AMC 10 · 2005 · #24
Grade 8 arithmeticPick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Name the two digits so the reversed number becomes an expression. Turning the digit-swap into algebra reveals a clean factorization, and then divisibility rules narrow the digits down to a single possibility.
Name the digits
Let the digits be a and b, so x = 10a + b and reversing gives y = 10b + a, with a and b each from 1 to 9.
A digit-swap is easy to handle once each digit has its own letter.
6.EE.B.6Introduce A VariableFactor the difference of squares
Factoring, x² - y² = (x - y)(x + y) = 9(a - b) · 11(a + b) = 99(a - b)(a + b).
Rewriting a difference of squares as two factors exposes the hidden 9 and 11.
Rewriting a difference of squares as two factors exposes the hidden prime factors.
▸ Why?
A difference of two squares is the two numbers added multiplied by the two subtracted.
▸ Why?
A perfect square needs every prime paired up, so a lone prime factor has to find a partner.
Force a perfect square
Since 99 = 3² · 11, the lone 11 needs a partner; a - b is at most 8, so the sum must supply it: a + b = 11.
A perfect square needs each prime factor paired up, so the lone 11 must find a partner.
8.EE.A.2Eliminate PossibilitiesPin down the digits
Then m² = 1089(a - b), so a - b is an odd perfect square; a - b = 9 would force a = 10, leaving a = 6, b = 5.
Only a square times 1089 stays a square, and parity leaves just one workable gap between the digits.
8.EE.C.8Eliminate PossibilitiesAdd up x, y, and m
So x = 65, y = 56, and m² = 1089 gives m = 33, hence x + y + m = 154, choice (E).
Once the digits are fixed, the three numbers just add straight to the total.
8.EE.A.2Convert To AlgebraWrite a two-digit number as 10a + b, and a difference of reversed squares always becomes 99 times (a - b)(a + b) -- then the primes tell you which digits fit.
- Name the digits
- Factor the difference of squares
- Force a perfect square
- Pin down the digits
- Add up x, y, and m