AMC 10 · 2014 · #20
Grade 6 number-theoryPick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
A number with k eights is far too big to multiply directly for the size of k we expect. So compute a few small cases and hunt for a pattern (Tool #5): the products 8·88, 8·888, and so on. Once the digit shape of the product is clear, name the count of eights with a variable k (Tool #4) to write the digit sum as a formula in k, then turn "digit sum =1000" into a simple equation and solve it (Tool #13).
Multiply the first few cases
Start small: multiply 8 by strings of 2, 3, 4, and 5 eights, keeping every product in full.
Doing a few concrete products by hand lets the shape of the answer show itself.
5.NBT.B.5Look For A PatternRead off the digit pattern
Lined up as 704, 7104, 71104, every product is a 7, then (k-2) ones, then 04.
The middle just gains one more 1 each time you add an 8, so the rule is easy to extend.
4.OA.C.5Look For A PatternWrite the digit sum in terms of k
Add those digits: 7+(k-2)+0+4, and the constants collapse to leave a digit sum of k+9.
Every extra 8 adds exactly one more 1 to the middle, so each step raises the digit sum by just 1.
Every extra digit adds exactly one more one to the middle, so each step raises the digit sum by one.
▸ Why?
A number is its digits sitting in fixed places, so a new place adds its own digit and nothing else.
▸ Why?
Each step adds the same fixed amount to the digit sum, which is what makes the rule easy to extend.
Solve for k
The digit sum is 1000, so k+9=1000 gives k=991, which is choice (D).
Once the digit sum is a tidy formula k+9, finding k is just undoing the +9.
6.EE.B.7Convert To AlgebraMultiply a few small cases, catch the digit pattern, and the giant problem shrinks to k+9=1000.
- Multiply the first few cases
- Read off the digit pattern
- Write the digit sum in terms of k
- Solve for k