Problem 16 · 2012 AMC 8
Medium
Algebra & Patterns
place-value-greedy
Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?
Show answer
Answer: C — 87431.
Show hints
Hint 1 of 2
A digit in the ten-thousands place is worth far more than the same digit in the ones place. So to make the sum biggest, the biggest digits must sit in the highest places — greed wins here.
Still stuck? Show hint 2 →
Hint 2 of 2
This is a place-value greedy argument: the two ten-thousands digits should be 9 and 8, the two thousands digits 7 and 6, and so on down. That forces each position to use a known pair — then just check which answer fits.
Show solution
Approach: biggest digits to biggest places (greedy), then match the pairs
- Since a higher place multiplies a digit by more, dropping a big digit into a high place adds the most to the sum. So assign greedily from the top: the two leading digits use {9, 8}, the next pair {7, 6}, then {5, 4}, then {3, 2}, then {1, 0}.
- That means whichever number you look at, its digits left-to-right must come one from each pair: position 1 from {9,8}, position 2 from {7,6}, position 3 from {5,4}, position 4 from {3,2}, position 5 from {1,0}.
- Test the choices against this stencil — only 87431 fits: 8∈{9,8}, 7∈{7,6}, 4∈{5,4}, 3∈{3,2}, 1∈{1,0}. ✓
- Why greedy is safe here: moving a larger digit to a higher place never hurts the sum, so the maximum must look exactly like this. The same "big digits to big places" logic handles most digit-arrangement extremes.
Mark:
· log in to save