The problem is proposed for the 5th grade, at page 6.
" Determine the natural number n for which:
2^{n+11}-2^{n+6}+2^{n+5}+2^{n+2}-2^n=2019."
ANSWER CiP
n=0;
Indeed: 2^{11}-2^6+2^5+2^2-2^0= 2048-64+32+4-1=2019.
Solution CiP
If the number n>0, then all the powers of 2 on the left side of the equation are even numbers, so the result of their sum is an even number, so it cannot be equal to 2019.
All that remains is to check the number n=0, and it is found that he verifies the equation.
\blacksquare
REMARKS CiP
1^R. We will use the following formula
2^k\;+\;2^k\;=\;2^{k+1},\;\;\;2^{k+1}-2^k=2^k,\;\;k=0,1,2,...
2^R. The number in base ten 2019 is written in base 2 as follows:
2019_{(10)}=111\;1110\;0011_{(2)}
\Leftrightarrow\;\;2019=2^{10}+2^9+2^8+2^7+2^6+2^5+2^1+2^0.
As it is known, this writing is unique, there is no one like it. Replacing the terms 2^k with the second formula in Remark 1^R, with the exception of 2^5, we have:
2019=(2^{11}-2^{10})+(2^{10}-2^9)+(2^9-2^8)+(2^8-2^7)+(2^7-2^6)\;\underline{+2^5}\;+(2^2-2^1)+(2^1-2^0).
If we open all the parentheses and keep in mind that x-y+y=x,\;\;x,y,z \in \mathbb{N}, we get:
2019=2^{11}-2^6+2^5+2^2-2^0.
3^R. The given equation can still be written, also with the formulas from the mentioned remark:
2^{n+10}+2^{n+10}-2^{n+6}+2^{n+5}+2^{n+1}+2^{n+1}-2^n=2019
\Leftrightarrow\;2^{n+10}+(2^{n+10}-2^{n+9})+(2^{n+9}-2^{n+8})+(2^{n+8}-2^{n+7})+(2^{n+7}-2^{n+6})+2^{n+5}+(2^{n+2}-2^{n+1})+(2^{n+1}-2^n)=2019
\Leftrightarrow\;2^{n+10}+2^{n+9}+2^{n+8}+2^{n+7}+2^{n+6}+2^{n+5}+2^{n+1}+2^n=2019,
and considering the unique writing in base 2, we obtain the only possibility n = 0.
\square\;\square\;\square
