Determine $4$ specific digits in $34!$

Find the values of $a,b,c,d\in\mathbb{N}$ such that

$$ 34!=295232799cd9604140847618609643ab0000000 $$

My Attempt:

The factorial of $34$ contains a $3$, so the RHS must be divisible by $3$. Similarly, it must be divisible by $7$, $11$, $13$, $19$ etc.

But I do not understand how can I calculate $a,b,c,d$ in this equation.

Solution 1:

$34!$ has 7 powers of 5, which explains the last 7 0's.

Since $34!$ has $17+8+4+2+1 = 32$ powers of 2, $34! / 10^{7}$ has $32-7=25$ powers of 2. Doing a divisibility by $2^{7} = 128$ on the last 7 digits, we get that $ab = 52$.

(Note: If we had the last 3 digits missing, we could do a divisibility by $2^{10} = 1024$ on the last 10 digits. This is a useful approach that isn't often mentioned.)

Now, since these are digits, $0 \leq c, d \leq 9$, so $-9 \leq c-d \leq 9$ and $0 \leq c+d \leq 18$.

Use the fact that $34!$ is a multiple of 9, to tell you the value of $c+d$. We get that $c+d = 3$ or $12$. Use the fact that $34!$ is a multiply of 11, to tell you the value of $c-d$. We get that $c-d = -3$ or $8$. Since $2c$ is an even number from 0 to 18, we conclude that $c=0, d=3$.

Solution 2:

So $34!=(\ldots) + b10^7+a10^8+c10^{27}+d10^{28}$. Any divisibility rule by $k$ with $k\mid10^{20}-1$ cannot help here because it would allow to increase $a$ at the cost of $c$ and $b$ at the cost of $d$. Unfortunately, this rules out $9$ and $11$. But $7$ and $13$ combined should be helpful (note that $7\cdot 13=10^2-10+1$).

Solution 3:

Ah, nice. Have you tried applying the divisibility criteria (in terms of decimal digits) for the various numbers you mention? Did I take it right that you have just covered those criteria?

Solution 4:

$34! = 295232799cd96041408476186096435ab000000$

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

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

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

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

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

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

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

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