Mathathon 2020 Problem 5 $-$ Hard

Let’s name the characters’ ages after their initials: $A,B,C,D,E.$

Look at this sentence.

Cody said to Alex:’I am ten years older than you.‘

If Cody’s the younger one, he is telling the $\color{#D61F06} truth$ . This leads to a contradiction so Cody is older, i.e. $C\gt A.$

Now we know that the statements

‘Brook is younger than Dusty’ and ‘Cody is ten years older than Alex’

are $\color{#3D99F6}false$ , ergo Brook is older than Dusty( $B\gt D$ ).

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Now since Brook’s older than Dusty, the statement

‘Dusty is nine years older than Erin’

Is $\color{#D61F06} true$ .
Now we know $C\gt B\gt D\gt E$ and $C\gt A$ .

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Erin said to Brook: ‘I'm seven years older than Alex.’

This is obviously true. So now we know $C\gt B\gt D\gt E\gt A$ .

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So these are the $\color{#D61F06} true$ statements:

Dusty said to Brook: ‘I'm nine years older than Erin.’
Erin said to Brook: ‘I'm seven years older than Alex.’
Alex said to Brook: ‘Your age is exactly 70% greater than mine.’
Brook said to Cody: ‘Erin is younger than you.’
Brook said to Cody: ‘The difference between your age and Dusty's is the same as the difference between Dusty's and Erin's.’

So these give us a set of equations:(the numbers in the brackets are the lines in the quote for reference) $\begin{cases} D=E+9,(1)\\ E=A+7,(2)\\ C=D+9=E+18=A+25,(1,2,5)\\ 170\%A=B.(3) \end{cases}$ Now look at those two equations: $C=A+25,170\%A=B,D+9=A+25(D=A+16)$ . Look at the inequality from the last section(the age sequence) to get $A+25\gt 170\%A\gt A+16.$ Since we assume the ages are all integers, A must be divisible by 10 for 170%A to be an integer. So $A=30$ , and everyone’s ages are $C=55,B=51,D=46,E=37,A=30.$ So the answer is $55+51+46+37+30=\boxed{\colorbox{#CEBB00}{219}}.$

$\text{As a beginning we can name the characters after their initials: A=Alex, B=Brook, C=Cody, D=Dusty, and E=Erin.}$

$\text{Then we can check if certain combinations are impossible.}$

$\color{#EC7300}{\textit{Is it possible that E is the youngest person?}}$

$\textit{No, because E would have to equal A+7, which is a contradiction if E is the youngest.}$

$\color{#EC7300}{\textit{Is it possible that D is the youngest person?}}$

$\textit{ No, because D would have to equal E+9, which is a contradiction if D is the youngest.}$

$\color{#EC7300}{\textit{Is it possible that C is the youngest person?}}$

$\textit{ No, because C would have to equal A+10, which is a contradiction if C is the youngest.}$

$\textbf{Chart of possible solutions}$

$\text{But we can ask more questions...}$

$\color{#EC7300}{\textit{Can Brook be the oldest? (btw. Alex is the youngest then because, as proved above, no one else can be the youngest.)}}$

$\textit{No. E would be greater than C, according to statement 4, but C would equal A+10 but E only A+7, which is a contradiction.}$

$\color{#EC7300}{\textit{Can Alex be older than Cody?}}$

$\textit{ No, A cannot be greater than C because statement 6 would have to be true but statement 6 explicitly states that C is A+10.}$

$\text{Therefore we know that A<C. Due to that we can deduce from claim 7 that D<B.}$

$\text{This also makes it impossible for Brook to be the youngest or the 2nd youngest.}$

$\textbf{Chart of possible solutions}$

$\text{ As D < B and D = E + 9 are both valid, Dusty can not be the youngest or the 2nd youngest. And Erin has to be younger than him.}$

$\text{In addition to that, Cody has to be the oldest as Brook has been proven to not be the oldest, Erin has to be younger than Dusty, and Dusty is supposed to be younger than Brook. And Alex has been proven to be the youngest.}$

$\textbf{This gives us only one possible solution.}$

$\textbf{Can it withstand all claims?}$

• 1.D=E+9
• 2.E=A+7
• 3.B= 1.7 times A
• 4.E<C
• 5.CD is not 6
• 6.C is nor A+10
• 7.D<B
• 8.CD=DE

$\text{There is not contradiction in these statements. Now we only need to solve for A.}$

