Question

ABAC is a number which has the property that the first two digits are 1
greater than the last two digits. Find a four digit number which has this
property and is a perfect square. The sum of digits of this number will
be​

Answer 1

Step-by-step explanation:

Let the number be ‘aabb’ , which is a perfect square number.

its value = 1000a+100a+10b+b ,

=1100a+11b =11(100a+b),………..(1)

As per our assumption 11(100a+b) is a square number .

as 11 is a factor of the square number , another 11 as a factor exists in (100a+b).

so (100a+b)is divisible by 11

or (99a+a+b) is divisible by 11

or, 11×9a+(a+b) is divisible by 11

or, (a+b) must be divisible by 11.

Our assumption was the number is aabb where a or b are simple digits like 1,2,3,4 ….etc .

now value of ‘a’ may be any digit from 1 to 9 , same as for ‘b’ . so maximum value of a+b =9+9=18 , and upto 18 there is only one number 11 which is fully divisible by 11.

So a+b=11

now from (1)

value of the number =11(100a+b)

=11(99a+a+b) =11(99a+11)

=11×11×(9a+1)

now (9a+1) will have also two same factors as we have selected the number is a square number.

value of “a” may be any digit from 1 to 9

When a=1 , 9a+1=9×1+1=10

When a=2, 9a+1=9×2+1=19..So on

so, value of 9a+1 willbe any of the under noted values

10 , 19 , 28 , 37 , 46 , 55 , 64 , 73 , 82

out of them only 64=8×8 i.e. product of same two numbers , and it is for the value of a =7

when a=7 ,

9a+1 =9×7+1=63+1=64=8×8 .

so a=7 , and then b=11–7=4

The perfect square number is aabb=7744 ans and it is only one four digit perfect square 

Answer 2

Step-by-step explanation:

let the digits at the ten and thousandth place be a and b .Then according to the problem

x² = 1000a + 100(b+1) + 10a + b

1010a + 101b = x² - 100

101(10a + b) = (x + 10)(x - 10)

101 is a prime number and (10a + b) is at most in the nineties. so 101 = X+10

thus X = 91

x²= 8281

sum = 19