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**Daniel123****Member**- Registered: 2007-05-23
- Posts: 663

The four digit integer aabb is a perfect square. Find a and b.

n^2 = 1000a + 100a + 10b + b = 1100a + 11b = 11(99a + a + b)

n^2 is divisible by 11, so it must also be divisible by 11^2. Therefore 11|(a+b), but since 1 < a + b < 18, a + b = 11.

I got this far, but wasn't really sure how to continue. My book's solution then says:

"Now since n^2 is a square, b cannot be 2,3,7 or 8. Also, b cannot be 5, since the square would end in 25, not 55"

I don't get this :S

Thanks.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

b is the last digit of aabb, which is a square. Squares cannot end in 2, 3, 7 or 8, so b is none of these. Also, when a square ends in 5, its second-last digit must be 2, but this contradicts the last two digits matching so b can't be 5 either.

Why did the vector cross the road?

It wanted to be normal.

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**Daniel123****Member**- Registered: 2007-05-23
- Posts: 663

mathsyperson wrote:

b is the last digit of aabb, which is a square. Squares cannot end in 2, 3, 7 or 8, so b is none of these. Also, when a square ends in 5, its second-last digit must be 2, but this contradicts the last two digits matching so b can't be 5 either.

Why can't squares end in 2, 3, 7 or 8? Why must the second last digit be 2 if the last is 5?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Sorry, I thought you knew that.

The last digit of x² is affected only by the last digit of x, so we only need to consider the squares of 0 to 9. These are:

0

1

4

9

16

25

36

49

64

81

None of these end in 2, 3, 7 or 8, so no square ends in 2, 3, 7 or 8.

Also, the only one of those that ends in 5 is 5². This means that if a square's last digit is 5 then its root also has a last digit of 5.

(10a+5)² = 100a² + 100a + 25.

ie. squaring something whose last digit is 5 gives you a multiple of 100 plus 25.

Why did the vector cross the road?

It wanted to be normal.

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