What is the sum of all digits in the telephone dial (0 to 9)?

Correct — A.
Rule: sum of 0 to n = n(n + 1)⁄2, and 0 adds nothing.

0 + 1 + 2 + … + 9 = 9 × 10⁄2 = 45

Check by pairing: (0 + 9) + (1 + 8) + (2 + 7) + (3 + 6) + (4 + 5)
= 5 × 9 = 45 → option (a).

Why the others are wrong

• (b)
  55 — 55 is 1 + 2 + … + 10. The digits stop at 9, so the 10 does not belong. Without it the total falls to 45.
• (c)
  50 — 50 is not a multiple of 9, but the digits pair into five sums of 9 each, so the total must be one. 45 is.
• (d)
  40 — 40 is 5 short of the total. The five pairs (0 + 9), (1 + 8), (2 + 7), (3 + 6), (4 + 5) make 9 each, which is 45.

The sum of the first n natural numbers is n(n + 1)⁄2. Adding 0 changes nothing, so 0 to 9 sums to 9 × 10⁄2 = 45.

The pairing trick is where the formula comes from: first plus last, second plus second-last, and so on, each pair giving the same total.

Key facts

• 1 + 2 + … + n = n(n + 1)⁄2.
• 0 + 1 + … + 9 = 45.
• 1 + 2 + … + 10 = 55.

Common traps

• Summing 1 to 10 out of habit and answering 55.
• Worrying whether 0 belongs on the dial, when it adds nothing to a sum.

No directly related past PYQ was found.