The number x is exactly divisible by 5 and the remainder obtained on dividing the number y by 5 is 1. what remainder will be obtained when (x + y) is divided by 5?

Let the second number be N.
Then, first number = 2N and third number = 4N
? 2N + N+ 4N/3 = 56
? 7N = 3 x 56
? N = 3 x 56/7 = 24
So, the smallest number is 24.

A number that ends in 2 , 3 , 7 or 8 is never a perfect square.

6² -1 = 35, which is divisible by both 5 and 7.

Let the number be x,
Then x - 4 = 21/x
? x² - 4x - 21 = 0
? x² - 7x + 3x - 21 = 0
? x(x - 7) + 3(x - 7) = 0
? (x - 7) (x + 3) = 0
? x = 7 (Neglecting = -3)

Let the number be x and (15-x)
Then, x² + (15 - x)²= 113
? x² - 15x + 56 = 0
? (x-7) (x-8) = 0
? x = 8 or x = 7
So, the numbers are 7 and 8

Let the value for ? be N
So N x 9 x 7 x 3 = a number with unit digit 1.
Clearly, the minimum value of N is 9.

Let second number be 3N, then, first one is 6N and the third one is 2N.
From question 3N + 6N + 2N = 132
? 11N = 132
? N = 12

? Second number = 3N = 3 x 12 = 36

Let unit digit = x and ten's digit = y
3(x + y)= 10y + x .....(i)
10y + x + 45 = 10x + y ....(ii)

2x - 7y = 0.9x -9y = 45
or x - y = 5 .... (iii)

Solving these equations, we get
x=7, y=2
? Required number = 27

Let the required fraction be p/q.
Then, (p +1) / (q +1) = 4
? p - 4q= 3 ....(i)

And, (p - 1) / (q -1) = 7
? p - 7q = -6 ....(ii)

Solving these equation we get p = 15, q=3

Let the number be a and b .
Then a² + b² = 68 ....(i)
and (a - b)² = 36
Now, (a - b)² = 36 ....(ii)
? a² + b² - 2ab = 36
? 68 - 2ab = 36
? 2ab = 32
? ab = 16 .