How Old?
If you transpose the digits of Abe's age, you get Bill's age. The difference between Abe's and Bill's ages is twice Claude's age. Bill is ten times as old as Claude.
What color is the bear? ... oh wait ... wrong problem. What are the ages?
What color is the bear? ... oh wait ... wrong problem. What are the ages?
posted by Zaux at
4:34 PM
16 Comments:
They are all 0
Cam
or just Claude is 0
Cam
A=54,B=45,C=4.5 works if fractional age of Claude is allowed
Cam
Haven't got the answer (and shan't tonight) but Abe and/or Bill must be over a 100 years old. Otherwise Cam is right.
... assuming integer ages.
LOL, 540, 450 and 45 does it(thanks Cam).
I guess Abe, Bill and Claude are trees.
How Old ?
Assume <100 years old
10x+y=A
10y+x=B
where x an y are integers from 0 to 9
A-B=9x-9
So if A>B
9x-9y=2C
x=2/9*C+y
B=10C
10y+x=10C
sub in x eq
10*y+2/9*C+y=10C
11y=(10-2/9)C
11y=88/9*C
y=8/9*C
so if C=9 y=8
then from 10y+x=10C we have 80+x=90
x=10 ??, x must be from 0 to 9
if fractional age of C allowed
C=9/8*y
10y+x=10C
x=10*(9/8*y)-10*y=10/8*y
try y=0 to 9 to find integer solutions <=9 for x
y=0 x=0 trivial
y=4 x=5
so A=54,B=45, C=4.5
Test if B>A
9x-9y=-2C
x=-2/9*C+y
B=10C
10y+x=10C
sub in x eq
10*y-2/9*C+y=10C
11y=(10+2/9)C
11y=92/9*C
y=92/99*C
if fractional age of C allowed
C=99/92*y
10y+x=10C
x=10*(99/92*y)-10*y=70/92*y
try y=0 to 9 to find integer solutions <=9 for x
y=0 x=0 trivial
no integer solutions
Answer:
If fractional age for Claude is allowed
Abe=54,
Bill=45,
Claude=4.5
Cam
Why stop at tress ? Why not make them stars ?
Abe 54*10^12, Bill 45*10^12, Claude 4.5*10^12
Cam
nothing wrong with someone being 4.5 years of age ... good job Cam... as usual
infinity, infinity and infinity.
Assuming Bill older than Abe:
10y + x - (10x + y) = 0.2(10y + x)
9y - 9x = 0.2(10y + x)
45y - 45x = 10y + x
35y = 46x
(5*7)y = (2*23)x
x = 35, y = 46 is the smallest solution (in the positive integers), so Bill cannot be older than Abe.
Therefore, Abe must be older than Bill =>
(10x + y) - (10y + x) = 0.2 (10y + x)
9x - 9y = 0.2 (10y + x)
45x - 45y = 10y + x
44x = 55y
4x = 5y
=> x = 5, y = 4 is the only possible solution if the ages are less than 100.
So Abe is 54, Bill is 45 and Claude is 4.5 is the only possible answer for Abe and Bill < 100.
small slip. I've assumed Abe and Bill have an integral age.
5.4, 4.5 and 0.45 also work. The question makes it vaguely clear (I jest) that Abe and Bill have ages that must be repesented with two digits.
At a pinch could say, .54, .45 and 0.045. I think 0.54, 0.45 would be pushing it a bit.
Hi Zaux. Despite been miffed at the non-integral value of Claude's age (I do accept that that is fair), I think it was a fine problem. Thank you for posting it.
I know the answer!! the bear is purple!!! lol to much math!
:P
...noooooo. The unicorns were purple. The bear was brown.
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