The largest n ∈ N, for which 7ⁿ divides 101!, is :

Q: Wait, why do we divide by $7^2$ after dividing by $7$? Don't we double count?

That is a brilliant observation! We are not double counting; we are accounting for the 'extra' factors of $7$. A number like $49$ ($7^2$) contains two factors of $7$. The first division, $\lfloor \frac{101}{7} \rfloor$, counts one factor of $7$ for every multiple of $7$. However, for $49$, it only counts one of its two factors. The second division, $\lfloor \frac{101}{49} \rfloor$, picks up that 'second' factor of $7$ that was missed. It is a beautiful, systematic way to ensure every single prime factor is accounted for.

Q: What if the number was not prime? What if the question asked for the largest $n$ such that $10^n$ divides $101!$?

If the base is composite, like $10$, you cannot use Legendre's Formula directly. You must first find the prime factorization of the base: $10 = 2 \times 5$. Then, you calculate the exponent of $2$ in $101!$ and the exponent of $5$ in $101!$ separately. Since $10^n$ requires both a $2$ and a $5$ to form a $10$, the limiting factor will be the prime with the smaller exponent. You would take the minimum of the two exponents.

Q: Why do we stop when the denominator is larger than the numerator?

Because the floor function $\lfloor x \rfloor$ returns $0$ for any $x$ where $0 101$, the fraction $\frac{101}{7^k}$ is less than $1$, so its floor is $0$. Since all subsequent powers of $7$ will also be greater than $101$, all further terms in the infinite sum will be $0$. Thus, the sum naturally terminates.

Question

The largest $n \in \mathbb{N}$, for which $7^n$ divides $101!$, is :

Options:

(A) 16

(B) 15

(C) 18

(D) 19

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks for the largest power of $7$ that divides $101!$. This is equivalent to finding the exponent of the prime $p=7$ in the prime factorization of $101!$.

We are not calculating the value of the factorial itself, but rather determining the frequency of the prime factor $7$ within the product of all integers from $1$ to $101$.

### The Sieve of Legendre

To solve this efficiently, we employ **Legendre's Formula**. This formula provides the exponent $E_p(n!)$ of a prime $p$ in the prime factorization of $n!$ using the following summation:

$$E_p(n!) = \lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor + \lfloor \frac{n}{p^3} \rfloor + \dots$$

The logic behind this is a systematic "sieve." We first count all multiples of $7$, then account for the "extra" factors of $7$ present in multiples of $7^2$, $7^3$, and so on.

### The Calculation

We apply the formula with $n = 101$ and $p = 7$.

First, we calculate the number of multiples of $7$ up to $101$:

$$\lfloor \frac{101}{7} \rfloor = 14$$

Next, we account for the multiples of $7^2 = 49$:

$$\lfloor \frac{101}{49} \rfloor = 2$$

These two values represent the numbers $49$ and $98$, which each contribute an additional factor of $7$ beyond the first one already counted.

Finally, we check for multiples of $7^3 = 343$:

$$\lfloor \frac{101}{343} \rfloor = 0$$

Since $343 > 101$, the sequence terminates here as all subsequent terms will also be $0$.

### The Grand Total

To find the total exponent, we sum the results of our calculations:

$$14 + 2 = 16$$

The prime number $7$ appears exactly $16$ times in the prime factorization of $101!$.

Therefore, the largest power of $7$ that divides $101!$ is **$7^{16}$**.

Animated Solution for Mathematics - Permutations and Combinations: The largest $n \in N$ , for which $7^{n}$ divides $101!$ , is :

Select Answer:

Visualized Solution

Understanding the Problem

Legendre's Formula

Identifying Parameters

Calculating the First Term

Calculating the Second Term

Checking Higher Powers

Final Summation

Conclusion

The Sigma Insight: Factorial Notation

Analyzing the Setup

The Sieve of Legendre

The Calculation

The Grand Total

Similar Questions

The largest value of $n$ , for which $4 0^{n}$ divides $60!$ , is

If $\frac{^{2 n + 1} P _{n - 1}}{^{2 n - 1} P _{n}} = 11 : 21$ , then $n^{2} + n + 15$ is equal to:

The value of $(2 \cdot^{1} P_{0} - 3 \cdot^{2} P_{1} + 4 \cdot^{3} P_{2} - ... up to 5 1^{t h} term) + (1! - 2! + 3! - ... up to 5 1^{t h} term)$ is equal to :

The missing value in the following figure is Use the logic which gives answer in single digit

Animated Solution for Mathematics - Permutations and Combinations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The largest n∈N, for which 7n divides 101!, is :