question_answer
1)
Find the values of the letters in each of the following and give reasons for the steps involved. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
Answer:
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: 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: 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 \[\therefore \] \[A=1\] is not possible. When \[A=5,\] Then \[\therefore \] \[A=5\] is not possible. When \[A=6,\] Then 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: 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 \[\therefore \] Unit's digit of \[3\times A\] is A. So, it is essential that \[A=5\]. Hence, we get \[\therefore \] \[A=5,\,\,B=0\] 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 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 \[\therefore \] \[A=5,\,B=0\] and \[C=2\] If \[B=5,\] then we have 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 \[\therefore \] \[A=2,\,B=5\] and \[C=1\]. 7. Here, there are two letters A and B whose values are to be found out. We have 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: \[\therefore \] \[A=7\] 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 \[90-19=71\] \[\Rightarrow \] \[A1=71\] \[\Rightarrow \] \[A=7\] 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 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. 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 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: Therefore, \[A=8\] and \[B=1\].