What is the smallest perfect square that ends with the four digits 9009?
In turn 4 people throw away three nuts from a pile and hide a quarter of the remainder finally leaving a multiple of 4 nuts. How many nuts were at the start?
What are the possible remainders when the 100-th power of an integer is divided by 125?
What is the remainder when 3 2001 is divided by 7?
Zi Heng Lim and Hagar El Bishlawi found a pattern when they raised 3 to a power, divided it by seven and found the remainder.
When you divide the exponent 2001 by the number 6, the answer will be 333 R3. The third number in the remainder pattern is 6, therefore 3 2001 divided by 7 has a remainder of 6, assuming that the pattern keeps repeating itself every six powers.
Can you prove that this pattern will keep repeating itself?
Pierce Geoghegan and Etienne Chan solved this problem using modulus arithmetic. This is Pierce's solution:
Initially we use the fact that 3 n (mod 7) = 3 x [3 n-1 (mod 7)]
so
3 1 = 3(mod 7) = 3 3 2 = 3 x 3(mod 7) = 2 3 3 = 3 x 2(mod 7) = 6 3 4 = 3 x 6(mod 7) = 4 3 5 = 3 x 4(mod 7) =5 3 6 = 3 x 5(mod 7) = 1
Now we have 3 6 = 1 mod 7 and it follows that 3 1998 = (3 6 ) 333 = 1 333 (mod 7) = 1
But 3 2001 = 3 1998 x 3 3 = 1 x 3 3 (mod 7) = 6 therefore the remainder is 6 when 3 2001 is divided by 7.