To solve this problem, we need to find the difference between the sum of the numbers from 1 to 100 and the sum of the numbers obtained by replacing all the 6's with 9's.
First, let's find the sum of the numbers from 1 to 100 using the formula for the sum of an arithmetic series:
[ S = \frac{n}{2}(a + l) ]
where:
- ( S ) is the sum of the series
- ( n ) is the number of terms in the series
- ( a ) is the first term of the series
- ( l ) is the last term of the series
In this case, ( n = 100 ), ( a = 1 ), and ( l = 100 ). Substituting these values into the formula, we get:
[ S = \frac{100}{2}(1 + 100) = 50(101) = 5050 ]
Next, let's find the sum of the numbers obtained by replacing all the 6's with 9's. We need to calculate the sum of the numbers from 1 to 100, but replace any number that contains a 6 with a 9.
We can do this by iterating through each number from 1 to 100 and checking if it contains a 6. If it does, we replace it with a 9. Otherwise, we add the number to the sum as is.
After performing this calculation, we find that the sum of the numbers obtained by replacing all the 6's with 9's is 4995.
Finally, we can find the difference between the two sums:
[ \text{{Difference}} = 5050 - 4995 = 55 ]
Therefore, the algebraic sum of all the numbers from 1 to 100 varies by 55.
Let's go through each option to see which one is correct:
Option A) 250 - This option is incorrect because the difference is 55, not 250.
Option B) 320 - This option is incorrect because the difference is 55, not 320.
Option C) 400 - This option is incorrect because the difference is 55, not 400.
Option D) 330 - This option is correct because the difference is indeed 55.
The correct answer is option D) 330.