Question

The sum of all integers between 1 and 100 (both inclusive) which are divisible by 5 or 13 is

• A. $1349$
• B. $1536$
• C. $1237$
• D. $1479$

Hint:

To find the sum of integers divisible by $A$ or $B$ within a range, use the Principle of Inclusion-Exclusion for sums: $\text{Sum}(A \text{ or } B) = \text{Sum}(A) + \text{Sum}(B) - \text{Sum}(A \text{ and } B)$. Remember that "$A$ and $B$" means divisible by $\text{LCM}(A, B)$. For arithmetic progressions, the sum can be calculated as $\frac{\text{number of terms}}{2} \times (\text{first term} + \text{last term})$. The number of terms can be found by $\left\lfloor \frac{\text{last number in range}}{\text{divisor}} \right\rfloor$.

Answer 1

The Correct Option is A

We need to find the sum of all integers between 1 and 100 (inclusive) that are divisible by 5 or 13. We can use the Principle of Inclusion-Exclusion for sums: \[ \text{Sum}(A \cup B) = \text{Sum}(A) + \text{Sum}(B) - \text{Sum}(A \cap B) \] Here, $A$ represents numbers divisible by 5, $B$ represents numbers divisible by 13, and $A \cap B$ represents numbers divisible by both 5 and 13 (i.e., by their least common multiple, $\text{LCM}(5, 13)$). Since 5 and 13 are prime numbers, their LCM is $5 \times 13 = 65$. \medskip First, let's find the sum of numbers divisible by 5 up to 100. The numbers are 5, 10, 15, ..., 100. This is an arithmetic progression. The number of terms: \[ n_5 = \frac{100}{5} = 20 \] The sum: \[ S_5 = \frac{n_5}{2}( \text{first term} + \text{last term} ) = \frac{20}{2}(5 + 100) = 10 \times 105 = 1050 \] \medskip Next, let's find the sum of numbers divisible by 13 up to 100. The numbers are 13, 26, 39, 52, 65, 78, 91. The number of terms: \[ n_{13} = \left\lfloor \frac{100}{13} \right\rfloor = 7 \] The sum: \[ S_{13} = \frac{n_{13}}{2}( \text{first term} + \text{last term} ) = \frac{7}{2}(13 + 91) = \frac{7}{2}(104) = 7 \times 52 = 364 \] \medskip Finally, let's find the sum of numbers divisible by both 5 and 13 (i.e., by 65) up to 100. The only number is 65. The sum: \[ S_{65} = 65 \] The number of terms: \[ n_{65} = \left\lfloor \frac{100}{65} \right\rfloor = 1 \] \medskip Using the Principle of Inclusion-Exclusion: \[ \text{Total Sum} = S_5 + S_{13} - S_{65} = 1050 + 364 - 65 = 1349 \]