Question:
The highest power of 3 contained in 100! is
The highest power of 3 contained in 100! is
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Simply divide the number by the prime repeatedly and sum the quotients (ignoring remainders) until the quotient is zero.
Updated On: Jul 3, 2026
- 47
- 48
- 24
- 40
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The Correct Option is B
Solution and Explanation
Step 1: Concept
To find the highest power of a prime $p$ in $n!$, we use Legendre's formula: $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor$.
Step 2: Meaning
Here $n = 100$ and $p = 3$. We need to calculate the sum of the floor values of $100$ divided by increasing powers of $3$ (3, 9, 27, 81).
Step 3: Analysis
$\lfloor 100/3 \rfloor = 33$; $\lfloor 100/9 \rfloor = 11$; $\lfloor 100/27 \rfloor = 3$; $\lfloor 100/81 \rfloor = 1$.
Step 4: Conclusion
Summing these values: $33 + 11 + 3 + 1 = 48$. Thus, the highest power of 3 in 100! is $3^{48}$. Final Answer: (B)
To find the highest power of a prime $p$ in $n!$, we use Legendre's formula: $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor$.
Step 2: Meaning
Here $n = 100$ and $p = 3$. We need to calculate the sum of the floor values of $100$ divided by increasing powers of $3$ (3, 9, 27, 81).
Step 3: Analysis
$\lfloor 100/3 \rfloor = 33$; $\lfloor 100/9 \rfloor = 11$; $\lfloor 100/27 \rfloor = 3$; $\lfloor 100/81 \rfloor = 1$.
Step 4: Conclusion
Summing these values: $33 + 11 + 3 + 1 = 48$. Thus, the highest power of 3 in 100! is $3^{48}$. Final Answer: (B)
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