7400
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1
2011-02-01T00:00:01
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<br />
Joseph recently noticed that this time next year he will be twice as old as Jasmine.<br />
<br />
I couldn't help noticing that when I am 66 at this time of year the combined future ages of Joseph and Jasmine will also be 66.<br />
<br />
Also, 6 years from now my future age will be a multiple of the sum of Joseph and Jasmine's future ages. Of course, all ages are non-negative integers.<br />
<br />
How old are Joseph, Jasmine and I now?<br />
<br />
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<mdoxml version="1.0"><span class="editorial">Thanks to Thomas Udale from Winchester College for this solution, which was the age of me and my two children at the time of writing the problem :-)</span><br></br>
<br></br>
Joseph's age = x<br></br>
Jasmine's age = y<br></br>
your age = z<br></br>
<br></br>
from sentence 1 : x+1=2(y+1)<br></br>
from sentence 2: 66 = x+y+2n = z+n<br></br>
from sentence 3: z+6=m(x+y+12)<br></br>
<br></br>
z &lt; 66<br></br>
z+6 &lt; 72<br></br>
m(x+y+12) &lt; 72<br></br>
m(x+y+12) = m(3y+13)<br></br>
<br></br>
m must be a positive integer, because it is the multiple, greater than 1<br></br>
<br></br>
if m = 4, then m(3y+13) &gt; 72<br></br>
<br></br>
if m = 2:<br></br>
then z+6 = 6y+26<br></br>
from equation 1 into equation 2 : z + n = 3y +1 +2n<br></br>
z = 6y+ 20 = 3y + 1 + n<br></br>
n = 3y + 19<br></br>
equation 2:<br></br>
3y + 2(3y+12) +1 = 66<br></br>
y = 3,<br></br>
x = 7,<br></br>
z = 38,<br></br>
<br></br>
if m = 3,<br></br>
following same as above,<br></br>
n = 6y + 38<br></br>
3y + 2(6y+38) + 1 = 66<br></br>
15y + 77 = 66,<br></br>
so y would be negative.</mdoxml>
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<mdoxml version="1.0"><br></br>
Let Joseph be $x$ years old now, Jasmine be $y$ years old now, and
me be $z$ years old now.<br></br>
<br></br>
From the given information, we find three simulataneous
equations.<br></br>
<br></br>
$x+1=2(y+1) \Rightarrow x=2y+2$<br></br>
$z+6=k(x+6+y+6) \Rightarrow z=k(3y+13)-6$<br></br>
$z+m=66=x+m+y+m \Rightarrow z=3y+1+m$<br></br>
where k and m are some positive integers.<br></br>
<br></br>
From this we deduce: $$66=3y+1+2m \Rightarrow m=\frac{65-3y}{2}
$$<br></br>
We know the two equations: $z=66-m=66-\frac{65-3y}{2}$ and
$z=k(3y+13)-6$<br></br>
<br></br>
Equating the two, we find: \begin{eqnarray*}
66-\frac{65-3y}{2} &=& k(3y+13)-6 \\ 27 &=&
y(6k-3) \end{eqnarray*}<br></br>
<br></br>
If $k< 3$ then y is not an integer. So $k=2 \Rightarrow
y=3$<br></br>
<br></br>
From this it is easy to find that x=7 and z=38.<br></br>
<br></br>
<br></br>
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Weekly Challenge 50: How old?
How old are these three people?