Answer:162
Step-by-step explanation:
we calculate 0 in every hundred.
in 100 to 200
101,102,103,104,105,106,107,108,109,110 total 10 zeros
and 120,130,140,150,160,170,180,190 total 8 zeros
total zeros=10+8=18
this will go with every hundred within 999
so from 201-300,301-400,401-500,501-600,601-700,701-800,801-900,901-999
there are eight more 100's
then total zeros= 18*9=162 zeros
There are 45 integers between 11 and 999999 which consist of only one digit being repeated. There are 831430 integers that contain at least one repeated digit.
Of the 729 numbers that satisfy the requirement of positive integers, 104 are divisible by 7.
9000 integers.
there are 10 integers
180
648
There are 9 integers.
None. But there are three integers there.
There are 30 such integers.
17
There are twelve instances where the integers from 1 to 200 contain the digit 1 at least twice:-11,101,110,111,121,131,141,151,161,171,181,191.
There are 22 integers between them.