Key Concepts and Formulas

• Princ of Inclusion-Exclusion (PIE) for Sums: To find the sum of elements in a set $U$ that satisfy a certain property (e.g., not divisible by any of a set of primes $p_1, p_2, \ldots, p_k$), we start with the sum of all elements in $U$ and then subtract the sums of elements divisible by each $p_i$, add back the sums of elements divisible by each pair $p_i p_j$, subtract sums divisible by each triplet $p_i p_j p_k$, and so on.
• Sum of Multiples of $d$ up to $N$: The sum of all positive integers $n \le N$ that are divisible by $d$ is given by $d + 2d + \ldots + \lfloor N/d \rfloor d$. This is an arithmetic progression with sum $d \cdot \frac{\lfloor N/d \rfloor (\lfloor N/d \rfloor + 1)}{2}$.
• Coprime Numbers: Two integers $a$ and $b$ are coprime (or relatively prime) if their greatest common divisor (H.C.F.) is 1. This means they share no common prime factors.

Step-by-Step Solution

The problem asks for the sum of all integers $n$ in the set $\{1, 2, \ldots, 100\}$ such that the H.C.F. of $n$ and 2040 is 1. This means we need to find the sum of integers $n$ in the given range that are coprime to 2040.

Step 1: Prime Factorization of 2040 First, we find the distinct prime factors of 2040. $2040 = 10 \times 204 = (2 \times 5) \times (2 \times 102) = (2 \times 5) \times (2 \times 2 \times 51) = (2 \times 5) \times (2^2 \times 3 \times 17) = 2^3 \times 3 \times 5 \times 17$. The distinct prime factors of 2040 are $p_1=2, p_2=3, p_3=5, p_4=17$. The condition H.C.F.($n$, 2040) = 1 means that $n$ must not be divisible by 2, 3, 5, or 17.

Step 2: Calculate the Total Sum of the Set The set is $U = \{1, 2, \ldots, 100\}$. The sum of all elements in this set is: $S = \sum_{n=1}^{100} n = \frac{100(100+1)}{2} = \frac{100 \times 101}{2} = 50 \times 101 = 5050$ This is the starting point for our PIE calculation.

Step 3: Calculate the Sum of Multiples of Each Distinct Prime Factor Let $A_d$ denote the sum of multiples of $d$ in the set $\{1, 2, \ldots, 100\}$. We use the formula: $A_d = d \cdot \frac{\lfloor 100/d \rfloor (\lfloor 100/d \rfloor + 1)}{2}$.

• Sum of multiples of 2: $A_2 = 2 \cdot \frac{\lfloor 100/2 \rfloor (\lfloor 100/2 \rfloor + 1)}{2} = 2 \cdot \frac{50 \times 51}{2} = 2550$.
• Sum of multiples of 3: $A_3 = 3 \cdot \frac{\lfloor 100/3 \rfloor (\lfloor 100/3 \rfloor + 1)}{2} = 3 \cdot \frac{33 \times 34}{2} = 3 \times 33 \times 17 = 1683$.
• Sum of multiples of 5: $A_5 = 5 \cdot \frac{\lfloor 100/5 \rfloor (\lfloor 100/5 \rfloor + 1)}{2} = 5 \cdot \frac{20 \times 21}{2} = 5 \times 10 \times 21 = 1050$.
• Sum of multiples of 17: $A_{17} = 17 \cdot \frac{\lfloor 100/17 \rfloor (\lfloor 100/17 \rfloor + 1)}{2} = 17 \cdot \frac{5 \times 6}{2} = 17 \times 15 = 255$.

The sum of these terms is $S_1 = A_2 + A_3 + A_5 + A_{17} = 2550 + 1683 + 1050 + 255 = 5538$.

Step 4: Calculate the Sum of Multiples of Products of Two Distinct Prime Factors We need to consider pairs of prime factors: (2,3), (2,5), (2,17), (3,5), (3,17), (5,17).

