Here is my file:

 
700       7912345678910
61234567891234567891
700       8012345678910
61234567891234567891

 
grep ^700 file1 | cut -c 11,12 | grep 79

grep '^700.......79' file1

awk ' { if(($0 ~ /^700/)&&(substr($0,11,2)==79)) print; } ' filename

That did it! I hadn't seen the 'if' statement used with 'substr' in awk before. Thanks for the lesson.

Try this short version (still longer than mjf's):

$ awk '/^700/ && $2~/^79/' file
700       7912345678910

Thank you, RudiC, for the alternative; that'll go into my library of examples. However, in this case, my source file sometimes has no white space before column 11, so I need to search the specific column range (for the sake of brevity, I simplified my example file). bipinajith's solution was the most accurate and worked on more than one unix OS.

OK, try this one:

awk '/^700/ && $11$12=="79"' FS="" file