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2000 AMC 8

Problem 14

Problem 14 · 2000 AMC 8 Hard
Number Theory last-digitmod-10

What is the units digit of 1919 + 9999?

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Answer: D — 8.
Show hints
Hint 1 of 2
Don't be scared by the huge exponents — a number's last digit only depends on the last digit of the base. Both 19 and 99 end in 9, so this is really 'what does 9 to a big power end in?'
Still stuck? Show hint 2 →
Hint 2 of 2
List the last digits of 9¹, 9², 9³…: 9, 1, 9, 1, … — a length-2 cycle. So the only thing that matters is whether the exponent is odd or even.
Show solution
Approach: reduce each base to its last digit, then find the power's cycle
  1. Last digit depends only on the base's last digit, so 19¹⁹ matches 9¹⁹ and 99⁹⁹ matches 9⁹⁹.
  2. Powers of 9 end in 9 (odd exponent) or 1 (even exponent). Both exponents 19 and 99 are odd, so each power ends in 9.
  3. 9 + 9 = 18, so the sum ends in 8.
  4. You'll see it again: for the last digit of any power, throw away everything but the base's units digit, then find its short repeating cycle (length 1, 2, or 4) and reduce the exponent against that cycle.
Another way — 9 is one less than 10:
  1. Since 9 = 10 − 1, a power of 9 is just a touch off a multiple of 10: 9^odd ends in the same digit as (−1)^odd = −1, i.e. a 9; 9^even ends like +1.
  2. Both exponents are odd, so both powers end in 9, and 9 + 9 = 18 → units digit 8.
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