We are 3 brothers, we like multiplying our ages each other.
m years later the result is m times than today
n years later the result is also n times than today
How old we are?
(m and n is not equal)
If an integer answer is needed, they can be
$5$, $6$ and $12$ years old,
while $n$ and $m$ are
$3$ and $4$.
Check:
now:
$5\times6\times12=360$
$3$ years later:
$(5+3)\times(6+3)\times(12+3)=8\times9\times15=1080=360\times3$
$4$ years later:
$(5+4)\times(6+4)\times(12+4)=9\times10\times16=1440=360\times4$
I think they are
$5$, $7.5$ and $10$ years old
$n$ and $m$ are
$2.5$ and $5$
Checking the results:
now:
$5\times7.5\times10=375$
$2.5$ years later:
$(5+2.5)\times(7.5+2.5)\times(10+2.5)=7.5\times10\times12.5=937.5=375\times2.5$
$5$ years later:
$(5+5)\times(7.5+5)\times(10+5)=10\times12.5\times5=1875=375\times5$
Here's my first python code/analysis to get the result. Ignore optimization (I just needed to practice my python :-) ) and assume integer ages
a=[]
b=[]
c=[]
for i in range(1,121):
a.append(i)
b.append(i)
c.append(i)
for a_age in a:
for b_age in b:
for c_age in c:
if (a_age>=b_age and a_age>=c_age and b_age>=c_age):
today = a_age*b_age*c_age
m=0
while m<120:
m = m+1
r1 = today*m
r2 = (m+a_age) * (m+b_age) * (m+c_age)
n=m
while (n<120):
n=n+1
r3 = today*n
r4 = (n+a_age) * (n+b_age) * (n+c_age)
if r1==r2 and r3==r4:
print ("found it ",a_age , " ", b_age ," ",c_age, " where m=",m, "and n=",n)
and the result comes as:
('found it ', 12, ' ', 6, ' ', 5, ' where m=', 3, 'and n=', 4)
we like multiplying our ages each other
EXACTLY mean? If they're x, y, z years old, that meansx*y*z
, right? $\endgroup$