These solutions, which use quite different methods, are by Ang Zhi Ping, age 16, River Valley High School, Singapore and Pierce Geoghegan 17, Tarbert Comprehensive, Ireland. Good work Ang Zhi and Pierce!

(a) Here is Ang Zhe Ping's solution: A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. A 4-digit base 10 number, aabb , whereby a and b are digits, could be written as:

aabb = 1000a+100a+10b+b =1100a + 11b =11(100a+b)

Hence, the number aabb is a multiple of 11 and since 11(100a+b) is a multiple of 11, and it has to be a square number, thus 100a+b is also divisible by 11, and (100a+b)/11 has also to be another perfect square. Now a and b are digits, thus, they are integers ranging from 0 to 9. Also, as 100a + b is divisible by 11 and 99a is divisible by 11, we must have a + b is divisible by 11. So the only possible values of 100a+b are: 200+9=209=11x19 and 308, 407, 506, 605, 704, 803 and 902.

Dividing these numbers by 11:

209 = 11x19;

308 = 11x28;

407 =11x37;

506 = 11x46;

605 = 11x55;

704 = 11x64;

803 = 11x73 and

902= 11x82.

The only number in this list which is itself a square number (11 times a square number) is 704 so the only number satisfying the required conditions is 7744.

As observed, 700+4=704=11x64, thus, satisfying the conditions that 100a+b is a multiple of 11, and (100a+b)/ 11 is another square (64=8 ²). The number is 7744, equal to 88 ² .

This solution comes from Pierce Geoghegan:

aabb = a(10³)+ a(10²) +b(10) + b = 11(b + 100a).

Let aabb = x². We know x²=0 (mod 11) and that implies x² = 0 ( mod 121) (since 11 is prime 11² must be a factor of x² ). So x² = 121y² for some integer y and since aabb < 10000 we know y < 10.

Testing reveals y=8 and aabb=7744.

(b) Prove that 11¹⁰−1 is divisible by 100.

Editors note: These solutions use the Remainder Theorem and Modulus Arithmetic. Can you prove the result using the Binomial Theorem, or yet another method?

This solution comes from Pierce Geoghegan. For the second question note: 11 = 11 (mod 25) ; 11² = −4  (mod 25) ; 11³ = −19 (mod 25) ; 11⁴ = 16  (mod 25) ; 11⁵ = 1  ( mod 25) ... (A)

Now 11¹⁰−1=(11⁵ + 1)(11⁵ − 1) and using (A)

(11⁵)− 1 = 0 (mod 25)

Also 11⁵ − 1=0 (mod 2) since 11⁵=1 (mod  2) and 11⁵ + 1=0 (mod 2) since 11⁵=1  (mod 2).

So (11⁵ + 1)(11⁵ − 1) = (0 (mod 2))(0 (mod 50)) = 0 (mod 100). QED

This is Ang Zhi Ping's solution.

Let a polynomial P(x) be x¹⁰−1 . Observe that when x=1 , P(x) is reduced to 0, (1¹⁰−1=0). Hence, (x−1) is a factor of P(x) .

Thus, 11¹⁰−1 can be factorized to (11−1)Q(x) , whereby Q(x) is the quotient. We know by now, 11¹⁰−1 is a multiple of 10. Taking x¹⁰−1=(x−1)Q(x) , the quotient Q(x) can be easily evaluated using synthetic division:

x¹⁰−1=(x−1)(x⁹ + x⁸ + x⁷ + x⁶ + x⁵ + x⁴ + x³ + x² + x + 1)

Hence

Q(x) = (x⁹+x⁸+x⁷+x⁶+x⁵+x⁴+x³+x²+x+1).

One can see that when x=1 , the remainder of Q(1)=10 . Thus, Q(x)=(x−1)R(x)+10 and (R(x) is another quotient. For 11¹⁰−1 , P(x)=(x−1)((x−1)R(x)+10) which gives P(11)=(11−1)((11−1)R(x)+10)=10(10R(x)+10)=100(R(x)+1).

Thus, 11¹⁰−1=100(R(x)+1) , and hence the number is a multiple of 100.