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(Headbang)

Printable View

- August 10th 2008, 12:54 AMvinodannunumber theory
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(Headbang) - August 10th 2008, 02:41 AMtopsquark
- August 10th 2008, 02:57 AMKiwi_DaveQ1
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. - August 10th 2008, 10:13 AMJaneBennet
- August 10th 2008, 11:06 AMJaneBennet
- August 10th 2008, 08:44 PMNonCommAlg
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