Question 1012016
Six consecutive odd numbers (or six consecutive even numbers)
could be represented by
{{{n}}} , {{{n+2}}} , {{{n+4}}} , {{{n+6}}} , {{{n+8}}} , and {{{n+10}}} .
The sum would be
{{{n+n+2+n+4+n+6+n+8+n+10=n+n+n+n+n+n+2+4+6+8+10=6n+30}}}
In this case,
{{{6n+30=132}}}-->{{{6n=132-30}}}-->{{{6n=102}}}-->{{{n=102/6}}}-->{{{n=17}}} .
So the numbers are 17, 19, 21, 23, 25, and 27.
The sum of all the tens digits is
{{{1+1+2+2+2+2=10}}} .
 
Another way:
If the sum is {{{132}}} , the average is {{{132/6=22}}} .
Consecutive odd integers form an arithmetic sequence,
and in an arithmetic sequence, the average (or mean) is also the median.
That means that {{{6/2=3}}} of the 6 consecutive odd numbers are more than {{{22}}}, with a {{{2}}} for a tens digit,
and the other {{{3}}} are less than {{{22}}} , including {{{21}}} {{{19}}} and {{{17}}} .
So {{{4}}} of the six odd numbers have a {{{2}}} as a tens digit, and the other {{{2}}} have a {{{1}}} as the tens digit.
So the sum of the tens digits is
{{{4*2+2*1=8+2=10}}} .