grep -E {,1} showing results that have more than 1 occurance

Question

I've been trying to figure out how to grep a line with only N occurrences of a character.

[root@example DIR]# grep -E "6{,1}" test.txt
6543
6625
6668
6868
6666
1161

What I want is for grep to print out the following:

[root@example DIR]# grep -E "6{,1}" test.txt
6543
1161

What am I missing?

Answer 1

With awk I'd:

$ awk '/6/&&!/6.*6/' file
6543
1161

It translates to grep like:

$ grep 6 file | grep -v 6.*6
6543
1161

Edit:

@Sundeep's clever idea to use 6 as a field separator and to count the fields (see comments):

$ awk -F6 'NF==2' file
6543
1161

^ his comment below.