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Age Ascertainment IV (Posted on 2013-09-18) Difficulty: 3 of 5
  1. Alice is as old as Bob will be when Alice is twice the age that Bob was when Alice’s age was half the sum of their present ages.
  2. The sum of the digits of Alice’s current age and the sum of the digits of Bob’s current age are equal.
If I told you if the difference between the digits in one of their ages is 3 or not - you will be able to tell me their ages.

What are the current ages of Alice and Bob, given that each of their ages is less than 100?

No Solution Yet Submitted by K Sengupta    
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Solution solution?? | Comment 1 of 2

Alice was half the sum of their present ages A - (A+B)/2 years ago.
At that time, Bob was B-(A - (A+B)/2).
Twice that age is 2*(B-(A - (A+B)/2)).
Alice will be that double age in 2*(B-(A - (A+B)/2)) - A years.
At that time, Bob will be B + 2*(B-(A - (A+B)/2)) - A.
Alice is that age now, so:
       A = B + 2*(B-(A - (A+B)/2)) - A
      
       2*A = B + 2*(B-(A - (A+B)/2))
           = 3*B - 2*A + A + B
           = 4*B - A
          
       3*A = 4*B   
      
For A as multiples of 4 (and B as multiples of 3):

            sod
 A  B      A   B
 4  3      4   3
 8  6      8   6
12  9      3   9
16 12      7   3
20 15      2   6
24 18      6   9
28 21     10   3
32 24      5   6
36 27      9   9 *
40 30      4   3
44 33      8   6
48 36     12   9
52 39      7  12
56 42     11   6
60 45      6   9
64 48     10  12
68 51     14   6
72 54      9   9 *
76 57     13  12
80 60      8   6
84 63     12   9
88 66     16  12
92 69     11  15
96 72     15   9

In only two cases, marked with *, are the sod's equal for the two ages. In one of these cases one of the ages (Alice's) has digits that differ by 3; in the other case neither has age digits that differ by 3. So either way telling whether or not that's the case would decide the ages.


  Posted by Charlie on 2013-09-18 12:30:43
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