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
6 years, 4 months ago

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2 votes

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Let the first two digits be a a , and the last two digits be b b . Then the number is N=1000a+100a+10b+b=1100a+11b=11(100a+b)N = 1000a + 100a + 10b + b = 1100a + 11b= 11(100a+b) . Note that 11 11 divides N N , and 1111 being a prime, for N N to be a perfect square 11 11 must also divide N11 \frac{N}{11} , i.e. 11 11 must divide 100a+b 100a + b . But note that 100a+b=99a+a+b=911a+a+b 100a + b= 99a + a+b = 9*11a + a+b . Hence for 11 11 to divide 100a+b 100a + b , 11 11 must divide a+b a+b , i.e. a+b a+b must be a multiple of 11 11 . Note that 1a9 1 \leq a \leq 9 , 0b9 0 \leq b \leq 9 . Thus 1a+b18 1 \leq a+b \leq 18 . The only multiple of 11 11 between 1 1 and 18 18 is 11 11 itself. Thus a+b=11 a+b= 11 . Now note that the quadratic residues modulo 10 10 are 0 0 , 1 1 , 4 4 , 5 5 , 6 6 , and 9 9 . Thus the only possible values of b b are4 4 , 5 5 , 6 6 , and 9 9 (note that b b can't be 0 0 or 1 1 since that would imply a=11 a= 11 or a=10 a= 10 respectively ) .

For b=4 b= 4 , we have a=7 a= 7 , and the number is 7744=882 7744 = 88^2 (a perfect square). For b=5 b= 5 we have a=6 a= 6 and the number is 6655 6655 (not a perfect square). For b=6 b= 6 we have a=5 a= 5 and the number is 5566 5566 (not a perfect square). For b=7 b= 7 we have a=4 a= 4 and the number is 4477 4477 (not a perfect square).

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

Sreejato Bhattacharya - 6 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=11a+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 aabb=121×(9a+1) \overline{aabb} = 121 \times (9a + 1 ) (using that a+b=11a+b = 11 ), so 9a+1 9a + 1 must be a perfect square. Checking a=1 a = 1 to 9, gives a=7 a = 7 as the only square.

Calvin Lin Staff - 6 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 ?

Nishant Sharma - 6 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 :)

Va Nd - 6 years, 4 months ago

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@Va Nd Oh I see.

Nishant Sharma - 6 years, 4 months ago

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An integer q q is called a quadratic residue mod n n if and only if there exists an integer x x such that x2q(modn) x^2 ≡ q (mod n) . To put it other way round, if a sequence of integers ri r_i is defined such that ri r_i is the remainder when i2 i^2 is divided by n n , then the quadratic residues mod n n are the distinct elements of the set [r1,r2,...,rn1] [ r_1, r_2, ..., r_{n-1} ] .

Sreejato Bhattacharya - 6 years, 4 months ago

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Isn't this an RMO problem?

Vikram Waradpande - 6 years, 4 months ago

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Yep, RMO 1991 or 1992 maybe.

A Brilliant Member - 6 years, 1 month ago

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