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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, 2 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 \( 7744 \). Sreejato Bhattacharya · 4 years, 2 months ago

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@Sreejato Bhattacharya 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. Calvin Lin Staff · 4 years, 2 months ago

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@Sreejato Bhattacharya I could understand everything except the "quadratic residues modulo 10" thing. Please explain what does that mean ? Nishant Sharma · 4 years, 2 months ago

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@Nishant Sharma 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 :) Va Nd · 4 years, 2 months ago

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@Va Nd Oh I see. Nishant Sharma · 4 years, 2 months ago

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@Nishant Sharma 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 \( [ r_1, r_2, ..., r_{n-1} ] \). Sreejato Bhattacharya · 4 years, 2 months ago

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Isn't this an RMO problem? Vikram Waradpande · 4 years, 2 months ago

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@Vikram Waradpande Yep, RMO 1991 or 1992 maybe. Paramjit Singh · 3 years, 11 months ago

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