Let K be the largest value such that $(2023!)^{2023!}$ is divisible by $42^{\mathrm{K}}$. How many consecutive 0 digits are at the end of \mathrm{K}$?
I tried to find the highest power of $42$ that divides $2023!$ Since $42 = 2 \times 3 \times 7$, we need to count the occurrences of 2, 3, and 7 in the prime factorization of $ 2023!$.
Counting the number of factors of 2 in $2023! =\left\lfloor\frac{2023{2}\right\rfloor +\left\lfloor\frac{2023}{2^2}\right\rfloor \left\lfloor\frac{2023{2^3}\right\rfloor + \ldots = 1011 + 505 + 252 + \ldots$.
Counting the number of factors of 3:
$\text{Number of factors of } 3 = \left\lfloor\frac{2023}{3}\right\rfloor + \left\lfloor\frac{2023}{3^2}\right\rfloor + \left\lfloor\frac{2023}{3^3}\right\rfloor + \ldots.$
Counting the number of factors of 7: $\text{Number of factors of } 7 = \left\lfloor\frac{2023}{7}\right\rfloor + \left\lfloor\frac{2023}{7^2}\right\rfloor + \ldots$
Then take the minimum of these counts, pair up to get the factors of 42: $K = \min(\text{Number of factors of 2}, \text{Number of factors of 3}, \text{Number of factors of 7}) $
Calculating these counts:
- Number of factors of 2: $ = 1011 + 505 + 252 + 126 + 63 + 31 + 15 + 7 + 3 + 1 = 2014 $
- Number of factors of 3:$ = 674 + 224 + 74 + 24 + 8 + 2 = 1006 $
- Number of factors of 7: $ = 289 + 41 + 5 = 335 $
Taking the minimum of these counts: $K = \min(2014, 1006, 335) = 335.$
So, the largest value of $ K $ is 335. Counting the consecutive zeros at the end of 335, is 2. Therefore, the answer is $2$. Is this right or wrong?
You have found out that $355$ is the largest value of $N$ with $42^N\mid2023!$. But the number in question is $2023!$ to the power of itself, and so $K=355\cdot2023!=71\cdot5\cdot2023!$. We now count how many $5$ factors are in $K$: $$1+\sum_{i=1}^\infty\lfloor2023/5^i\rfloor=1+404+80+16+3=504$$ Since this is smaller than the number of $2$ factors in $K$ – $2014$ – the answer is $504$.