Find the sum of the natural numbers to 200which are exactly divisible by [a] 3 [b]5?

Answer 1

Sum of natural numbers from 1 to n = n/2 + [2a + (n – 1) × d]
where n = total numbers, a = first number and d = common difference.
a) Numbers are from 1 to 200 and divisible by 3
Total numbers from 1 to 200 which are divisible by 3 are = 200 / 3 = 66
a = first number divisible by 3 is = 3 and d = common difference = 3
substituting these values in the formula,
Sum = 66/2 [ 2 × 3 + (66 – 1) × 3]
= 33 [ 6 + 195 ] = 33 × 201 = 6633
b) a) Numbers are from 1 to 200 and divisible by 3
Total numbers from 1 to 200 which are divisible by 5 are = 200 / 5 = 40
a = first number divisible by 5 is = 5 and d = common difference = 5
substituting these values in the formula,
Sum = 40/2 [ 2 × 5 + (40 – 1) × 5]
= 20 [ 10 + 195 ] = 20 × 205 = 4100

Answer 2

5

Answer 3

Hint #1

a) You need to calculate the number of numbers divisible by 3 from 1 to 200 - what is 200 divided by 3 (only the whole part).

b) Do the same for 5's to get 40.

Hint #2

a) Look at "pairs" which sum to 201 (3+198, 6+195, ... 99 + 102) how many pairs are there? Multiply that by 201 to get your answer.

b) Do the same for 205 (200+5, 195+10, ... 105 +100).

Answer 4

a) That would be:
3 + 6 + 9 + ... + 198
= 3 * (1 + 2 + 3 + ... + 66)

b) That would be:
5 + 10 + 15 + ... + 200
= 5 * (1 + 2 + 3 + ... + 40)

I'm sure you know the formula for adding up (1 + 2 + 3 + ... + n) already. (If you don't, it's time to get off the Internet and open a book.)