8th Class Mathematics Playing with Numbers

  • 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

    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:

                    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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