Question

f(y)=y repeated y times,for example f(3)=333,f(5)=55555.
then a=f(2001)+f(2002)+f(2003)+(2004)+(2005)+.......(2012)+(2013)+(2014)+(2015).
what is the remainder upon division of a by 3 ?

Answer 1

Answer on Question #84836 – Math – Combinatorics | Number Theory Question

$f(y) = y$ repeated $y$ times, for example $f(3) = 333, f(5) = 55555$.

then $a = f(2001) + f(2002) + f(2003) + f(2004) + f(2005) + \ldots$ $f(2012) + f(2013) + f(2014) + f(2015)$.

what is the remainder upon division of $a$ by 3?

Solution

If $x$ has decimal expansion $x_1 x_2 \ldots x_n$, then

$$x (\mathrm{mod} 3) = x_1 + x_2 + \ldots + x_n (\mathrm{mod} 3)$$

Indeed,

$$\begin{array}{l} x = 10^{k_1} x_1 + 10^{k_2} x_2 + \ldots + x_n \\ = (10^{k_1} - 1) x_1 + (10^{k_2} - 1) x_2 + \ldots + (10^{k_{n-1}} - 1) x_{n-1} + (x_1 + x_2 + \ldots + x_n) \end{array}$$

Since $10^k - 1$ is divisible by 3 for any $k$,

$$x (\mathrm{mod} 3) = x_1 + x_2 + \ldots + x_n (\mathrm{mod} 3)$$

This means that $f(y)(\mathrm{mod}3) = y + y + \ldots + y = y * y = y^2(\mathrm{mod}3)$

(here there are $y$ summands)

Then $a(\mathrm{mod}3) = 2001^2 + 2002^2 + 2003^2 + \ldots + 2015^2 (\mathrm{mod}3)$

We have:

$$2001 = 3 * 667,$$

from which

$$2001 = 0 (\mathrm{mod} 3),$$ $$2001^2 = 0^2 (\mathrm{mod} 3), 2002^2 = 1^2 (\mathrm{mod} 3), 2003^2 = 2^2 (\mathrm{mod} 3), \ldots$$

Then

$$\begin{array}{l} a (\mathrm{mod} 3) = \\ = 0^2 + 1^2 + 2^2 + 0^2 + 1^2 + 2^2 + 0^2 + 1^2 + 2^2 + 0^2 + 1^2 + 2^2 + 0^2 + 1^2 + 2^2 (\mathrm{mod} 3) = \\ = (0^2 + 1^2 + 2^2) \cdot 5 (\mathrm{mod} 3) = 25 (\mathrm{mod} 3) = 1 (\mathrm{mod} 3) \end{array}$$

Answer: 1.

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