The sum of all numbers divisible by 3 from 1 to 100 is 1683.
Understanding the Calculation
This problem involves finding the sum of an arithmetic series. The numbers divisible by 3 between 1 and 100 form an arithmetic progression (AP): 3, 6, 9, ..., 99.
- First term (a): 3
- Common difference (d): 3
- Last term (l): 99
We can find the number of terms (n) using the formula for the nth term of an AP: l = a + (n-1)d
Solving for n:
99 = 3 + (n-1)3
96 = 3(n-1)
32 = n-1
n = 33
There are 33 numbers divisible by 3 between 1 and 100.
Now, we can find the sum (S) using the formula for the sum of an arithmetic series: S = n/2 * (a + l)
S = 33/2 (3 + 99)
S = 33/2 102
S = 33 * 51
S = 1683
Therefore, the sum of all numbers divisible by 3 from 1 to 100 is 1683. This is confirmed by the provided reference stating that S22=1683. Note that the reference uses a different method, but the result is the same.
Alternative Approaches
While the arithmetic series method is efficient, other approaches exist. One could use a loop in a programming language to iterate through numbers and sum those divisible by 3. However, the arithmetic series method provides a more elegant and mathematically concise solution.
