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What is the sum of all digits which appear in all the integers from 10 to 100?

Published in Mathematics 2 mins read

The sum of all digits that appear in the integers from 10 to 100 is 856.

Let's break this down:

We need to consider all numbers from 10 up to and including 100. We can look at the tens place and the ones place separately.

Tens Place:

  • From 10-19, the tens digit is always 1. This appears 10 times. The sum is 1 * 10 = 10.
  • From 20-29, the tens digit is always 2. This appears 10 times. The sum is 2 * 10 = 20.
  • From 30-39, the tens digit is always 3. This appears 10 times. The sum is 3 * 10 = 30.
  • From 40-49, the tens digit is always 4. This appears 10 times. The sum is 4 * 10 = 40.
  • From 50-59, the tens digit is always 5. This appears 10 times. The sum is 5 * 10 = 50.
  • From 60-69, the tens digit is always 6. This appears 10 times. The sum is 6 * 10 = 60.
  • From 70-79, the tens digit is always 7. This appears 10 times. The sum is 7 * 10 = 70.
  • From 80-89, the tens digit is always 8. This appears 10 times. The sum is 8 * 10 = 80.
  • From 90-99, the tens digit is always 9. This appears 10 times. The sum is 9 * 10 = 90.
  • For 100, the tens digit is 0. This appears 1 time. The sum is 0 * 1 = 0.
    The sum of the tens digits is 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 0 = 450.

Ones Place:

  • For each decade (10-19, 20-29, etc.), the ones digits cycle through 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The sum of these digits is 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45.
  • This cycle repeats 9 times, from 10-99, so the sum of the ones digits from 10-99 is 45 * 9 = 405.
  • For 100, the ones digit is 0.

Hundreds Place:

  • For 100, the hundreds digit is 1. This appears 1 time. The sum is 1.

Total Sum:

The total sum of all digits is 450 (tens) + 405 (ones) + 1 (hundreds) = 856.

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