Find the natural numbers 'n' such that 'n!' ends with exactly 31 zeroes?a110b115c120d125

1. Break Down the Problem

To find the natural numbers $n$ such that $n!$ ends with exactly 31 zeroes, we need to determine how many trailing zeroes are in $n!$. The number of trailing zeroes in a factorial is determined by the number of times the factor 10 appears in its prime factorization, which is influenced by the number of factors of 5 (since there are always more factors of 2 than 5).

2. Relevant Concepts

The formula to find the number of trailing zeros $Z(n)$ in $n!$ is given by:

$Z(n) = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \ldots$

3. Analysis and Detail

We need to solve for $n$ such that $Z(n) = 31$. We calculate $Z(n)$ for various values of $n$:

• For $n = 125$: $Z(125) = \left\lfloor \frac{125}{5} \right\rfloor + \left\lfloor \frac{125}{25} \right\rfloor + \left\lfloor \frac{125}{125} \right\rfloor$ $= 25 + 5 + 1 = 31$

• For $n = 120$: $Z(120) = \left\lfloor \frac{120}{5} \right\rfloor + \left\lfloor \frac{120}{25} \right\rfloor$ $= 24 + 4 = 28$

• For $n = 115$: $Z(115) = \left\lfloor \frac{115}{5} \right\rfloor + \left\lfloor \frac{115}{25} \right\rfloor$ $= 23 + 4 = 27$

• For $n = 110$: $Z(110) = \left\lfloor \frac{110}{5} \right\rfloor + \left\lfloor \frac{110}{25} \right\rfloor$ $= 22 + 4 = 26$

Since $Z(125) = 31$, we check if there are any other $n$ values under 125 that yield 31 zeroes:

• For $n = 126$: $Z(126) = \left\lfloor \frac{126}{5} \right\rfloor + \left\lfloor \frac{126}{25} \right\rfloor$ $= 25 + 5 = 30$

• For $n = 127$: $Z(127) = \left\lfloor \frac{127}{5} \right\rfloor + \left\lfloor \frac{127}{25} \right\rfloor$ $= 25 + 5 = 30$

• For $n = 128$ to $n = 129$: The counts still yield 30.

4. Verify and Summarize

The only confirmed $n$ value that gives exactly 31 trailing zeroes is $n = 125$.

Final Answer

The natural number $n$ such that $n!$ ends with exactly 31 zeroes is: 125 (Option d).