BLOG 2 -PLAYING WITH NUMBERS - LETTERS FOR DIGITS IN MULTIPLICATION-

BLOG 2 -PLAYING WITH NUMBERS - LETTERS FOR DIGITS IN MULTIPLICATION-

GOOD MORNING BOYS..

                                                 LETTERS FOR DIGITS (PART II)

LEARNING OUTCOMES 

 By the end of this lesson you will be able  to find the " DIGIT" to the " LETTER" IN THE ARITHMETIC PRODUCT  PROBLEM and hence solve the puzzle 

HERE are two rules we follow while doing such puzzles

(1)Each ALPHABET in the puzzle must stand for just one digit.Each DIGIT  must be represented by just ONE DIGIT.


(2)The first digit of a number cannot be ZERO.Thus, we write the number  fifty nine as " 59". and NOT AS "059" or "0059"

CLICK HERE  SEE THE VIDEO ON  INTRODUCTION ON PLAYING WITH NUMBERS FOR MULTIPLICATION 


  NOW LET US SOLVE  NCERT QUESTIONS BASED ON IT

Q3 :   

Find the value of the letter in the following and give reasons for the steps involved. 


 Answer : 

The multiplication of A with A itself gives a number whose ones digit is A again.

 This happens only when A = 1, 5, or 6.  

If A = 1, then the multiplication will be 11 × 1 = 11.

However, here the tens digit is given as 9. 

Therefore, A = 1 is not possible.

 Similarly, if A = 5, then the multiplication will be 15 × 5 = 75. 

Thus, A = 5 is also not possible.  

If we take A = 6, then 16 × 6 = 96. 

Therefore, A should be 6.  

The multiplication is as follows. 

Hence, the value of A is 6.   

Q5 :   

Find the values of the letters in the following and give reasons for the steps involved. 

Answer : 

The multiplication of 3 and B gives a number whose ones digit is B again.  

Hence, B must be 0 or 5.  

Let B is 5.  

Multiplication of first step = 3 × 5 = 15  

1 will be a carry for the next step. 

We have, 3 × A + 1 = CA  

This is not possible for any value of A. 

Hence, B must be 0 only. If B = 0, then there will be no carry for the next step

.  

We should obtain, 3 × A = CA  

That is, the one's digit of 3 × A should be A. This is possible when A = 5 or 0. 

However, A cannot be 0 as AB is a two-digit number.  

Therefore, A must be 5 only. The multiplication is as follows.   

 Hence, the values of A, B, and C are 5, 0, and 1 respectively

Q6 :  

                      

Find the values of the letters in the following and give reasons for the steps involved.  

Answer : 

The multiplication of B and 5 is giving a number whose ones digit is B again. 

This is possible when B = 5 or B = 0 only.  

In case of B = 5, the product, B × 5 = 5 × 5 = 25  

2 will be a carry for the next step.  

We have, 5 × A + 2 = CA, which is possible for A = 2 or 7  The multiplication is as follows.   

If B = 0, 

B × 5 = B

0 × 5 = 0  

There will not be any carry in this step. 

In the next step, 5 × A = CA  

It can happen only when A = 5 or A = 0  

However, A cannot be 0 as AB is a two-digit number. 

Hence, A can be 5 only. The multiplication is as follows

Hence, there are 3 possible values of A, B, and C.  

(i)     5, 0, and 2 respectively  

(ii)   2, 5, and 1 respectively  

(iii) 7, 5, and 3 respectively  

Q7 :   

Find the values of the letters in the following and give reasons for the steps involved.  

Answer : 

The multiplication of 6 and B gives a number whose one's digit is B again.  

It is possible only when B = 0, 2, 4, 6, or 8  

If B = 0, then the product will be 0. Therefore, this value of B is not possible.  

If B = 2, then B × 6 = 12 and 1 will be a carry for the next step. 

6A + 1 = BB = 22  6A = 21 and hence, any integer value of A is not possible.  

If B = 6, then B × 6 = 36 and 3 will be a carry for the next step.

  

6A + 3 = BB = 66  6A = 63 and hence, any integer value of A is not possible. 

If B = 8, then B × 6 = 48 and 4 will be a carry for the next step.

  

6A + 4 = BB = 88  6A = 84 and hence, A = 14. However, A is a single digit number. Therefore, this value of A is not possible. 

If B = 4, then B × 6 = 24 and 2 will be a carry for the next step.  

6A + 2 = BB = 44    6A = 42 and hence, A = 7  The multiplication is as follows.   

Hence, the values of A and B are 7 and 4 respectively.   

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