Number system - sum of two digit numbers

Question

The sum of four two digit numbers is $221$. None of the eight digits is 0 and none of them are same. Which of the following is not included among the eight digit ?

$$(a) \;\;1 \\ (b)\;\; 2 \\ (c)\;\; 3\\ (d)\;\; 4$$

Is there any shortcut to solve this question as I got the answer which is $(d)\;\; 4$ by trial and error method. Please suggest

Answer 1

It is enough to know what is the last number modulo $9$.

Since $10a+b \equiv a+b \pmod 9$, $221 \equiv 5 \pmod 9$, and $1+2+3+4+5+6+7+8+9 = 45 \equiv 0 \pmod 9$, if the unused digit is $x$, then the sum of all the used digits is $221 \equiv 5$, and it also is $45-x \equiv -x$.

So $x \equiv -5 \equiv 4 \pmod 9$, and therefore the unused digit is $4$.

Answer 2

Hint:

$xy+ab+cd+ef=221$

$9(x+a+c+e)+y+b+d+f+(x+a+c+e)=221$

$\equiv(x+a+c+e)+y+b+d+f \equiv 5 \mod 9 $

Now you know which number has to be excluded.

Answer 3

Suppose the numbers are $a_1b_1, \ a_2b_2, \ a_3b_3, \ a_4b_4.$

Then from $$\begin{array}{cc} a_1b_1& \\ a_2b_2& \\a_3b_3& \\ a_4b_4 &+ \\ \hline \\221 \end{array} $$ we conclude that only three possibilities can happen.
Either
$b_1+b_2+b_3+b_4=31$ and $a_1+a_2+a_3+a_4=19\,,$
$b_1+b_2+b_3+b_4=21$ and $a_1+a_2+a_3+a_4=20$ or
$b_1+b_2+b_3+b_4=11$ and $a_1+a_2+a_3+a_4=21\,.$
Since $a_1,\ a_2,\ a_3,\ a_4,\ b_1,\ b_2,\ b_3,\ b_4$ are all different and $1+2+\ldots+9=45$ we conclude that only the second possibility can happen and $4$ is not included in the digits $a_i,b_i$ ($a_1+a_2+a_3+a_4+b_1+b_2+b_3+b_4=41$).

Answer 4

Hint $\ $ Adding in the number $\rm\,0\,d\,$ formed by the missing digits $\rm\,0,d\,$ and casting nines shows

$\rm\qquad\quad mod\ 9\!:\ \ 2\!+\!2\!+\!1\!+\!d\, \equiv\, 0\!+\!1\!+\!2\!+\cdots + 9\,\equiv\, 0,\ $ i.e. $\rm\ 5+d\equiv 0\ \Rightarrow\ d\equiv -5\equiv 4$