Masters Of Age Theory

When Ash and Bill first met, their squares had only three digits, so the oldest they could have been was 31. That means that there were only a few possible ages the two of them could have been. Those being : 14 & 31 (196 & 961), 13 & 31 (169 & 961), 16 & 25 (256 & 625), 12 & 21 (144 & 441), 13 & 14 (169 & 196).

Now the square of the sum of each of these pair of ages are respectively: 2025, 1936, 1681, 1089, 729.

Bill was 16 when they first met and Ash is 81 now.

Least and largest numbers with a square of 3 digits=11 and 31 respectively. Numbers between 11 and 31 (inclusive) having similar digits in their square :( 13,14,31) ,(16,25),(12,21) . Taking squares of : 13+14=729 14+31=2025 13+31=1936 16+25=1681 12+21=1089 Only 1681 satisfies the given criteria.

There are only two numbers i.e 25 & 16 whose squares have the same digits 625 and 256. So summing them and squaring gives us 41^2 = 1681. So 16.81 is the answer.

25^2=625;16^2=256; 6,2&5 are in different order in two squared numbers.16+25=41;41^2=1681;16 is one of the ages of them when they first met.so the ans is 16.81.

When bill and ash meet, their ages square has only three digit, and which are common in sq. of both. First two no. represent bill's age when they first met and other two digit represent ash's present age, so their should be a comparable difference b/w the two. So, their are only two possible outcomes i.e 16 & 25{256 & 625} or 12 & 21{144 &441}. square's = 1681 & 1089 respectively. Hence, Bill was 16 when they first met and now.Ash is 81

16^2=256 and 25^2=625. Then the square sum of both age is: 41^2=1681 where the first 2 digits is the age of B when they first met. The last 2 digits is A 's age now

Bills age when they first met is 16 and Ash age is 25. (16)^2=256 & (25)^2=625 which have same digits arranged differently. 16+25=31. (41)^2=1681. 16 is age of bills then and 81 age of ash now.