Answer :
Final answer:
The sum of odd numbers from 1 to 2019, inclusive, is given by n², which turns out to be 1020100 for this case. This number is divisible by both 100 and 101. Therefore, the answer to the question is option 'C', which states that the sum is divisible by both 100 and 101.
Explanation:
The sum of the series of odd numbers can be given by n² where n is the number of terms. In the case of odd numbers between 1 and 2019, we have 1010 terms (2019 / 2 = 1009,5 --> we take the next integer --> 1010). So the sum would be 1010² = 1020100.
Now we need to check if 1020100 is divisible by 100 and 101. It's clear that 1020100 is divisible by 100 as its last two digits are zero. Let's check divisibility by 101. For that, you can use a calculator. Dividing 1020100 by 101 we get 10100 which is an integer, thus 1020100 is also divisible by 101.
So the sum of odd numbers from 1 to 2019 (both inclusive) is divisible by both 100 and 101, so the correct answer is 'C'.
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The sum of the odd numbers from 1 to 2019 both inclusive, is divisible by both 100 and 101. The correct option is C.
What can se say about the sum of these odd numbers?
To determine whether the sum of odd numbers from 1 to 2019 is divisible by 100, 101, both, or neither, we can use the arithmetic progression formula for the sum of an arithmetic series:
Sum = (n/2) * (first term + last term)
Where:
- n = number of terms
- first term = the first term in the series
- last term = the last term in the series
In this case, the series is the odd numbers from 1 to 2019. The first term is 1, the last term is 2019, and the common difference between consecutive terms is 2 (since they are odd).
Number of terms, n = (last term - first term) / common difference + 1
n = (2019 - 1) / 2 + 1
n = 1010
Sum = (1010/2) * (1 + 2019)
Sum = 510 * 2020
Sum = 1020200
Now let's check the divisibility:
A) 1020100 is divisible by 100 because it ends with two zeros.
B) 1020200 is divisible by 101:
1020100/101 = 10,100
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