Animated Solution for Mathematics - Sequence and Series: The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is :

A Journey into Arithmetic Progressions

Welcome, future engineer. Today, we aren't just solving a problem; we are peeling back the layers of number theory to reveal the elegant structure hidden beneath.

When you look at a problem like this—finding the sum of two-digit numbers with specific remainders—it is easy to feel overwhelmed. You might be tempted to start scribbling down numbers, hoping to catch them all.

But I want you to pause. Take a breath. In the JEE Advanced arena, we don't hunt for numbers; we build the mathematical machinery that captures them for us.

Defining the Universe

First, let's define our universe. We are strictly confined to the world of two-digit positive numbers.

On our number line, this is the interval $[10,99]$. This is our boundary, our fence.

Anything outside this, we ignore. Anything inside, we must account for. This is the first step of any great problem-solver: defining the constraints of the system.

The Power of Euclid's Lemma

Now, let's look at the condition: numbers that leave a remainder of $2$ or $5$ when divided by $7$. This is where Euclid's Division Lemma becomes our best friend.

It tells us that any integer can be expressed as $7n+r$. Here, our remainder $r$ is either $2$ or $5$.

This splits our problem into two distinct, beautiful arithmetic progressions. We aren't looking for a random collection of numbers; we are looking for two orderly sequences:

1. The sequence defined by $7n+2$ 2. The sequence defined by $7n+5$

By separating them, we turn one chaotic problem into two structured, solvable ones. This is the essence of mathematical strategy: divide and conquer.

The Inequality Trap

Many students stumble here. They try to guess the first and last terms. But why guess when you can calculate?

We use the inequality $10\le 7n+r\le 99$ to find the exact range of $n$. For our first case, $7n+2$, we solve:

Subtracting $2$ from all sides, we get $8\le 7n\le 97$. Dividing by $7$, we find $1.14\le n\le 13.85$.

Since $n$ must be an integer, $n$ can be any value from $2$ to $13$. This gives us exactly $13-2+1=12$ terms.

The first term is $7(2)+2=16$, and the last is $7(13)+2=93$.

We repeat this logic for the second case, $7n+5$:

Subtracting $5$, we get $5\le 7n\le 94$. Dividing by $7$, we find $0.71\le n\le 13.42$.

Here, $n$ ranges from $1$ to $13$. This gives us $13-1+1=13$ terms. The first term is $7(1)+5=12$, and the last is $7(13)+5=96$.

The Grand Finale

Now, we have our two sequences. We don't need to add them one by one. We use the powerful sum formula for an arithmetic progression: $S=\frac{N}{2}(a+l)$, where $N$ is the number of terms, $a$ is the first term, and $l$ is the last term.

For the first sequence:

For the second sequence:

Finally, we combine our results. The total sum is simply the sum of these two parts:

Look at that result. $1356$. It didn't come from luck or brute force; it came from understanding the structure of the numbers.

You didn't just find an answer; you mapped a territory. Keep this mindset as you move forward.

Whether it's calculus, mechanics, or algebra, always look for the underlying pattern. That is how you master JEE Advanced.