Problem 20 · 1993 AJHSME
Stretch
Number Theory
borrowing-patterndigit-sum
When 1093 − 93 is expressed as a single whole number, the sum of the digits is
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Answer: D — 826.
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Hint 1 of 2
You can't write out 10⁹³ — but you don't have to. Shrink the problem: what does 10⁴ − 93 look like? Then 10⁵ − 93? A pattern will jump out, and big exponents just stretch it.
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Hint 2 of 2
Subtracting from 1000…0 forces a chain of borrows that turns the middle into all 9s. Find how many 9s appear (in terms of the exponent), then add the leftover digits.
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Approach: spot the borrowing pattern from a small case
- Test small: 10⁴ − 93 = 10000 − 93 = 9907, and 10⁵ − 93 = 99907. The borrowing eats the zeros into a run of 9s, ending in 07. The pattern: 10ⁿ − 93 is (n − 2) nines followed by '07'.
- For n = 93 that's 91 nines, then a 0 and a 7. Digit sum = 91 × 9 + 0 + 7 = 819 + 7 = 826.
- Why this transfers: when an exponent is too big to compute, simulate the SAME operation on a tiny exponent, read the digit pattern, then scale it up. Numbers like 10ⁿ minus a small amount always collapse into a predictable string of 9s — borrowing is regular.
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