Let S = {(m, n) : m, n ∈ {1, 2, 3, ..., 50}}. If the number of elements (m, n) in S such that 6ᵐ + 9

Q: Why can't I just plug in values for $m$ and $n$ to see if it works? Isn't that faster?

Brute force is the enemy of the JEE Advanced aspirant! While you might find a few pairs, you will inevitably miss the pattern and waste precious time. Modular arithmetic is not just a shortcut; it is the mathematical language that reveals the underlying structure of the problem. By using it, we solve for all 2500 pairs in a single stroke of logic.

Q: Why did you use $9 \equiv -1 \pmod 5$ instead of $9 \equiv 4 \pmod 5$?

Great question! While both are mathematically correct, using $-1$ makes the exponentiation trivial. $(-1)^n$ is either $1$ or $-1$, which is incredibly easy to handle. If you used $4^n$, you would have to deal with powers of 4, which is much more cumbersome. Always choose the modular representative that makes your algebra cleaner!

Q: How do you know there are exactly $k-1$ pairs for $m+n=k$ so quickly?

Visualize the line $m+n=k$ on a coordinate plane. Since $m$ and $n$ must be at least 1, the smallest value $m$ can take is 1 (making $n = k-1$) and the largest value $m$ can take is $k-1$ (making $n = 1$). Since $m$ can be any integer from 1 to $k-1$, there are exactly $k-1$ such pairs. It is a beautiful geometric interpretation of a simple algebraic constraint.

Question

Let $S = \{(m, n) : m, n \in \{1, 2, 3, \dots, 50\}\}$. If the number of elements $(m, n)$ in S such that $6^m + 9^n$ is a multiple of 5 is $p$ and the number of elements $(m, n)$ in S such that $m + n$ is a square of a prime number is $q$, then $p + q$ is equal to ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a set $S$ containing all ordered pairs $(m, n)$ where $m, n \in \{1, 2, \dots, 50\}$. This defines a grid of $50 	imes 50 = 2500$ points.

Our objective is to determine two values, $p$ and $q$, and calculate their sum. We begin by defining $p$ as the number of pairs $(m, n)$ such that $6^m + 9^n$ is a multiple of 5.

### The Modular Dance

To find $p$, we solve the congruence:
$$6^m + 9^n \equiv 0 \pmod{5}$$

Since $6 \equiv 1 \pmod{5}$, it follows that $6^m \equiv 1^m \equiv 1 \pmod{5}$ for any positive integer $m$. This simplifies our expression significantly.

Next, we observe that $9 \equiv -1 \pmod{5}$. Therefore, $9^n \equiv (-1)^n \pmod{5}$. Substituting these into our congruence, we obtain:
$$1 + (-1)^n \equiv 0 \pmod{5}$$

This condition holds if and only if $(-1)^n \equiv -1 \pmod{5}$. This occurs precisely when $n$ is an **odd integer**.

Since $m$ can be any of the 50 integers in the range $[1, 50]$ and $n$ must be an odd integer in the same range, we have 50 choices for $m$ and 25 choices for $n$ (the odd numbers $\{1, 3, \dots, 49\}$). Thus:
$$p = 50 	imes 25 = 1250$$

### The Combinatorial Hunt

Now, we determine $q$, the number of pairs $(m, n)$ such that $m + n$ is the square of a prime number. The primes are $2, 3, 5, 7, 11, \dots$, and their squares are $4, 9, 25, 49, 121, \dots$.

Given that $1 \le m, n \le 50$, the sum $m + n$ must satisfy $2 \le m + n \le 100$. Consequently, we only consider the squares $4, 9, 25,$ and $49$.

For a fixed sum $k = m + n$, the number of pairs $(m, n)$ with $1 \le m, n \le 50$ is given by $k-1$ provided that $k \le 51$. Since all our target sums ($4, 9, 25, 49$) are less than or equal to 50, the count for each sum $k$ is simply $k-1$:

*   For $k = 4$: $4 - 1 = 3$ pairs.
*   For $k = 9$: $9 - 1 = 8$ pairs.
*   For $k = 25$: $25 - 1 = 24$ pairs.
*   For $k = 49$: $49 - 1 = 48$ pairs.

Summing these counts, we find $q$:
$$q = 3 + 8 + 24 + 48 = 83$$

### The Grand Finale

Having determined both components of the problem, we perform the final addition:
$$p + q = 1250 + 83 = 1333$$

The final result is **1333**.

Enter Numerical Value:

Visualized Solution

Understanding the Set $S$

Condition for $p$ : $6^{m} + 9^{n}$ is a Multiple of $5$

Simplifying $6^{m} (mod 5)$

Simplifying $9^{n} (mod 5)$

Solving for $n$

Calculating $p$

Condition for $q$ : $m + n$ is a Square of a Prime

Counting Pairs for $m + n = 4$ and $m + n = 9$

Counting Pairs for $m + n = 25$ and $m + n = 49$

Calculating $q$ and Final Sum $p + q$

The Sigma Insight: Fundamental Principle of Counting

Analyzing the Setup

The Modular Dance

The Combinatorial Hunt

The Grand Finale

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Enter Numerical Value:

Visualized Solution

Understanding the Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6m(mod5)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 9n(mod5)

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n is a Square of a Prime

Counting Pairs for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n=4 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n=9

Counting Pairs for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n=25 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n=49

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q and Final Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p+q