p, q, r stand for three different digits and pq, qr, pqr etc stand for two, three digit numbers.?

Question

If qr multiplied by qr = pqr, find the values of p, q, r.

Answer 1

One possible choice is p = 6, q = 2, r = 5

qr = 25
pqr = 625

qr x qr = 25 x 25 = pqr = 625

I'm not positive that's the only answer, but that's one.

Answer 2

qr is actually the number (10q + r) in the decimal
system and pqr is actually (100p + 10q + r).

So the equation to solve is :
(10q + r)^2 = 100p + 10q + r

or, 100q^2 + 20qr + r^2 = 100p + 10q + r (Equ.A)

For the last digit of r^2 to be the same as r,
r must be 0, 1, 5 or 6, as sahsjing says.

1) Let r = 0 in Equ.A.
Then, 100q^2 = 100p + 10q
Here we see on the RHS that q must be 0 or a
multiple of 10, both of which are untrue. So, r ≠ 0.

2) Let r = 1 in Equ.A.
Then, 100q^2 + 20q + 1 = 100p + 10q + 1
or, 100q^2 + 10q = 100p
10q must be a multiple of 100, so either q = 0 or a
multiple of 10, both of which are again untrue. So, r ≠ 1.

3) Let r = 5 in Equ.A.
Then, 100q^2 + 100q + 25 = 100p + 10q + 5
or, 100q^2 + 90q + 20 = 100p
Here, 90q + 20 must be a multiple of 100.
The only value that q can be is 2.
So, 100q^2 + 90q + 20 = 400 + 180 + 20 = 600
and 600 = 100p, which means p = 6.
Thus, qr = 25 and pqr = 625.

4) Let r = 6 in Equ.A.
Then, 100q^2 + 120q + 36 = 100p + 10q + 6
or, 100q^2 + 110q + 30 = 100p
Here, 110q + 30 must be a multiple of 100.
This can only be true for q = 7.
But if q = 7, then 100q^2 + 110q + 30 = 5700,
which means 5700 = 100p, or p = 57, which is untrue.
Therefore, r ≠ 6.

So there is only the one solution.

Answer 3

25^ = 625 is the only answer because no other two digit number squared has such a property.

Reasons: Since the two digit number squared gives the same last digit, the only possible last digits are 0, 1, 5, and 6, of which 0, 1, and 6 can be easily excluded.