p1. Find all real numbers that satisfy the equation $$(1 + x^2 + x^4 + .... + x^{2014})(x^{2016} + 1) = 2016x^{2015}$$ p2. Let $A$ be an integer and $A = 2 + 20 + 201 + 2016 + 20162 + ... + \underbrace{20162016...2016}_{40\,\, digits}$ Find the last seven digits of $A$, in order from millions to units. p3. In triangle $ABC$, points $P$ and $Q$ are on sides of $BC$ so that the length of $BP$ is equal to $CQ$, $\angle BAP = \angle CAQ$ and $\angle APB$ is acute. Is triangle $ABC$ isosceles? Write down your reasons. p4. Ayu is about to open the suitcase but she forgets the key. The suitcase code consists of nine digits, namely four $0$s (zero) and five $1$s. Ayu remembers that no four consecutive numbers are the same. How many codes might have to try to make sure the suitcase is open? p5. Fulan keeps $100$ turkeys with the weight of the $i$-th turkey, being $x_i$ for $i\in\{1, 2, 3, ... , 100\}$. The weight of the $i$-th turkey in grams is assumed to follow the function $x_i(t) = S_it + 200 - i$ where $t$ represents the time in days and $S_i$ is the $i$-th term of an arithmetic sequence where the first term is a positive number $a$ with a difference of $b =\frac15$. It is known that the average data on the weight of the hundred turkeys at $t = a$ is $150.5$ grams. Calculate the median weight of the turkey at time $t = 20$ days.
Problem
Source:
Tags: algebra, geometry, combinatorics, number theory, indonesia juniors
10.11.2021 12:21
10.11.2021 14:57
11.11.2021 04:39
parmenides51 wrote: p2. Let $A$ be an integer and $A = 2 + 20 + 201 + 2016 + 20162 + ... + \underbrace{20162016...2016}_{40\,\, digits}$ Find the last seven digits of $A$, in order from millions to units. $10^0 \cdots \cdots (2+0+1+6)\times \frac{40}{4}=90 \cdots \cdots \boxed{0}$ $10^1 \cdots \cdots (2+0+1+6)\times 9+(2+0+1) +9 =93 \cdots \cdots \boxed{3}$ $10^2 \cdots \cdots (2+0+1+6)\times 9+(2+0) +9 =92 \cdots \cdots \boxed{2}$ $10^3 \cdots \cdots (2+0+1+6)\times 9+(2) +9 =92 \cdots \cdots \boxed{2}$ $10^4 \cdots \cdots (2+0+1+6)\times 9 +9 =90 \cdots \cdots \boxed{0}$ $10^5 \cdots \cdots (2+0+1+6)\times 8+(2+0+1) +9 =84 \cdots \cdots \boxed{4}$ $10^6 \cdots \cdots (2+0+1+6)\times 8+(2+0) +8 =82 \cdots \cdots \boxed{2}$ So, the last seven digits of $A$ are $\boxed{2402230}$.
11.11.2021 15:38
S1
However, the $x=-1$ should be discarded because it is a mere artifact coming from the $X^2$, which option is moreover not compatible with the original equation (presence of the odd power "2015"). @PhiloMath below: it has not been prescribed by the family doctor that only AM-GM should be used for arriving at a result.
11.11.2021 16:37
19.11.2021 21:03
iniffur wrote: S1
This is not a correct solution Note: @above has the correct solution.
13.07.2022 07:46
no, the solution thatTalmon proposed is right. Note that x cannot equal to zero, so we can divide both sides by x^2015. Notice that, it'll become what Talmon mean.
13.07.2022 12:51
@BackToSchool Can you please explain what you did and why,in a bit more explanative way,I don't get it
14.07.2022 00:58
19.07.2022 03:51
DanReynolds wrote: @BackToSchool Can you please explain what you did and why,in a bit more explanative way,I don't get it Here is a solution for you.
19.07.2022 17:06
I probably made a mistake somewhere so if someone could check it that would be great. Or if there is a less bashy non PIE solution.