What is the average of all three-digit numbers divisible by 23?

Correct — B. Multiples of 23 form an AP, and an AP's average is (first + last) ÷ 2.

First three-digit multiple: 23 × 4 = 92 is too small, so 23 × 5 = 115
Last three-digit multiple: 23 × 43 = 989, since 23 × 44 = 1012

Average = (115 + 989) ÷ 2
= 1104 ÷ 2 = 552 → option (b)

Why the others are wrong

• (a)
  752 — 752 is 200 above the middle of the list. An AP averages to its middle value, (115 + 989) ÷ 2 = 552, and 752 would need first + last = 1504.
• (c)
  652 — 652 would need first + last = 1304. The smallest and largest three-digit multiples of 23 are 115 and 989, which add to 1104 and average 552.
• (d)
  452 — 452 would need first + last = 904, less than the largest term 989 on its own. The terms run from 115 to 989, so their middle is 552.

Numbers divisible by a fixed d form an arithmetic progression with common difference d. The average of any AP equals (first term + last term) ÷ 2, because the terms pair off symmetrically about the middle.

So the work is locating the two ends: the smallest and largest three-digit multiples of 23. The number of terms (43 − 5 + 1 = 39) is not needed for the average.

Key facts

• Three-digit multiples of 23 run from 115 (23 × 5) to 989 (23 × 43).
• There are 43 − 5 + 1 = 39 of them.
• The average of an AP is (first + last) ÷ 2.

Common traps

• Starting the list at 92, which is 23 × 4 but has two digits.
• Averaging 100 and 999, the ends of the three-digit range, instead of the ends of the list of multiples: that gives 549.5.

No directly related past PYQ was found.