Two parts of one question.

Prove that when a perfect square number is divided by 7, the remainder cannot be possibly be 3, 5, 6.

Therefore, is it possible using 2, 3, 4, 5, 6 to form a 5 digit number without repeating digits, and that the number is a perfect square?

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A perfect square number divided by 7 leaving a remainder will be of the form `(7x+k)^2` where k can assume the values from 0 to 6.

For k=0,

`(7x+0)^2 =49x^2` when divided by 7 leaves a remainder of 0.

For k=1,

`(7x+1)^2 =49x^2 +14x+1=7(7x^2+2x)+1,` when divided by 7 leaves a remainder of 1.

For k=2,

`(7x+2)^2 =49x^2+28x +4=7(7x^2+4x)+4,` when divided by 7 leaves a remainder of 4.

For k=3,

` (7x+3)^2 =49x^2+42x+9=7(7x^2+6x+1)+2,`

when divided by 7 leaves a remainder of 2.

For k=4,

`(7x+4)^2 =49x^2 +56x+16=7(7x^2+8x+2)+2` when divided by 7 leaves a remainder of 2.

For k=5,

`(7x+5)^2 =49x^2+70x+25=7(7x^2+10x+3)+4,` when divided by 7 leaves a remainder of 4.

For k=6,

`(7x+6)^2 =49x^2+84x+36=7(7x^2+12x+5)+1,` when divided by 7 leaves a remainder of 1.

Hence, it is evident that when a perfect square number is divided by 7, the remainder cannot assume values of 3, 5 or 6

** Any**integer number can be expressed as

`7x +a` where `a` is integer `in (0,6)`

Since * all *perfect squares are the square of particular integer numbers then

**perfect squares `P` can be expressed as**

*all*`P = (7x+a)^2`

Now, expanding this out we get

`P = 49x^2 + 14ax + a^2 = 7(7x^2 + 2ax) + a^2`

The remainder upon dividing `P` by 7 will then be `a^2` modulo 7.

Since ` a` can only take the values 0,1,2,3,4,5,6 then `a^2` can only take the values 0,1,4,9,16,25,36 and `a^2` *modulo 7 *can only take the values

0,1,4,2,2,4,1.

Therefore the remainder upon dividing a perfect square `P` by 7 can only take the values 0,1,2,4 unlike other integers that can also take the values 3,5,6.

**Answer as proved above.**

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