🇺🇸 AMC 8 ⇄ switch contest
1994 AJHSME

Problem 25

Problem 25 · 1994 AJHSME Stretch
Number Theory small-casespattern

Find the sum of the digits in the answer to

9999…9994 nines×4444…4494 fours

where a string of 94 nines is multiplied by a string of 94 fours.

Show answer
Answer: A — 846.
Show hints
Hint 1 of 2
94 nines is far too big to multiply out — but a number THIS regular always hides a pattern. Shrink the problem: try 9×4, then 99×44, then 999×444, and watch what the digit sum does each time you add one more nine.
Still stuck? Show hint 2 →
Hint 2 of 2
This is the 'solve a tiny version first' technique: nail the rule on 1, 2, 3 nines, confirm it's growing by the same amount each step, then leap straight to 94 without ever doing the giant multiplication.
Show solution
Approach: spot the pattern from small cases, then leap to 94
  1. Do the small versions and total the digits: 9·4 = 36 (digit sum 9), 99·44 = 4356 (sum 18), 999·444 = 443556 (sum 27). Each extra nine adds exactly 9 to the digit sum, so the rule is: digit sum = 9 × (number of nines).
  2. With 94 nines, the digit sum is 9 × 94 = 846.
  3. Why it grows by a steady 9: look at 999·444 = 443556 — it's a run of 4s, then a 3, then a run of 5s, then a 6. Adding one more nine lengthens the 4-run AND the 5-run by one digit each, so the digit sum climbs by 4 + 5 = 9 every time. With 94 nines the product is 93 fours, a 3, 93 fives, a 6, giving 4·93 + 3 + 5·93 + 6 = 9·94 = 846. The reusable move: when a number is too huge to compute, test the smallest cases, check the step between them is constant, then leap to the big case.
  4. Sanity check on the pattern: the all-nines factor is a multiple of 9, so the product is too — and a multiple of 9 always has a digit sum that's a multiple of 9. Our 846 = 9 × 94 passes, while the only non-multiple-of-9 choice (1072) can be ruled out on sight. (Several choices ARE multiples of 9, so this test confirms 846 is legal but doesn't pick it alone — the small-case rule is what pins down the exact value.)
Another way — rewrite the nines as a round number minus 1:
  1. A string of 94 nines equals 10⁹⁴ − 1, so the product is (10⁹⁴ − 1)·(444…4) = (444…4 with 94 zeros tacked on) − (444…4). Subtracting the 94-digit string of 4s from that shifted copy borrows down the line and lands on 93 fours, a 3, 93 fives, a 6 — exactly the shape the small cases predicted.
  2. Its digit sum is 4·93 + 3 + 5·93 + 6 = 372 + 3 + 465 + 6 = 846. Turning 99…9 into 10ⁿ − 1 is the standard trick that explains WHY these repeated-digit products come out so regular.
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