2010 AIME I Problem 10

Attempt Problem 10 of the 2010 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AIME I solutions, or check the answer key.

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10.

Let NN be the number of ways to write 20102010 in the form 2010=a3103+a2102+a110+a0, \begin{aligned} 2010 &= a_3 \cdot 10^3 + a_2 \cdot 10^2 \\ &\quad {}+ a_1 \cdot 10 + a_0, \end{aligned} where the aia_i's are integers, and 0ai99.0 \le a_i \le 99. An example of such a representation is 11031 \cdot 10^3 +3102+ 3 \cdot 10^2 +67101+ 67 \cdot 10^1 +40100.+ 40 \cdot 10^0. Find N.N.

Answer: 202
Concepts:number basedigitsbijection
Difficulty rating: 2840
Solution:

Write each coefficient as ai=10bi+cia_i = 10b_i + c_i with digits bi,ci{0,1,,9};b_i, c_i \in \{0, 1, \ldots, 9\}; every integer 0ai990 \le a_i \le 99 splits this way uniquely. Setting m=b3b2b1b0m = b_3 b_2 b_1 b_0 and n=c3c2c1c0n = c_3 c_2 c_1 c_0 (read as base-1010 numbers), the condition becomes 2010=10m+n.2010 = 10m + n.

Conversely, any nonnegative integers m,nm, n with 10m+n=201010m + n = 2010 satisfy m201m \le 201 and n2010,n \le 2010, so each has at most four digits; those digits recover the bib_i and ci,c_i, hence the ai.a_i. So representations correspond exactly to choices of m{0,1,,201}m \in \{0, 1, \ldots, 201\} with n=201010m,n = 2010 - 10m, and N=202.N = 202.

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