ad is a number => a \(\ne\) 0

a is a square number => a = 1;4;9

Case 1: a = 1

=> ad = 1d is a square number => d = 6

=> cd = c6 is a square number => c = 1;3

=> abcd = 1b16 or abcd = 1b36

=> b = 9 ( abcd = 1936 = \(44^2\))

Case 2: a = 4 => d = 9, c = 4

=> abcd = 4b49 (This doesn't make sense)

Case 3 a = 9 => \(d\in\varnothing\)

SO a = 1, b = 9, c = 3, d = 6

1 student exchanged with 7 other students: 7 cell phone numbers were entered into phones

8 student that exchanged with 7 other students: 7 x 8 cell phone numbers were entered into phones

But every exchanged is counted twice

=> The number that the cell phone numbers were entered into phones is:

            \(\dfrac{7x8}{2}\) = 28

6 pencils and 4 pens cost 4.30

4 pencils and 6 pens cost 5.20

=> 10 pencils and 10 pens cost 4.30 + 5.20 = 9.50

=> 5 pencils and 5 pens cost 9.50 : 2 = 4.75

*We have: 1 + 2 - 3 - 4 = - 4

                5 + 6 - 7 - 8 = - 4

                  ...

                1985 + 1986 - 1987 - 1988 = - 4

=> A = (1 + 2 - 3 - 4) + (5 + 6 - 7 - 8) + ... + (1985 + 1986 - 1987 - 1988) +1989

         = - 4 x 497 + 1989 = -1988 + 1989 = 1

* There are 995 odd numbers, 994 even numbers

=> A \(\ne\) 0

=> the smallest possible positive integer that A can get is 1

6 pencils and 4 pens cost 4.30

4 pencils and 6 pens cost 5.20

=> 10 pencils and 10 pens cost 4.30 + 5.20 = 9.50

=> 5 pencils and 5 pens cost 9.50 : 2 = 4.25