac{6}{5} \right \rfloor = 1$

So there are $6+1 = 7$ zeros at the end of $34!$. Hence $$\color{red}{b = 0}$$

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THEOREM: Compute the following

• $N = 5q_1 + R_1$

• $q_1 = 5q_2 + R_2$

• $q_2 = 5q_3 + R_3$

• ...

• $q_{n-1} = 5q_n + R_n$

where $0 \le R_i < 5$ for all $i$ and $0 \le q_n < 5$.

Then the first non zero digit in $N!$ is

$U(N!) = 2^P \times Q! \times R_1! \times R_2! \times R_3! \dots \times R_n! \pmod{10}$

Where

• $P = q_1 + q_2 + \cdots + q_n$
• $Q = q_n$

We compute \begin{align} 34 &= 5(6) + 4 \\ 6 &= 5(1) + 1 \\ 1 &= 5(0) + 1 \\ \end{align}

$P = 6 + 1 = 7$

$Q = 0$

\begin{align} U(34!) &= 2^7 \times 0! \times 4! \times 1! \times 1! \pmod{10} \\ &= 8 \times 0! \times 4 \times 1! \times 1! \pmod{10} \\ &= 2 \end{align}

So $$\color{red}{a = 2}$$

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$34! = 2\; 95\; 23\; 27\; 99\; \color{red}{cd}\; 96\; 04\; 14\; 08\; 47\; 61\; 86\; 09\; 64\; 35\; 20\; 00\; 00\; 00$

Clearly $99 \mid 34!$ So, when we cast out $99's$, we should get $0$. Pairing off the numbers in $34!$ from right to left, skipping $cd$, and adding modulo $99$, we get

$ 2 + 95 + 23 + 27 + 99 + 96 + 04 + 14 + 08 + 47 + 61 + 86 + 09 + 64 + 35 + 20 + 00 + 00 + 00 \pmod{99} = 96$

So $cd = 99 - 96 = 03$

Hence

$$ \color{red}{c = 0} $$

$$ \color{red}{d = 3} $$

Answer 2

$34!=295232799039604140847618609643520000000$.

Therefore $a=2, b=0, c=0, d=3$.

Answer 3

This is a partial solution...

Use all the divisibility rules.

But first, you can notice that $32!=2^31\times5^7\times\cdots$, so there are $7$ zeros at the end of the number. So $b=0$, and $a\ne 0$.

$34!$ is divisible by $9$, so:

$$4+a+c+d=0\pmod 9.$$

It's divisible by $7$, so:

$$000-000+5a0-643+609-618+847-140+604-cd9+799+327-952+2=0\pmod 7,$$

so

$$835+5a0-cd9=0\pmod 7.$$

It's divisible by $11$, so:

$$2-2+5-2+9-7+2-3+9-9+7-9+d-c+4-0+6-0+4-1+7-4+1-7+4-8+8-1+6-9+0-6+3-4+6-0+a-5=0\pmod {11},$$

so

$$6+a-c+d=0\pmod {11}.$$

It's also divisible by $13$, so you can look at the alternate sum of three digits from the left.

And so on...

Try to look at this article about divisibility rules.

I hope this helps!