NIMCET 2008 — Mathematics PYQ

We are given that $7^m+7^n$ must be divisible by $5$. In terms of modular arithmetic, this is written as:

Step 2: Simplify the base using modulo 5

Since $7\equiv 2(\mathrm{mod}5)$, we can rewrite the equation as:

Step 3: Analyze the powers of 2 modulo 5

Let's look at the cycle of $2^k(\mathrm{mod}5)$:

• $2^1\equiv 2(\mathrm{mod}5)$

• $2^2\equiv 4(\mathrm{mod}5)$

• $2^3\equiv 8\equiv 3(\mathrm{mod}5)$

• $2^4\equiv 16\equiv 1(\mathrm{mod}5)$

• $2^5\equiv 2(\mathrm{mod}5)$ (The cycle repeats every 4 powers)

The possible remainders are $\{2,4,3,1\}$. Each of these remainders occurs exactly $100/4=25$ times for $m,n\in \{1,2,\dots ,100\}$.

Step 4: Find pairs $(2^m,2^n)$ that sum to $0(\mathrm{mod}5)$

For $2^m+2^n\equiv 0(\mathrm{mod}5)$, the possible pairs of remainders $(r_m,r_n)$ are:

1. $(1,4)$ because $1+4=5\equiv 0(\mathrm{mod}5)$

2. $(4,1)$ because $4+1=5\equiv 0(\mathrm{mod}5)$

3. $(2,3)$ because $2+3=5\equiv 0(\mathrm{mod}5)$

4. $(3,2)$ because $3+2=5\equiv 0(\mathrm{mod}5)$

Step 5: Calculate the number of ordered pairs

Each remainder ($1,2,3,$ or $4$) is produced by exactly 25 different values of $m$ (or $n$) in the range $1$ to $100$.

• For case (1, 4): There are $25$ choices for $m$ and $25$ choices for $n$. Total = $25\times 25=625$.

• For case (4, 1): There are $25$ choices for $m$ and $25$ choices for $n$. Total = $25\times 25=625$.

• For case (2, 3): There are $25$ choices for $m$ and $25$ choices for $n$. Total = $25\times 25=625$.

• For case (3, 2): There are $25$ choices for $m$ and $25$ choices for $n$. Total = $25\times 25=625$.

Step 6: Sum the possibilities

Total ordered pairs = $625+625+625+625=2500$.

Conclusion:

The total number of ordered pairs $(m,n)$ is 2500.

Correct Option: (c)