2018 AIME I Problem 2

Attempt Problem 2 of the 2018 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 2018 AIME I solutions, or check the answer key.

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2.

The number nn can be written in base 1414 as abc,\underline{a}\,\underline{b}\,\underline{c}, can be written in base 1515 as acb,\underline{a}\,\underline{c}\,\underline{b}, and can be written in base 66 as acac,\underline{a}\,\underline{c}\,\underline{a}\,\underline{c}, where a>0.a \gt 0. Find the base-1010 representation of n.n.

Answer: 925
Concepts:number baseplace valueDiophantine Equation
Difficulty rating: 2180
Solution:

Writing out the place values, n=196a+14b+cn = 196a + 14b + c =225a+15c+b= 225a + 15c + b =222a+37c,= 222a + 37c, where aa and cc are base-66 digits with 1a51 \le a \le 5 and 0c5,0 \le c \le 5, and 0b13.0 \le b \le 13.

Equating the last two expressions gives b=22c3a.b = 22c - 3a. Substituting into 196a+14b+c=225a+15c+b196a + 14b + c = 225a + 15c + b (which says 13b=29a+14c13b = 29a + 14c) yields 13(22c3a)=29a+14c,13(22c - 3a) = 29a + 14c, so 272c=68a,272c = 68a, that is a=4c.a = 4c. The digit bounds force c=1,c = 1, a=4,a = 4, and then b=2212=10,b = 22 - 12 = 10, which is a valid digit in bases 1414 and 15.15.

Therefore n=2224+371=925.n = 222 \cdot 4 + 37 \cdot 1 = 925. Indeed 925=1964+1410+1,925 = 196 \cdot 4 + 14 \cdot 10 + 1, confirming the base-1414 form. The answer is 925.925.

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