Solution Strategy

The problem asks for the total change in the sum of numbers from 1 to 100 when every digit '6' is replaced by a '9'. We can calculate this by summing the individual increases for each number where a replacement occurs. The amount of increase depends on the place value of the digit '6'.

Step 1: Calculate the Increase for '6' in the Units Place

First, we find all numbers between 1 and 100 that have a '6' in the units place.

• These numbers are: 6, 16, 26, 36, 46, 56, 66, 76, 86, 96.
• There are a total of 10 such numbers.
• For each of these numbers, replacing the '6' with a '9' in the units place increases its value by $9 - 6 = 3$.
• The total increase from the units place is $10 \times 3 = 30$.

Step 2: Calculate the Increase for '6' in the Tens Place

Next, we find all numbers between 1 and 100 that have a '6' in the tens place.

• These numbers are: 60, 61, 62, 63, 64, 65, 66, 67, 68, 69.
• There are a total of 10 such numbers.
• For each of these numbers, replacing the '6' with a '9' in the tens place increases its value by $90 - 60 = 30$.
• The total increase from the tens place is $10 \times 30 = 300$.

Step 3: Calculate the Total Variation

The total variation in the sum is the sum of all the individual increases. This method correctly accounts for the number 66, as the increase from its units digit is counted in Step 1 and the increase from its tens digit is counted in Step 2.

Total Variation = (Increase from units place) + (Increase from tens place)

Total Variation = $30 + 300 = 330$.

The algebraic sum of all the numbers varies by 330.