• Sum of multiples of $2 \times 3 = 6$: $A_6 = 6 \cdot \frac{\lfloor 100/6 \rfloor (\lfloor 100/6 \rfloor + 1)}{2} = 6 \cdot \frac{16 \times 17}{2} = 6 \times 8 \times 17 = 816$.
• Sum of multiples of $2 \times 5 = 10$: $A_{10} = 10 \cdot \frac{\lfloor 100/10 \rfloor (\lfloor 100/10 \rfloor + 1)}{2} = 10 \cdot \frac{10 \times 11}{2} = 10 \times 5 \times 11 = 550$.
• Sum of multiples of $2 \times 17 = 34$: $A_{34} = 34 \cdot \frac{\lfloor 100/34 \rfloor (\lfloor 100/34 \rfloor + 1)}{2} = 34 \cdot \frac{2 \times 3}{2} = 34 \times 3 = 102$.
• Sum of multiples of $3 \times 5 = 15$: $A_{15} = 15 \cdot \frac{\lfloor 100/15 \rfloor (\lfloor 100/15 \rfloor + 1)}{2} = 15 \cdot \frac{6 \times 7}{2} = 15 \times 21 = 315$.
• Sum of multiples of $3 \times 17 = 51$: $A_{51} = 51 \cdot \frac{\lfloor 100/51 \rfloor (\lfloor 100/51 \rfloor + 1)}{2} = 51 \cdot \frac{1 \times 2}{2} = 51 \times 1 = 51$.
• Sum of multiples of $5 \times 17 = 85$: $A_{85} = 85 \cdot \frac{\lfloor 100/85 \rfloor (\lfloor 100/85 \rfloor + 1)}{2} = 85 \cdot \frac{1 \times 2}{2} = 85 \times 1 = 85$.

The sum of these terms is $S_2 = A_6 + A_{10} + A_{34} + A_{15} + A_{51} + A_{85} = 816 + 550 + 102 + 315 + 51 + 85 = 1919$.

Step 5: Calculate the Sum of Multiples of Products of Three Distinct Prime Factors We need to consider triplets of prime factors: (2,3,5), (2,3,17), (2,5,17), (3,5,17).

• Sum of multiples of $2 \times 3 \times 5 = 30$: $A_{30} = 30 \cdot \frac{\lfloor 100/30 \rfloor (\lfloor 100/30 \rfloor + 1)}{2} = 30 \cdot \frac{3 \times 4}{2} = 30 \times 6 = 180$.
• Sum of multiples of $2 \times 3 \times 17 = 102$: Since $102 > 100$, there are no multiples of 102 in $\{1, \ldots, 100\}$. So, $A_{102} = 0$.
• Sum of multiples of $2 \times 5 \times 17 = 170$: Since $170 > 100$, $A_{170} = 0$.
• Sum of multiples of $3 \times 5 \times 17 = 255$: Since $255 > 100$, $A_{255} = 0$.

The sum of these terms is $S_3 = A_{30} + A_{102} + A_{170} + A_{255} = 180 + 0 + 0 + 0 = 180$.

Step 6: Calculate the Sum of Multiples of Products of Four Distinct Prime Factors We need to consider the product of all four prime factors: (2,3,5,17).

• Sum of multiples of $2 \times 3 \times 5 \times 17 = 510$: Since $510 > 100$, there are no multiples of 510 in $\{1, \ldots, 100\}$. So, $A_{510} = 0$.

The sum of these terms is $S_4 = A_{510} = 0$.

Step 7: Apply the Principle of Inclusion-Exclusion The sum of numbers $n \in \{1, \ldots, 100\}$ such that H.C.F.($n$, 2040) = 1 is given by: $\text{Desired Sum} = S - S_1 + S_2 - S_3 + S_4$ $\text{Desired Sum} = 5050 - 5538 + 1919 - 180 + 0$ $\text{Desired Sum} = -488 + 1919 - 180$ $\text{Desired Sum} = 1431 - 180$ $\text{Desired Sum} = 1251$

Common Mistakes & Tips

• Prime Factorization Accuracy: Ensure all distinct prime factors of 2040 are correctly identified. Any omission will lead to an incorrect result.
• Sum of Multiples Formula: Double-check the application of the arithmetic series sum formula for multiples. The $\lfloor N/d \rfloor$ term is crucial for determining the count of multiples.
• Alternating Signs in PIE: Pay close attention to the alternating signs in the PIE formula: subtract for single factors, add for pairs, subtract for triplets, and so on.
• Products Exceeding N: If a product of prime factors $d$ is greater than 100, the sum of its multiples ($A_d$) within the set is 0. This simplifies calculations for higher-order products.

Summary

To find the sum of integers in $\{1, 2, \ldots, 100\}$ that are coprime to 2040, we used the Principle of Inclusion-Exclusion. We first found the distinct prime factors of 2040 (2, 3, 5, 17). Then, we calculated the total sum of the set and systematically subtracted sums of multiples of individual prime factors, added back sums of multiples of products of two prime factors, subtracted sums of multiples of products of three prime factors, and so on. This process correctly accounts for numbers that share common factors with 2040, isolating the sum of those that do not.

The final answer is \boxed{1251}.