And this was after a few.
Two aged Number theory specialists sit at a bar.
Anthony said, "It just occurred to me that when we first met, the square of your age contained the same three digits as the square of my age, just in a different order."
Bradley said, "If you take the square of the sum of our ages when we met and split it into two 2-digit numbers, you'd have my age then and your age now"
1234 split into two 2 digit numbers results in 12 and 34.
let B = Bradley's age now and A = Anthony's age now, B/A = x/y where x and y are positive co prime integers.
c^2 = xy+y
If, on the day he met Bradley, Anthony added c to his age, 2c being smaller than Bradley's age (then). what answer would he arrive at?
The answer is 16.
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1 solution
Quite easy. The numbers 256 and 625 satisfy the requirements. Hence Bradley was 16 then and Anthony is 81 now Their present ages are 72 and 81. (81-25= 56 years have passed, hence Bradley is 16+56= 72 years old now). Hence, B/A= 8/9. Hence, c= 9 or -9. c cannot be 9, as 2c=18 is not less than 16. Hence, c=-9; and the result is obtained by adding Anthony's age then with c: 25+ (-9)= 16..