Question

Find the four digit number whose last two digits is the double of first 2 digits

Answer 1

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

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

=1100a+11b =11(100a+b),………..(1)

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.

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 ,

as a+b is divisible by 11, then only possible value of 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

value of a may be any digit from 1 to 9

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 ,b=11–7=4

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

Answer 2

Ahoy buddy !☺

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A four digit number whose last two digits is the double of first two digits is 2244.

Looking carefully at this number- 2244;

We came to a conclusion that its last two digits i.e., 44 is the double of its first two digits i.e., 22

'cause the double of 22 i.e., 22×2 is 44.

Similarly,
There can bE other such numbers like 1122, 3366, 4488, 1530, etc.

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Hope it helps you, buddy ! ☺