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# Number Theory problem

A four digit number has the following properties:

(i) it is a perfect square,

(ii) its first two digits are equal to each other,

(iii) its last two digits are equal to each other .

Find all such four-digit numbers.

Till now I have been able to find only one such number. I don't know if I am correct or not. Any idea ?

Note by Nishant Sharma
4 years, 4 months ago

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Let the first two digits be $$a$$, and the last two digits be $$b$$. Then the number is $$N = 1000a + 100a + 10b + b = 1100a + 11b= 11(100a+b)$$. Note that $$11$$ divides $$N$$, and $$11$$ being a prime, for $$N$$ to be a perfect square $$11$$ must also divide $$\frac{N}{11}$$, i.e. $$11$$ must divide $$100a + b$$. But note that $$100a + b= 99a + a+b = 9*11a + a+b$$. Hence for $$11$$ to divide $$100a + b$$, $$11$$ must divide $$a+b$$, i.e. $$a+b$$ must be a multiple of $$11$$. Note that $$1 \leq a \leq 9$$ , $$0 \leq b \leq 9$$. Thus $$1 \leq a+b \leq 18$$. The only multiple of $$11$$ between $$1$$ and $$18$$ is $$11$$ itself. Thus $$a+b= 11$$. Now note that the quadratic residues modulo $$10$$ are $$0$$, $$1$$, $$4$$, $$5$$, $$6$$, and $$9$$. Thus the only possible values of $$b$$ are$$4$$, $$5$$, $$6$$, and $$9$$ (note that $$b$$ can't be $$0$$ or $$1$$ since that would imply $$a= 11$$ or $$a= 10$$ respectively ) .

For $$b= 4$$, we have $$a= 7$$, and the number is $$7744 = 88^2$$ (a perfect square). For $$b= 5$$ we have $$a= 6$$ and the number is $$6655$$ (not a perfect square). For $$b= 6$$ we have $$a= 5$$ and the number is $$5566$$ (not a perfect square). For $$b= 7$$ we have $$a= 4$$ and the number is $$4477$$ (not a perfect square).

Hence there is only one number that satisfies the given conditions and it is $$7744$$.

- 4 years, 4 months ago

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This is good, though there isn't a need to appeal to quadratic residues (for those who don't know it).

Once you know that the cases are $$a+b = 11$$, there are 9 values to check for squares, and this can be done easily.

Alternatively if you are too lazy to find square roots, note that $$\overline{aabb} = 121 \times (9a + 1 )$$ (using that $$a+b = 11$$ ), so $$9a + 1$$ must be a perfect square. Checking $$a = 1$$ to 9, gives $$a = 7$$ as the only square.

Staff - 4 years, 4 months ago

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I could understand everything except the "quadratic residues modulo 10" thing. Please explain what does that mean ?

- 4 years, 4 months ago

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It means the last digit of a perfect square number can only be 0,1,4,5,6,9. You can prove it yourself, it is quite easy. Best wishes :)

- 4 years, 4 months ago

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Oh I see.

- 4 years, 4 months ago

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An integer $$q$$ is called a quadratic residue mod $$n$$ if and only if there exists an integer $$x$$ such that $$x^2 ≡ q (mod n)$$. To put it other way round, if a sequence of integers $$r_i$$ is defined such that $$r_i$$ is the remainder when $$i^2$$ is divided by $$n$$, then the quadratic residues mod $$n$$ are the distinct elements of the set $$[ r_1, r_2, ..., r_{n-1} ]$$.

- 4 years, 4 months ago

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Isn't this an RMO problem?

- 4 years, 4 months ago

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Yep, RMO 1991 or 1992 maybe.

- 4 years, 1 month ago

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