Find the
(i) sum of integers from 1 to 500 that are multiples of 2 as well as 5,
(ii) sum of integers from 1 to 500 that are multiples of 2 as well as 5 (inclusive),
(iii) sum of integers from 1 to 500 that are multiples of 2 or 5.
(i) 12750, (ii) 12750, (iii) 75250
(i) and (ii)
To be a multiple of both 2 and 5, a number must be a multiple of 10 (because 10 is the least common multiple of 2 and 5).
So the numbers we are adding are: 10, 20, 30, ..., 500.
This is an Arithmetic Progression (AP) with:
Number of terms: \(n = \dfrac{l - a}{d} + 1 = \dfrac{500 - 10}{10} + 1 = 50\).
Formula for sum of AP: \(S = \dfrac{n}{2}(a + l)\).
So, \(S = \dfrac{50}{2}(10 + 500) = 25 \times 510 = 12750\).
Thus, answer for (i) and (ii) = 12750.
(iii)
Now we need the sum of numbers that are multiples of 2 or 5.
This means we add all multiples of 2 and all multiples of 5, but be careful: numbers that are multiples of 10 will be counted twice (once in 2’s table, once in 5’s table). So we must subtract them once. This method is called the Inclusion–Exclusion Principle.
Step 1: Sum of multiples of 2
Multiples of 2 up to 500 are: 2, 4, 6, ..., 500.
There are \(250\) such numbers (since \(500/2 = 250\)).
Sum = \(2(1 + 2 + 3 + ... + 250)\).
We know formula: \(1 + 2 + ... + n = \dfrac{n(n+1)}{2}\).
So, sum = \(2 \times \dfrac{250 \times 251}{2} = 250 \times 251 = 62750\).
Step 2: Sum of multiples of 5
Multiples of 5 up to 500 are: 5, 10, 15, ..., 500.
There are \(100\) such numbers (since \(500/5 = 100\)).
Sum = \(5(1 + 2 + ... + 100) = 5 \times \dfrac{100 \times 101}{2}\).
So, sum = \(5 \times 5050 = 25250\).
Step 3: Subtract sum of multiples of 10
Multiples of 10 are already included in both the above sums, so they are counted twice.
We already calculated this in (i): sum = 12750.
Step 4: Apply Inclusion–Exclusion
Total sum = (sum of multiples of 2) + (sum of multiples of 5) − (sum of multiples of 10).
Total = \(62750 + 25250 − 12750 = 75250\).
So, answer for (iii) = 75250.