A nice puzzle to end this year!
Max and his younger friend Judy loved Mathematics and liked looking for patterns in numbers. One day, Judy had something amazing to talk about at the table.
"Look at this, Max!" Judy excitedly remarked "If you take the square of the sum of our ages, you'll get a four digit number whose first 2 digits are the age I was when we first met"
"This will really blow your socks off then, Judy!" Max adds "If you cross off the 2nd digit of that 4 digit result, you get the square of the age you were back then!"
"I never realized that!" Judy responds "And the digit you crossed off is exactly the difference of our ages!"
"Isn't Math great?" Max says with a smile on his face
How old are Max and Judy?
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1 solution
Let x be Max's age and y be Judy's age.
Thus the first line says x>y (Statement 1)
"Look at this, Max!" Judy excitedly remarked "If you take the square of the sum of our ages, you'll get a four digit number whose first 2 digits are the age I was when we first met"
Therefore, (x+y)^2 is between 999 and 10000 (since it's a four digit number)
Put the biggest 3 digit perfect square number in place for 999:
961<(x+y)^2<10000
Taking the square root:
31<x+y<100 (Statement 2)
Use the letters A, B, C, and D to represent the digits of (x+y)^2
(x+y)^2 = ABCD
AB is Judy's age when they met according to the last part.
"This will really blow your socks off then, Judy!" Max adds "If you cross off the 2nd digit of that 4 digit result, you get the square of the age you were back then!"
When you cross out B from ABCD you get ACD
So according to this part: (AB)^2=ACD
Since (AB)^2 is a 3 digit number:
(AB)^2<1000
Putting 1024 in place of 1000 since 1024 is the smallest perfect square NOT less than 1000:
(AB)^2<1024
AB<32 (by taking the square root)
Since AB is an integer AB is less than 32 by at least 1
Thus, AB*100 (or AB00) is less than 3200 by at least 100
Therefore, since CD<100 (being a 2 digit number), then ABCD aka (x+y)^2 will still be less than 3200 (since adding CD to AB00 still won't be enough for the at least 100 difference to get to 3200).
ABCD<3200
(x+y)^2<3200
Putting 3249 in the place of 3200 since 3249 is the smallest perfect square not less than 3200:
(x+y)^2<3249
x+y<57 (by taking the square root)
By combining this result with Statement 2 [31<x+y<100]:
31<x+y<57
This leaves only 25 possibilities for x+y. Checking using trial and error, we see 37 is the only possibility which fits [37^2=1369; comparing with ABCD, AB=13 and ACD=169; (AB)^2=13^2=169=ACD, just like we wanted]
x+y=37 and the second digit being crossed out (B) is 3 [Statement 3]
"I never realized that!" Judy responds "And the digit you crossed off is exactly the difference of our ages!"
According to Judy's claim and Statement 3, the difference of their ages is 3.
But does this mean x-y=3 OR y-x=3?
Statement 1 says x>y.
Therefore this must mean x-y=3
Combining this with Statement 3 we have the following system of equations in x and y:
x+y=37 and x-y=3
Adding these equations together gives 2x=40, therefore x=20
Substituting 20 in for x gives 20+y=37, therefore y=17
So the final answer is: Max is 20 and Judy is 17