murali_vikrala

• Jul 10th, 2006

 

using the arithmetic progression(A.P):nth term Tn=a+(n-1)d. where a is first term,d is the common difference.numbers divisible by both 2 and 3 are the lcm of 2 and 3 i.e. 6.therefore the terms would be 102,108,114,120,126,132,138,144,150,156,162,168,174,180,186,192,198.we need not write down the complete series,we just need to know the first three terms and the last term,then 102,108,114,............,...,198 and applying the above formula we get, 198=102+(n-1)6 [here six because t2-t1=108-102=6,similarly t3-t2=114-108=6]96=(n-1)616=(n-1) i.e.n=17.therefore 17 terms are possible

tanmai

• Sep 24th, 2015

 

Between 200 and 300 first number can be divided by 2 is 100 and the last is 200
and first number diveded by 3 is 102 and the last 198
numbers divisible by 2 is n and numbers divisible by 3 is = m
100 + (n - 1) 2 = 200
2n - 2 = 100
n - 1 = 50 ==> n = 51
and 102 + (m - 1) 3 = 198
34 + m - 1 = 66
17 + m - 1 = 33
m = 17 + 51 = 68
Total of such numbers = 17

atul rajput

• Aug 4th, 2016

 

100 to 200 devide by 9