1.find last two digits of 2^1000
2.(a,a+k)/k for a,knot equel to 0
3.(a,b)=1 and c>0 prove that an integer sugh that (a+bx,c)=1
Start writing down the last two digits of the sequence of numbers .
You will find that after every 20 entries the last two digits start to repeat. This is not surprising because the last two digits of are 16.
Now you don't have to go all of the way up to 1000 because there is a clear pattern.
If we seek then:
if n = 20X+1 then the last two digits are 52 (X>1)
For example the last digit of is 52.
You can find the last two digits of in a similar way.
this is not true because and are dependent. the problem is strangely tricky! here's my proof which is constructive: we'll consider two cases:
Case 1 every prime divisor of divides : let and suppose is a prime divisor of if then which is impossible because then thus
Case 2 some prime divisors of do not divide : let be the product of all prime divisors of that do not divide now let be a prime divisor of and suppose that
subcase1: if then by (1): then since we must have which is impossible by the definition of
subcase 2: if does not divide then by the definition of we'll have and thus by (1): which is a contradiction! thus