The sum of all the elements in the set {n ∈ {1, 2, ..., 100} : H.C.F. of n and 2040 is 1} is equal t

Q: Wait, why do we only consider odd numbers? Can't $n$ be even?

That is a fantastic question! Look at the prime factorization of $2040$. It is $2^3 \times 3 \times 5 \times 17$. Because $2$ is a prime factor of $2040$, any even number $n$ will share a factor of $2$ with $2040$. Thus, $\text{H.C.F.}(n, 2040)$ would be at least $2$, which violates the condition $\text{H.C.F.}(n, 2040) = 1$. Therefore, $n$ must be odd.

Q: Why do we add the intersections back in the Inclusion-Exclusion Principle?

Think of it as a correction mechanism. When we subtract the sum of odd multiples of $3$ and the sum of odd multiples of $5$, we have subtracted the odd multiples of $15$ twice. To fix this over-subtraction, we must add the sum of the multiples of $15$ back in once. It is all about ensuring every number is accounted for exactly once!

Q: Is there a formula for the sum of numbers coprime to $m$?

Yes, there is! For numbers up to $n$, the sum is $\frac{1}{2} n \phi(n)$ (for $n > 1$). However, in JEE problems, we often deal with specific ranges like $1$ to $100$. Using the Principle of Inclusion-Exclusion is often safer and more intuitive than memorizing formulas, as it helps you visualize the 'bad' numbers you are removing.

Question

The sum of all the elements in the set $\{n \in \{1, 2, \dots, 100\} : 	ext{H.C.F. of } n 	ext{ and } 2040 	ext{ is } 1\}$ is equal to ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the sum of all integers $n$ in the range $1 \le n \le 100$ such that $	ext{H.C.F.}(n, 2040) = 1$, we must first determine the prime factorization of $2040$.

Performing the prime factorization, we obtain:
$$2040 = 2^3 	imes 3 	imes 5 	imes 17$$

For $n$ to be coprime to $2040$, it must not be divisible by any of the prime factors $2, 3, 5,$ or $17$.

### Phase 1: The Odd Universe

Since $n$ cannot be divisible by $2$, $n$ must be an odd number. The set of odd numbers up to $100$ is $\{1, 3, 5, \dots, 99\}$.

There are $50$ such numbers. The sum of the first $k$ odd numbers is given by $k^2$. Therefore, the sum of our initial universe is:
$$S_{total} = 50^2 = 2500$$

### Phase 2: The Inclusion-Exclusion Dance

We must subtract the sum of odd multiples of $3, 5,$ and $17$ from $S_{total}$. We apply the **Principle of Inclusion-Exclusion** to account for overlaps.

**1. Sum of odd multiples of 3:**
The set is $\{3, 9, 15, \dots, 99\}$. This is $3 	imes (1 + 3 + 5 + \dots + 33)$. Since there are $17$ terms:
$$S_3 = 3 	imes 17^2 = 3 	imes 289 = 867$$

**2. Sum of odd multiples of 5:**
The set is $\{5, 15, 25, \dots, 95\}$. This is $5 	imes (1 + 3 + 5 + \dots + 19)$. Since there are $10$ terms:
$$S_5 = 5 	imes 10^2 = 500$$

**3. Sum of odd multiples of 17:**
The set is $\{17, 51, 85\}$.
$$S_{17} = 17 + 51 + 85 = 153$$

### Phase 3: Correcting for Overlaps

We must now add back the intersections that were subtracted twice:

**1. Intersection of 3 and 5 (Multiples of 15):**
The set is $\{15, 45, 75\}$.
$$S_{15} = 15 + 45 + 75 = 135$$

**2. Intersection of 3 and 17 (Multiples of 51):**
The set is $\{51\}$.
$$S_{51} = 51$$

**3. Intersection of 5 and 17 (Multiples of 85):**
The set is $\{85\}$.
$$S_{85} = 85$$

Note that the intersection of all three ($3 	imes 5 	imes 17 = 255$) is greater than $100$, so it contributes $0$ to the sum.

### Final Calculation

The sum of the 'bad' numbers is calculated as:
$$S_{bad} = (S_3 + S_5 + S_{17}) - (S_{15} + S_{51} + S_{85})$$
$$S_{bad} = (867 + 500 + 153) - (135 + 51 + 85)$$
$$S_{bad} = 1520 - 271 = 1249$$

Subtracting the 'bad' sum from our total universe sum:
$$S_{final} = 2500 - 1249 = 1251$$

The final sum of all integers $n \le 100$ coprime to $2040$ is **$1251$**.

Animated Solution for Mathematics - Sets and Relations: The sum of all the elements in the set ${n \in {1, 2, \dots, 100} : H.C.F. of n and 2040 is 1}$ is equal to ____.

Enter Numerical Value:

Visualized Solution

Prime Factorization of $2040$

Strategy: Focus on Odd Numbers

Sum of All Odd Numbers

Sum of Odd Multiples of $3$

Sum of Odd Multiples of $5$

Sum of Odd Multiples of $17$

Calculating Intersections

Principle of Inclusion-Exclusion

Final Calculation

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

Similar Questions

Let $A = {n \in N : H.C.F. (n, 45) = 1}$ and Let $B = {2 k : k \in {1, 2, \dots, 100}}$ . Then the sum of all the elements of $A \cap B$ is ________.

Let $A = {n \in N ∣ n^{2} \leq n + 10, 000}$ , $B = {3 k + 1 ∣ k \in N}$ and $C = {2 k ∣ k \in N}$ , then the sum of all the elements of the set $A \cap (B - C)$ is equal to

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Animated Solution for Mathematics - Sets and Relations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of all the elements in the set {n∈{1,2,…,100}:H.C.F. of n and 2040 is 1} is equal to ____.

Enter Numerical Value:

Visualized Solution

Prime Factorization of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2040

Strategy: Focus on Odd Numbers

Sum of All Odd Numbers

Sum of Odd Multiples of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3

Sum of Odd Multiples of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5

Sum of Odd Multiples of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 17

Calculating Intersections

Principle of Inclusion-Exclusion

Final Calculation

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Sets and Relations: The sum of all the elements in the set ${n \in {1, 2, \dots, 100} : H.C.F. of n and 2040 is 1}$ is equal to ____.

Prime Factorization of $2040$

Sum of Odd Multiples of $3$

Sum of Odd Multiples of $5$

Sum of Odd Multiples of $17$

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Let $A = {n \in N ∣ n^{2} \leq n + 10, 000}$ , $B = {3 k + 1 ∣ k \in N}$ and $C = {2 k ∣ k \in N}$ , then the sum of all the elements of the set $A \cap (B - C)$ is equal to

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