• # question_answer 1)                 Find the values of the letters in each of the following and give reasons for the steps involved. 1.                 3 A + 2 5 B 2                 2.                            4 A + 9 8 C B 3                 3.                            1 A $\times$ A 9 A                 4.                            A B + 3 7 6 A 5.                            A B $\times$ 3 C A B 6.                            A B $\times$ 5 C A B 7.                            A B $\times$ 6 B B B 8.                            A 1 + 1 B B 0 9. 2 A B + A B 1 B 1 8 10. 1 2 A + 6 A B A 0 9

1.            Here, there are two letters A and B whose values are to be found out. Let us see the sum in unit's column. It is $A+5$and we get 2 from this. So A has to be 7 $(\because \,\,A+5=7+5=12)$ Now, for sum in ten's column, we have $1+3+2=B$ $\Rightarrow$               $B=6$ Therefore, the puzzle is solved as shown below:

 3 7 + 2 5 6 2
That is, $A=7$ and $B=6$. 2.            Here, there are three letters A, B and C whose values are to be found out. Let us see the sum in unit's column. It is $A+8$and we get 3 from this. So A has to be 8 $(\because \,\,A+8=5+8=13)$                 Now, for the sum in ten's column, we have $1+4+9=14$ $\Rightarrow$               $B=4,\,\,C=1$ Therefore, the puzzle is solved as shown below:  4 5 + 9 8 1 4 3
That is, A $A=5,\,B=4$ and $C=1$. 3.            $\because$ Unit's digit of $A\times A$ is A $\therefore$  $A=1$ or $A=5$ or $A=6$ When $A=1,$ Then  1 1 $\times$ 1 1 1
$\therefore$  $A=1$ is not possible.                 When $A=5,$ Then  1 5 $\times$ 5 7 5
$\therefore$ $A=5$ is not possible. When $A=6,$ Then  1 6 $\times$ 6 9 6
which is admissible. $\therefore$  $A=6$. 4.            Here, there are two letters A and B whose values are to be found out. $B+7$gives A and        $A+3$ gives 6. The possible values are: $0+7=7$ $\Rightarrow$               $A=7$ but $7+3\ne 6$. So, not acceptable. $1+7=8$ $\Rightarrow$               $A=8$ but $8+3\ne 6$. So, not acceptable. $2+7=9$ $\Rightarrow$               $A=9$ but $9+3\ne 6$.  So, not acceptable. $3+7=10$ $\Rightarrow$               $A=0$ but $1+0+3\ne 6$. So, not acceptable. $4+7=11$ $\Rightarrow$               $A=1$ but $1+1+3\ne 6$. So, not acceptable. $5+7=12$ $\Rightarrow$               $A=2$. Also $1+2+3=6$. So, $B=5$ works and then we get A as 2. Therefore, the puzzle is solved as shown below:  2 5 + 3 7 6 2
That is, $A=2$ and $B=5$. 5.            Here, there are three letters A, B and C whose values are to be found out. $\because$     Unit's digit of $3\times B$ is B. So, it is essential that $B=0$ Hence, we get  A 0 $\times$ 3 C A 0
$\therefore$ Unit's digit of $3\times A$ is A. So, it is essential that $A=5$. Hence, we get   5 0 $\times$ 3 1 5 0
$\therefore$  $A=5,\,\,B=0$ and $C=1$ 6.            Here, there are three letters A, B and C, whose values are to be found out.        $\because$     Unit's digit of $5\times B$ is B             $\therefore$  $B=0$ or $B=5$ If $B=0,$ then we have  A 0 $\times$ 5 C A 0
Now, $5\times A=A\,\,\Rightarrow \,\,A=0$ or 5 But $A\ne 0$ as in the answer there is a third letter too. If $A=5,$ then we have  5 0 $\times$ 5 2 5 0
$\therefore$  $A=5,\,B=0$ and $C=2$ If $B=5,$ then we have  A 5 $\times$ 5 C A 5
Now, $5\times A+2=A$ $\Rightarrow$               $A=2$ | in the form of $5\times 2+2=12$ $\therefore$ Unit's digit is 2 which is equal to A. Then, we have  2 5 $\times$ 5 1 2 5
$\therefore$  $A=2,\,B=5$ and $C=1$. 7.            Here, there are two letters A and B whose values are to be found out. We have  A B $\times$ 6 B B B
The possible values of BBB are 111,   222,   333, etc. Let us divide these numbers by 6, we get $111\div 6=18,$ remainder 3 $\therefore$  111 is not acceptable. $222\div 6=37,$ remainder 0. Hence, the quotient is not of the form A2. $\therefore$ 222 is not acceptable. $333\div 6=55,$ remainder 3. $\therefore$ 333 is inadmissible. $444\div 6=74,$ remainder 0. Here the quotient 74 is of the form A4 which clearly works well. Hence, the puzzle is solved as shown below:  7 4 $\times$ 6 4 4 4
$\therefore$  $A=7$ and $B=4$. 8.            Here, there are two letters A and B whose values are to be found out. Let us see the sum in one's column, we have $1+B$ gives 0, i.e., a number whose unit's digit is 0. $\therefore$  $B=9$ | $\because$ B is itself a one digit number Then, we have the puzzle as  A 1 + 1 9 9 0
$90-19=71$ $\Rightarrow$         $A1=71$ $\Rightarrow$               $A=7$ Therefore, $A=7$ and $B=9$. 9.            We are to find out A and B. Let us see the sum in unit's column $B+1$ gives 8 $\Rightarrow$               $B=7$ | $\because$ B is in itself a one digit number Then, the puzzle becomes  2 A 7 + A 7 1 7 1 8
Now, see the sum in ten's column. $A+7$ gives 1, i.e., a number whose unit's digit is 1. $\therefore$  $A=4$ | $\because$ A itself is a one digit number Hence, the puzzle is solved as shown below.  2 4 7 + 4 7 1 7 1 8
Here,$A=4,\,B=7$. 10.          We are to find out the values of A and B. Let us see the sum in ten's column. It is $2+A$and we get 0 from this. $\therefore$  $A=8$ | $\because$ A in itself is a one digit number The puzzle then becomes  1 2 8 + 6 8 B 8 0 9
Now, see the sum in one's column. It is $8+B$and we get 9 from this. $\therefore$  $B=1$. Hence, the puzzle is finally solved as shown below:  1 2 8 + 6 8 1 8 0 9
Therefore, $A=8$ and $B=1$.

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