math question?

Question

how many integers from -3000 to 3000 inclusive, are divisible by 3??
can you explain also please...

Answer 1

ok as you can see the first integer divisable by three in that range is -3000. the last one is 3000.
as the first one devisable by 3 is -3000 the second should be
-3000+3 and so on, . you have to find out how many times you have to add 3 till it comes to 3000. so let's think it's x times
-3000+3*x=3000
3*x=6000
x=2000
So you have 2000 integers in between -3000 and 3000 that are devisable byn 3.
as -3000 and 3000 are devisable by 3 too. you have 2 more.
So the answer is 2002

Answer 2

Use the arithmetic series ...with first term as -3000 and last term
3000....and common difference is 3
so we get -3000, -2997 , -2994 , ............, 0 , ........2997 ,3000
nth term of AP series is given by
tn = a + ( n - 1) d ...where a is first term and d is common differnce
3000 = -3000 + ( n -1 ) 3 (add 3000 on both sides)
6000 = ( n -1 )3 ( divide both side by 3)
2000 = ( n - 1) (add 1 on both side

n =2001
so there are 2001 terms

hope this will help you ..