JEE Mathematics 2023 Question with Solution

Given: We need the number of natural numbers $n$ such that

$$3^n-3 \equiv 0 \pmod{7}$$

with $10\le n\le 100$.

Find: The number of values of $n$ satisfying the condition.

Reduce the condition modulo $7$:

$$3^n-3 \equiv 0 \pmod{7}$$

So,

$$3^n \equiv 3 \pmod{7}$$

Now find the pattern of powers of $3$ modulo $7$:

$$\begin{aligned} 3^1 &\equiv 3 \pmod{7}\\ 3^2 &\equiv 2 \pmod{7}\\ 3^3 &\equiv 6 \pmod{7}\\ 3^4 &\equiv 4 \pmod{7}\\ 3^5 &\equiv 5 \pmod{7}\\ 3^6 &\equiv 1 \pmod{7} \end{aligned}$$

Thus, the powers of $3$ modulo $7$ are periodic with period $6$.

From the cycle,

$$3^n \equiv 3 \pmod{7} \iff n \equiv 1 \pmod{6}$$

So let

$$n=6k+1$$

subject to

$$10\le 6k+1\le 100$$

This gives

$$9\le 6k\le 99$$

Hence,

$$1.5\le k\le 16.5$$

Therefore,

$$k=2,3,4,\dots,16$$

The number of such integers is

$$16-2+1=15$$

Therefore, the required number of elements is $15$.

Given: We need $3^n \equiv 3 \pmod{7}$ for $10\le n\le 100$.

Find: How many such values of $n$ exist.

Since the remainders of powers of $3$ modulo $7$ repeat every $6$ terms, the condition holds exactly when

$$n \equiv 1 \pmod{6}$$

Now list the numbers in the range matching this form:

$$13,19,25,\dots,97$$

This is an arithmetic progression with common difference $6$. Its number of terms is

$$\frac{97-13}{6}+1=15$$

Therefore, the required number of values is $15$.