Find the sum of all 3 digit whole numbers that are divisible by 13.
MATHS
YES
To find the sum of all 3-digit whole numbers that are divisible by 13, we need to first determine the range of numbers between 100 and 999 that are divisible by 13.
The first step is to find the smallest 3-digit number that is divisible by 13. We can do this by dividing 100 by 13 and rounding up to the nearest whole number. The smallest multiple of 13 greater than or equal to 100 is 104.
Next, we find the largest 3-digit number that is divisible by 13. Dividing 999 by 13 gives us 76.84, so we round down to the nearest whole number. The largest multiple of 13 less than or equal to 999 is 988.
Now, we have the range of numbers between 104 and 988 that are divisible by 13. We can calculate the sum of this arithmetic series using the formula:
Sum = (first term + last term) * number of terms / 2.
The first term is 104, the last term is 988, and the common difference is 13. We can substitute these values into the formula to get the sum:
Sum = (104 + 988) * (number of terms) / 2.
To find the number of terms, we can use another formula for the nth term of an arithmetic series:
nth term = first term + (n - 1) * common difference.
Let's solve for n:
988 = 104 + (n - 1) * 13.
884 = (n - 1) * 13.
67.15 = n - 1.
n ≈ 67.15 + 1.
n ≈ 68.15.
Since the number of terms must be a whole number, we round up to 69.
Substituting n = 69 into the formula for the sum, we have:
Sum = (104 + 988) * 69 / 2.
Calculating the sum:
Sum = 1092 * 69 / 2.
Sum = 75228 / 2.
Sum = 37614.
Therefore, the sum of all 3-digit whole numbers that are divisible by 13 is 37,614.
we have an arithmetic sequence with
a0 = 1st 3-digit number divisible by 13
a0 = 104
d = 13 (why?)
last term is largest divisible by 13 = 988
how many terms? (988-104)/13 + 1 = 69
S = 69/2(104+988) = 37674