Let x and y be distinct integers where 1 ≤ x ≤ 25 and 1 ≤ y ≤ 25. Then, the number of ways of choosi

Q: Wait, why do we treat $(x \in R_1, y \in R_4)$ and $(x \in R_4, y \in R_1)$ as separate cases?

Great question! In this problem, we are choosing an ordered pair $(x, y)$. Since $x$ and $y$ are distinct, picking $x=1$ and $y=4$ is a different selection than picking $x=4$ and $y=1$. By treating them as separate cases, we ensure we count every unique ordered pair correctly.

Q: Why is the $R_0$ case $5 \times 4$ instead of $5 \times 5$?

The condition states that $x$ and $y$ must be distinct integers. In the set $R_0 = \{5, 10, 15, 20, 25\}$, there are $5$ possible values. If we choose $x$ from this set, we have $5$ options. Once $x$ is fixed, $y$ cannot be equal to $x$, leaving us with only $4$ remaining choices for $y$. Hence, $5 \times 4 = 20$.

Q: What if the range was different, say up to 26?

If the range were $1$ to $26$, the remainder sets would no longer be symmetric. $R_1$ would contain an extra element ($26 \equiv 1 \pmod 5$), changing the counts in each set. Always check the size of your remainder sets first—symmetry is a luxury, not a guarantee!

Question

Let $x$ and $y$ be distinct integers where $1 \leq x \leq 25$ and $1 \leq y \leq 25$. Then, the number of ways of choosing $x$ and $y$, such that $x + y$ is divisible by 5, is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a vast grid of numbers from $1$ to $25$. You are tasked with finding pairs $(x, y)$ such that their sum is a multiple of $5$.

If you try to list them out, you will quickly find yourself drowning in a sea of arithmetic. Mathematics is not about brute force; it is about finding the hidden structure, which in this case is **modular arithmetic**.

### Partitioning the Universe

When we divide any integer by $5$, the remainder can only be $0, 1, 2, 3,$ or $4$. Let us partition our set $\{1, 2, \dots, 25\}$ into five kingdoms, which we call remainder sets $R_0, R_1, R_2, R_3,$ and $R_4$.

$R_0$ holds numbers like $5, 10, 15, 20, 25$. $R_1$ holds $1, 6, 11, 16, 21$, and so on.

Notice the elegance: each set contains exactly $5$ elements. This symmetry is our greatest weapon.

### The Combinatorial Dance

For $x + y$ to be divisible by $5$, the sum of their remainders must be $0$ or $5$. This gives us a few specific scenarios to dance through.

First, consider the case where both $x$ and $y$ come from $R_0$. Their sum will be a multiple of $5$. Since $x$ and $y$ must be distinct, we have $5$ choices for $x$ and $4$ choices for $y$, yielding:

$$5 	imes 4 = 20 	ext{ ways}$$

Now, consider the cross-remainder pairs. If $x$ is in $R_1$ (remainder $1$) and $y$ is in $R_4$ (remainder $4$), their sum is $1 + 4 = 5$, which is divisible by $5$.

Since these sets are disjoint, any $x \in R_1$ and $y \in R_4$ are guaranteed to be distinct. We have $5$ choices for $x$ and $5$ for $y$, giving $5 	imes 5 = 25$ ways.

Because order matters, we must also count the case where $x \in R_4$ and $y \in R_1$, which gives us another $25$ ways.

We repeat this logic for the pair $(R_2, R_3)$. Choosing $x \in R_2$ and $y \in R_3$ gives $25$ ways, and swapping them ($x \in R_3, y \in R_2$) gives another $25$ ways.

### The Grand Total

We have methodically covered every possibility: $20$ ways from $R_0$, and four sets of $25$ ways from the cross-remainder pairs. Adding these up:

$$20 + 25 + 25 + 25 + 25 = 120$$

It is a beautiful result. By shifting our perspective from the raw numbers to their remainders, a chaotic counting problem transforms into a simple, symmetric calculation.

The final answer is **120**. Keep this in your toolkit, future engineer—whenever you see divisibility, look for the remainders.

Animated Solution for Mathematics - Permutations and Combinations: Let $x$ and $y$ be distinct integers where $1 \leq x \leq 25$ and $1 \leq y \leq 25$ . Then, the number of ways of choosing $x$ and $y$ , such that $x + y$ is divisible by 5, is

Enter Numerical Value:

Visualized Solution

Problem Constraints: $x, y \in [1, 25]$

Modular Arithmetic Approach

Defining Remainder Sets $R_{i}$

Condition: $x + y \equiv 0 (mod 5)$

Case 1: $x, y \in R_{0}$

Case 2: $x \in R_{1}, y \in R_{4}$

Case 3: $x \in R_{4}, y \in R_{1}$

Case 4: $x \in R_{2}, y \in R_{3}$

Case 5: $x \in R_{3}, y \in R_{2}$

Total Number of Ways

The Sigma Insight: Fundamental Principle of Counting

Analyzing the Setup

Partitioning the Universe

The Combinatorial Dance

The Grand Total

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Enter Numerical Value:

Visualized Solution

Problem Constraints: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x,y∈[1,25]

Modular Arithmetic Approach

Defining Remainder Sets .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ri​

Condition: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x+y≡0(mod5)

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x,y∈R0​

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈R1​,y∈R4​

Case 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈R4​,y∈R1​

Case 4: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈R2​,y∈R3​

Case 5: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈R3​,y∈R2​