Does this number even exist?

A four digit number has the following properties:-

a) It is a perfect Square

b)It's first two digits are equal to each other

c) It's last two digits are equal to each other.

Find the sum of all four digit numbers which satisfy.

Details and assumptions

If you think that no such numbers exist, Input 0.


The answer is 7744.

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1 solution

Mehul Arora
Jun 17, 2015

Let the number be a a b b \overline {aabb}

We know 10 0 2 = 10000 100^2=10000 , Hence, The number has to be a two digit number

From the divisibility rules of 11, we get to know that this number is divisible by 11.

We also know from the given criteria, That the number is a perfect square.

Hence, The number has to be in the form of:- 1 1 2 × a 2 11^2 \times a^2 Where 0 > a > = 9 0>a>=9

We list out all such numbers.

121 × 1 = 121 121 \times 1=121

121 × 4 = 484 121 \times 4=484

121 × 9 = 1089 121 \times 9=1089

121 × 16 = 1936 121 \times 16=1936

121 × 25 = 3025 121 \times 25=3025

121 × 36 = 4356 121 \times 36=4356

121 × 49 = 5929 121 \times 49=5929

121 × 64 = 7744 121 \times 64=7744 < Satisfies the Criteria

121 × 81 = 9801 121 \times 81=9801

Hence, The only number which satisfies the conditions is 7744.

Answer:- 7744 \boxed {7744}

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