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Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?

1. $3$
2. $2$
3. $4$
4. $0$
5. $1$

### 1 comment

Option 5 is correct

four dight number will be like in this way(Any perfect square must end in 0,1,4,6,9,)

1100,1111,1144,1155 ,1166.,1199,

2200,2211, 2244.2255,.2266,2299,

3300,3311,3344,3355.,3366,3399,

4400,4411, 4444,4455.,4466,4499,

5500,5511, 5544,5555,5566,5599

,6600,6611,6644,6655,6666,6699

,7700,7711,7744(88^2), 7755,7766, 7799,

8800,8811,8844,8855,8866, 8899.

9900 ,9911,9944,9955,9966, 9999 here only 1 number which is perfact squre so answer is option 5) : 1

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2.3k points

4 digit numbers should won't include those starting with 0.

Any perfect square must end in 1, 4, 5, 6, 9. 0. So, this can be used here.
edited by
o yes , sorry....

updated :)

1
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