## Subject IV (7 points)

Consider the set of numbers formed only from the digits 1 and 2, with at most 2016 digits.

a) Calculate how many numbers have 5 digits.

b) Calculate how many numbers are in the set.

c) Show that there exists a number in the set that is divisible by \(3^6\).

Note:

All subjects are mandatory.

Working time 2 hours.

## Grading and correction guidelines for 5th grade
## Subject IV

Consider the set of numbers formed only by the digits 1 and 2, with at most 2016 digits.

a). Calculate how many numbers have 5 digits.

b). Calculate how many numbers are in the set.

c). Show that there exists a number in the set divisible by \(3^6\).

## Solution

a) A number with 5 digits of this form has 2 possibilities for each digit, 1 and 2, so there are \(2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^5 = 32\) numbers. \(2 p\)

b) For each number \(1 \leq k \leq 2016\), there are \(2^k\) numbers with \(k\) digits in the set.

Thus, the number of elements in the set is

\[ N = 2 + 2^2 + 2^3 + \ldots + 2^{2016} \]

\(.1 p\)

Adding 2, we get

\[ N + 2 = (2 + 2) + 2^2 + \ldots + 2^{2016} = 2^2 + 2^2 + 2^3 + \ldots + 2^{2016} = (2^3 + 2^3) + \ldots + 2^{2016} = \ldots \]

\[ = 2^{2016} + 2^{2016} = 2^{2017} \]

Thus, \( N = 2^{2017} - 2 \).

\(.2 p\)

c). Consider the 2016 numbers formed only by the digit 1. Since \(2016 = 2^5 \cdot 9 \cdot 7 > 3^3 \cdot 3^2 \cdot 3 = 3^6\), at least two of these numbers give the same remainder when divided by \(3^6\).

\(1 p\)

The difference between these two numbers is of the form \(111 \ldots 11000 \ldots 0 = 11 \ldots 1 \cdot 10^t\) and is divisible by \(3^6\). Since 3 and 10 have no common divisors, \(3^6\) divides the number formed only by the digit 1, which belongs to the set \(A\).