An important question from arithmetic progression chapter as it was already asked in previous year paper of 2011 in which we are to find the two digit numbers divisible by 6.
RS Aggarwal, Class 10, chapter 5A, question no 42.
Share
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
The least 3 digit no. is 100
The least 3 digit no divisible by 6: 6×12=102
The least 4 digit no is 1000
The largest 3 digit no. divisible by 6: 6×166=996
Using identity l=a+(n−1)d , where l is the last term, a is the first term, n is total no of terms and d is the common difference
Putting, l=996, a=102 and d=6
996=102+(n−1)6
996=102+6n−6
996=96+6n
6n=900
n=150
so, there are 150, 3 digits numbers which are divisible by 6