2018 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 25 Hard

For a positive integer n and nonzero digits a, b, and c, let A_n be the n-digit integer each of whose digits is equal to a; let B_n be the n-digit integer each of whose digits is equal to b, and let C_n be the 2n-digit (not n-digit) integer each of whose digits is equal to c. What is the greatest possible value of a+b+c for which there are at least two values of n such that C_n-B_n=A_n^2? (2018 AMC 10A Problem, Question#25)

  • A.

    12

  • B.

    14

  • C.

    16

  • D.

    18

  • E.

    20

Answer:D

Observe A_n=a(1+10+\cdots +10^{n-1})=a\cdot \frac{10^n-1}{9};

similarly B_n=b\cdot \frac{10^n-1}{9} and C_n=c\cdot \frac{10^{2n}-1}{9}. The relation C_n-B_n=A_n^2 rewrites as c\cdot \frac{10^{2n}-1}{9}-b\cdot \frac{10^n-1}{9}=a^2\cdot \left(\frac{10^n-1}{9}\right)^2.  Since n>0, 10^n>1 and we may cancel out a factor of \frac{10^n-1}{9} to obtain c\cdot (10^n+1)-b=a^2\cdot \frac{10^n-1}{9}. This is a linear equation in 10^n. Thus, if two distinct values of n satisfy it, then all values of n will. Matching coefficients, we need c=\frac{a^2}{9} and c-b=-\frac{a^2}{9}\Rightarrow b=\frac{2a^2}{9}. To maximize a+b+c=a+\frac{a^2}{3}, we need to maximize a. Since b and c must be integers, a must be a multiple of 3. If a=9 then b exceeds 9.

However, if a=6 then b=8 and c=4 for an answer of \boxed{\rm (D)~18}.

Immediately start trying n=1 and n=2. These give the system of equations 11c-b=a^2 and 1111c-11b=(11a)^2 (which simplifies to 101c-b=11a^2). These imply that a^2=9c, so the possible (a,c) pairs are (9,9), (6,4), and (3,1). The first puts b out  of range but the second makes b=8. We now know the answer is at least 6+8+4=18.

We now only need to know whether a+b+c=20 might work for any larger n. We will always get equations like 100001c-b=11111a^2 where the c coefficient is very close to being nine times the a coefficient. Since the b term will be quite insignificant, we know that once again a^2 must equal 9c, and thus a=9,c=9 is our only hope to reach 20. Substituting and dividing through by 9, we will have something like 100001-\frac{b}{9}=99999. No matter what n really was, b is out of range (and certainly isn't 2 as we would have needed).

The answer then is \boxed{\rm (D)~18}.

Notice that \left( 0.\overline{3}\right)^{2}=0.\overline{1} and \left( 0.\overline{6}\right)^{2}=0.\overline{4}. Setting a=3 and c=1, we see b=2 works for all possible values of n. Similarly, if a=6 and c=4, then b=8 works for all possible values of n. The second solution yields a greater sum of \boxed{\rm (D)~18}.

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