$\text{First equation:}$

$A+16 < 1.7 \cdot A$ $16 <0.7 \cdot A$ $22.85 < A$

$\text{Second equation:}$

$1.7 \cdot A < A + 25$ $0.7 \cdot A < 25$ $A < 35.7$

$\text{We know that all ages are integers. But 1.7 x n can only be an integer if n is a multiple of 10.}$

$\text{As there is only one multiple of 10 in the interval 23<A<35, we know that A=30.}$

$A = 30$ $E = A + 7 = 30 + 7 = 37$ $D = A + 16 = 30 + 7 = 46$ $B = A \cdot 1.7 = 30 \cdot 1.7 = 51$ $C = A + 25 = 30 + 25 = 55$

$\text{Now we only need to add those up...}$

$A + E + D + B + C = 30 + 37 + 46 + 51 + 55 = 219$

$\color{forestgreen}{\text{The sum of their ages is 219.}}$

Let the age of Alex, Brook, Cody, Dusty, and Erin be $A,B,C,D$ and $E$

Above statements can be translated as follows : $D<B\Leftrightarrow D=E+9..........[1]$ $E<B\Leftrightarrow E=A+7..........[2]$ $A<B\Leftrightarrow B=\dfrac{17}{10}A..........[3]$ $B<C\Leftrightarrow E<C..........[4]$ $C<D\Leftrightarrow |C-D|=6..........[5]$ $C<A\Leftrightarrow C=A+10..........[6]$ $C<A\Leftrightarrow B<D..........[7]$ $B<C\Leftrightarrow |C-D|=|E-D|..........[8]$ In $[6]$ there are $3$ cases

Case $1$ : $C<D$ is $\green{true}\Rightarrow C\cancel{=}A+10$

$\Rightarrow$ statement $[6]$ is $\red{contradicted}$

Case $2$ : $C=A+10$ is $\green{true}\Rightarrow C\cancel{<}D$

$\Rightarrow$ statement $[6]$ is $\red{contradicted}$

Case $3$ : $C<A$ and $C=A+10$ both are $\red{false}$

$\Rightarrow$ statement $[6]$ is $\green{correct}$

Therefore the only possible case is the third case.

So, $C>A\Rightarrow D<B\Rightarrow D>E\Rightarrow E<B\Rightarrow E>A$ From $[6],[7],[1]$ and $[2]$ we get $C>B>D>E>A..........[\alpha]$ And $E-A=7..........[I]$ $D-E=9..........[II]$ $C-D=9..........[III]$ $[I]+[II]\Rightarrow D-A=7+9=7+(10-1)=17-1=16..........[IV]$ $[III]+[IV]\Rightarrow C-A=16+9=16+(10-1)=26-1=25..........[V]$ Subtracting $A$ in $[\alpha]$ $C-A>B-A>D-A>0..........[\beta]$ From $[3],[\beta],[IV]$ and $[V]$ we get $25>\dfrac{17}{10}A-A>16>0$ $25>\dfrac{7}{10}A>16>0\quad\blue{\because \dfrac{17}{10}A-A=\dfrac{17A-10A}{10}=\dfrac{7}{10}A}$ Now as $(B-A)\in Z^+;B-A=7\times \dfrac{1}{10}A\Rightarrow 7|(B-A)$

Now as $21$ is the only multiple $7$ between $16$ and $25$ $\Rightarrow B-A=21$ $\dfrac{7}{10}A=21\quad \blue{\because B-A=\dfrac{7}{10}A\space see\space [3]}$ $A=21\times\dfrac{10}{7}=10\times\dfrac{21}{7}=10\times3=30$ $\boxed{A=30}$ From $[3]$ $B=\dfrac{17}{10}A=\dfrac{17}{1\cancel{0}}\times3\cancel{0}=17\times3=51\Rightarrow\boxed{B=51}$ From $[I]$ $E-A=7\Rightarrow E=7+A\Rightarrow E=30+7=37\Rightarrow \boxed{E=37}$ From $[II]$ $D-E=9\Rightarrow D=9+E=9+37=46\Rightarrow \boxed{D=46}$ From $[III]$ $C-D=9\Rightarrow C=9+D=9+46=55\Rightarrow \boxed{C=55}$ Therefore, $\blue{A}\green{+}\red{B}\green{+}\blue{C}\green{+}\red{D}\green{+}\blue{E}=\blue{30}\green{+}\red{51}\green{+}\blue{55}\green{+}\red{46}\green{+}\blue{37}=\boxed{219}$