My Son and Me
Math brain teasers require computations to solve.
If you reverse the digits of my age, you
have the age of my son. A year ago, I was
twice his age. How old are we both now?
Answer
Father: 73; Son: 37
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Comments
Sane  
Mar 12, 2005
| I think this is the furthest you can go with an equation. I tried forever but I don't think you can go any further without doing plain trial and error.
let (a) and (b) represent the digits of my age, then (y) and (z) represent the digits of my son's age, where (10a + b) is my age and (10y + z)
is my son's age:
1) [if you reverse the digits of my age, you have the age of my son]
10b + a = 10y + z
2) [a year ago, I was twice his age]
10a + b - 1 = (10y + z - 1) * 2
10a + b - 1 = 20y + 2z - 2
10a + b = 20y + 2z - 1
3) [exclude 10y + z to set up for substitution]
[10a + b] / 2= [20y + 2z - 1] / 2
5a + b/2 = 10y + z - (.5)
5a + b/2 + 1/2 = 10y + z
4) [combine equation 1 and 3]
5a + b/2 + 1/2 = 10b + a
(4a + 1/2) / 4 = (9.5b)/4
a + 0.125 = 9.5b/4
a = 9.5b/4 - 0.125
a = 2.375b - 0.125
The only two integers between 0 and 9 that can make that above statement true is:
a = 7
b = 3
Therefore the ages are 37 and 73. |
jntrcs
Apr 01, 2006
| zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz i liked the puzzle but that explanation wa ssooooooooooooo over my head guess it would help if i had already taken algebra but man that's confusing why did you type that up? |
stil 
Apr 23, 2006
| Notation notes: an uppercase letter represents a digit as a symbol and a lowercase letter represents the same digit's value. AB = 10*a + b.
The father's age is currently AB and the son's is BA. therefore: a>b.
Last year (AB-1)=2*(BA-1). Expanding 10*a + b - 1 = 20*b + 2*a - 2.
8*a = 19*b - 1.
a = (19*b -1) / 8.
(b can not be an even integer else the product 19*b would be even and subtracting 1 would give an odd number, any of which would not be divisible by 8.)
a = (19/ *b - 1/8.
a = 2.375*b - 0.125.
(For b=>1, a is at least twice b, so b can not be as great as 5.)
With 0, 2, and 4 previously eliminated, 1 and 3 are the only integers to consider, and it clearly is not 1. |
stil
Apr 23, 2006
| Notation notes: an uppercase letter represents a digit as a symbol and a lowercase letter represents the same digit's value. AB = 10*a + b.
The father's age is currently AB and the son's is BA. therefore: a>b.
Last year (AB-1)=2*(BA-1). Expanding 10*a + b - 1 = 20*b + 2*a - 2.
8*a = 19*b - 1.
a = (19*b -1) / 8.
(b can not be an even integer else the product 19*b would be even and subtracting 1 would give an odd number, any of which would not be divisible by 8.)
a = (19/*b - 1/8.
a = 2.375*b - 0.125.
(For b=>1, a is at least twice b, so b can not be as great as 5.)
With 0, 2, and 4 previously eliminated, 1 and 3 are the only integers to consider, and it clearly is not 1. |
stil
Apr 23, 2006
| a = (19/*b - 1/8.
has accidental emoticon obscuring a = ( 19/8 )*b - 1/8. |
Jimbo
Apr 30, 2007
| Stil, if you change the subject of your equation you get a = (8a+1)/19. So 8a+1 has to be divisible by 19. Testing quickly 8a+1 for a = 1,2,3,4,5,6,7 we find 8x7+1=57 (3x19) is divisible by 19. Therefore a=7, b=3. |
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