Three students write on the blackboard next to each other three two-digit squares. In the end, they observe that the 6-digit number thus obtained is also a square. Find this number!
Let the number be denoted by N^2, where N is a positive integer.
Then by the problem, N^2 must have the form abcd
Since, each of ab,cd and ef is a 2-digit perfect square, we must have:
ab=x^2, cd=y^2 and ef=z^2,
Then, (ab,cd,ef)€ (16,25,36,49,64,81)
=> (x,y,z) € (4,5,6,7,8,9)
Now, N^2=abcdef
= 10^4*x^2 +10^2*y^2+z^2
Knowing that 4<= x,y,z <=9, we obtain:
((10^2)*x)^2 < 10^4*x^2 +1616
<= 10^4*x^2 +10^2*y^2+z^2
<= 10^4*x^2 +8181
< 10^4*x^2 +2200x +11^2, since we know that x>=4
=(10^2*x+11)^2
Accordingly, 10^2*x +1 < N < 10^2*x+11
If N is a multiple of 10, then ef=00, a contradiction.
Therefore, N = 10^2*x+t (say), whenever 1<=t<=9
Then, t^2 =ef=z^2=> t=z
Also, 2xz=2tx=cd=y^2
Accordingly, xz must equal precisely twice a perfect square.
Therefore x, z cannot belong to (5,6,7) ss none of the 3 pairs multiples to a perfect square.
So, x,z € (4,8,9)
Now, 2xz=y^2
Accordingly, 2xz=81=9^2 is a contradiction, with its rhs being odd.
So, 2xz<81=> (x,z)€(4,8)
Hence, (x,z)=(4,8),(8,4)
Then, 2xz=y^2 gives:
y^2=2*4*8
=> y=8
Hence, N^2 = abcdef= 166464 OR 646416
This checks out, since:
V(166464)=408, a positive integer, and:
V(646416)=804, a positive integer.
Consequently, the 6-digit number written on the blackboard was EITHER 166464 OR 646416.
Edited on January 8, 2022, 9:40 pm
Analytical Puzzle Solution
