Question in Image Below
Solution
22) Fact
Perfect Squares can only have last digits 0, 1, 4, 5, 6, 9
Perfect Square CANNOT have these last digits: 2, 3, 7, 8
If (i) and (ii) were true, a number with last digit 2 is a perfect square, impossible!
If (ii) and (iii) are true, a number with last digit 1 – 48 has last digit 3 is a perfect square, contradiction again!
The only possibility is that statement (i) and (iii) must both be true
Statement (ii) is false
A + 41 = x^2 and A – 48 = y^2
Difference
x^2 – y^2 = 89 , x > y
(x – y)(x + y) = 89 and 89 is a prime number
(x – y) = smaller factor = 1
and
(x + y) = 89
Adding these:
(x – y) + (x + y) = 1 + 89
2x = 90
x = 45
Put back in (x – y) = 1
45 – y = 1
y = 44
Finally
A + 41 = x^2
Becomes
A + 41 = 45^2
A = 1984
Check
(i) A + 41 is a Perfect Square
1984 + 41 = 2025 = 45^2 (True)
(ii) Last digit of A = 1984 is 1 (False)
(iii) A – 48 is a Perfect Square
1984 – 48 = 1936 = 44^2 (True)
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