Question 694287
Yes.
As a matter of fact, that is the way to multiply a two digit number times 11.
It is very easy, as long as the sum of those digits is less than 10.
You just write the two digits, leaving some space in the middle, and then write the sum in that middle space.
{{{a}}} = first (hundreds) digit
{{{b}}} = third (ones) digit
Then, second (tens) digit = {{{a+b<=9}}},
and the value of the 3-digit number is
{{{100a+10(a+b)+b=100a+10a+10b+b=(100a+10a)+(10b+b)=10a(10+1)+b(10+1)=(10a)*11+11b=11(10a+b)}}}
{{{10a+b}}} is the 2-digit number with {{{a}}} as a first (tens) digit and {{{b}}} as a second (ones) digit.
The 3-digit number in the problem is {{{11}}} times the two digit number ab.