Find the number of zeros at the end of the 120! + 230! + 40! + 450! + 60! + 670!1 point28201669

To find the number of zeros at the end of the expression $120! + 230! + 40! + 450! + 60! + 670!$, we need to determine the number of trailing zeros in each factorial individually and then identify the smallest count since the addition of these large numbers will not change the number of trailing zeros.

Step 1: Break Down the Problem

1. Calculate the number of trailing zeros in each factorial: $120!$, $230!$, $40!$, $450!$, $60!$, and $670!$.
2. Identify the minimum number of trailing zeros among all calculated values.

Step 2: Relevant Concepts

The number of trailing zeros in $n!$ can be found using the formula: $Z(n) = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \ldots$

Step 3: Calculate the Number of Trailing Zeros

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

2. For $230!$: $Z(230) = \left\lfloor \frac{230}{5} \right\rfloor + \left\lfloor \frac{230}{25} \right\rfloor + \left\lfloor \frac{230}{125} \right\rfloor = 46 + 9 + 1 = 56$

3. For $40!$: $Z(40) = \left\lfloor \frac{40}{5} \right\rfloor + \left\lfloor \frac{40}{25} \right\rfloor = 8 + 1 = 9$

4. For $450!$: $Z(450) = \left\lfloor \frac{450}{5} \right\rfloor + \left\lfloor \frac{450}{25} \right\rfloor + \left\lfloor \frac{450}{125} \right\rfloor = 90 + 18 + 3 = 111$

5. For $60!$: $Z(60) = \left\lfloor \frac{60}{5} \right\rfloor + \left\lfloor \frac{60}{25} \right\rfloor = 12 + 2 = 14$

6. For $670!$: $Z(670) = \left\lfloor \frac{670}{5} \right\rfloor + \left\lfloor \frac{670}{25} \right\rfloor + \left\lfloor \frac{670}{125} \right\rfloor = 134 + 26 + 5 = 165$

Step 4: Verify and Summarize

The calculated trailing zeros are:

• $Z(120) = 28$
• $Z(230) = 56$
• $Z(40) = 9$
• $Z(450) = 111$
• $Z(60) = 14$
• $Z(670) = 165$

The minimum value among all calculated trailing zeros is $9$ (from $40!$).

Final Answer

The number of zeros at the end of $120! + 230! + 40! + 450! + 60! + 670!$ is $\boxed{9}$.