696 Number Theory
750.Observe that 2002= 103 + 103 + 13 + 13 , so that
20022002 = 20022001 · 2002 =(
( 2002 )^667
) 3
( 103 + 103 + 13 + 13 )
=( 10 · 2002667 )^3 +( 10 · 2002667 )^3 +( 2002667 )^3 +( 2002667 )^3.
This proves the first claim. For the second, note that modulo 9, a perfect cube can be
only±1 or 0. Therefore, the sum of the residues modulo 9 of three perfect cubes can be
only 0,±1,±2, or±3. We verify that
20022002 ≡ 42002 ≡( 43 )^667 · 4 ≡ 1 · 4 ≡ 4 (mod 9).It is easy now to see that 2002^2002 cannot be written as the sum of three cubes.
(communicated by V.V. Acharya)
751.Denote the perfect square byk^2 and the digit that appears in the last four positions
bya. Thenk^2 ≡a· 1111 (mod 10000). Perfect squares end in 0, 1, 4, 5, 6, or 9, soa
can only be one of these digits.
Now let us examine case by case. Ifa=0, we are done. The casesa∈{ 1 , 5 , 9 }can
be ruled out by working modulo 8. Indeed, the quadratic residues modulo 8 are 0,1, and
4, while asaranges over the given set,a·1111 has theresidues 7 or 3.
The casesa=2 or 4 are ruled out by working modulo 16, since neither 4· 1111 ≡
12 (mod 16)nor 6· 1111 ≡ 10 (mod 16)is a quadratic residue modulo 16.
752.Reducing modulo 4, the right-hand side of the equation becomes equal to 2. So the
left-hand side is not divisible by 4, which means thatx=1. Ify>1, then reducing
modulo 9 we find thatzhas to be divisible by 6. A reduction modulo 6 makes the left-
hand side 0, while the right-hand side would be 1+(− 1 )z=2. This cannot happen.
Therefore,y=1, and we obtain the unique solutionx=y=z=1.
(Matematika v Škole(Mathematics in Schools), 1979, proposed by I. Mihailov)
753.Note that a perfect square is congruent to 0 or to 1 modulo 3. Using this fact we can
easily prove by induction thatan≡ 2 (mod 3)forn≥1. Since 2· 2 ≡ 1 (mod 3), the
question has a negative answer.
(Indian International Mathematical Olympiad Training Camp, 2005)
754.By hypothesis, there exist integerstandNsuch thataN+b =tk. Choosem
arbitrary positive integerss 1 ,s 2 ,...,sm, and consider the number
s=(as 1 +t)k+∑mj= 2(asj)k.Then
s≡tk≡aN+b≡b(moda).