, 2013

Number theory Difficulty 5.1 AIME, harder Prove it Slovenia

When the third grade pupil Benjamin calculated the sum $1+2+3+\ldots+2012$ , he forgot to add some terms, and he got an incorrect sum that was divisible by $2011$ . When Anika calculated the sum $A = 1+2+3+\ldots+2013$ , she forgot to add the same terms as Benjamin, and she got an incorrect sum $N$ that was divisible by $2014$ . What is the ratio $N/A$ of the two sums?

Solution

Let us denote the sum of the terms omitted by Benjamin by $x$ . Since $1 + 2 + 3 + \ldots + 2012 = \frac{2012 \cdot 2013}{2} = 1006 \cdot 2013$ , Benjamin's result was equal to $1006 \cdot 2013 - x$ . So, there exists a non-negative integer $m$ , such that $1006 \cdot 2013 - x = 2011m$ .

Since $A = 1 + 2 + 3 + \ldots + 2013 = \frac{2013 \cdot 2014}{2} = 2013 \cdot 1007$ , Anika's result was equal to $N = 2013 \cdot 1007 - x$ . So, there exists a non-negative integer $n$ , such that $2013 \cdot 1007 - x = 2014n$ .

Expressing $x$ from both equalities and comparing the results we get $1006 \cdot 2013 - 2011m = 2013 \cdot 1007 - 2014n$ , or $2014n - 2011m - 2013 = 0$ . This last equality can be rearranged to $2011(n-m) = 2013-3n$ .

Since $2014n = 2013 \cdot 1007 - x \le 2013 \cdot 1007$ , we get $n \le \frac{2013 \cdot 1007}{2014} < 1007$ . So, $-1008 < 2013 - 3n \le 2013$ . At the same time $2013-3n$ is divisible by $3$ and the equality implies it is also divisible by $2011$ . This is only possible for $2013 - 3n = 0$ or $n = 671$ .

Hence, $\frac{N}{A} = \frac{2014n}{2013 \cdot 1007} = \frac{2014 \cdot 671}{2013 \cdot 1007} = \frac{2}{3}$ .

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