Class 08 CRYPTOGRAM OR CRYPTARITHMETIC

A mathematical puzzle written by using alphabets in place of digits is called a Cryptogram. Every letter of the English alphabet stands for a different digit throughout the problem. 

Solve this interesting Cryptogram.

Solution: 

If left hand digit in the sum is a single digit, it must be 1. ∴ M = 1  

2. S + M should be at least 10.

M = 1, therefore, S can either be 9 or 8. 

S = 9 if there is no carry over from E + 0. 

S = 8, if there is a carry over from E + 0.

If S = 8 or 9, S + M can be 10 or 11.

11 can be rejected because then O will be 1

O cannot take the same value as M.

O = 0 .. we have S = 9

M = 1,0 = 0

3. E + 0 = N and O = 0

But, E cannot be equal to N.

 There must be a carry over from previous column. 

N must be = E + 1. Let E = 5 :: N = 6

4. N + R = E

6 + R = ends at 5 = 15 

R = 9 (or 8, if carry over from previous column is there) 

R cannot be 9 (QS = 9). Therefore, R = 8

5. The digit left with us are 2, 3, 4, 7D cannot be taken as 2 or 3 or 4 because, D + 5 has to be greater than 10 (1 is to be carried over to next column)

 D = 7

And the final solution is:

So, we have S = 9, E = 5, N = 6, D = 7, M = 1, 0 = 0, R = 8, Y = 2

Do Yourself:

Replace each of the letters with one of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, so that the additions are correct

Is there more than one possible answer?

 2. Replace each of the letters by one of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 so that the subtractions are correct.

Is there more than one possible answer?

 Class 08 Activity – Lucky Numbers

Find all the Lucky Numbers between 1 and 100 as follows.

Procedure 1.

Write down all the numbers  from 1 to 100.  

2. Circle the first number 

The first row will then be 1 2 3 4 5 6 8 9 10

The first number not circled is 2.

Counting from 1, cross out every 2nd number.

The first four rows will then be 

 The first number not circled or crossed out is 3.

Circle this number.

Counting from 1, cross out every 3rd number not crossed out.

The first four rows will then be 

The first number not circled or crossed out is 7.Circle this number.

Counting from 1, cross out every 7th number not crossed out.

The first four rows will then be

5. Continue like this until all the numbers are either circled or crossed out.

The circled numbers are Lucky Numbers.

Investigate Lucky Numbers.

You could look for patterns.

You could investigate the sums or differences of Lucky Numbers.

You could find all the Lucky Numbers which are smaller than 1000.

Class 08 Activity – Algebraic Expressions4

 Class 08 Activity – Algebraic Expressions4

Objective: 

To solve linear equations in one variable by activity method.

Materials Required: 

Thick paper strips of two colours (red and green) of dimensions x cm x 1 cm, thick paper squares of two colours (red and green) of dimensions 1 cm x 1 cm.

Procedure: 

Let us solve the linear equation in one variable 5x-3 = 3x + 5.

Let us represent the linear equation 5x-3 = 3x + 5 using strips and squares.

2. Subtracting (removing) three green strips from L.H.S. as well as from R.H.S.

5x-3x-3 = 3x-3x + 5

2x-3 = 5

The strips and squares in fig. 2 represent 2x-3 = 5

3. Now add three green squares on L.H.S. as well on R.H.S.

(b)2x-3+ 3 = 5 + 3

2x = 8

x = 4

Observations:

Now if we substitute x = 4 in the given linear equation 

5x-3 = 3x + 5, 

we get 5x4-3 = 3 x 4 + 5

20-3 = 12 + 5

17 = 17

Hence, x = 4 is the required solution of the given linear equation which has been solved experimentally and verified.

 Class 08 Activity – Algebraic Expressions3

Objective: 

To verify the identity x²-y² = (x + y) (x-y).

Materials Required: 

Some thick sheets of paper (cardboard), scissors, geometry box, sketch pen, pencil, etc.

 Procedure: 

Let us verify the identity x²-y² = (x + y) (x-y) by taking r = 7, y = 4.

On a thick sheet of paper (cardboard), draw a square of side 7 cm. Cut it out.

Area of this piece = 7 x 7 cm² = 7² cm²

2. From one of its corners, cut out a square piece of side 4 cm, as shown in the figure.

Area of the remaining shape = (7² – 4²) cm²

3. Now, cut the remaining shape along the dotted line and rearrange the two rectangles as shown. 

Clearly, the resulting figure is a rectangle of dimensions (7 + 4) cm x (7-4) cm. So, area of this shape = (7 + 4) x (7-4) cm².

But, this-rectangular piece is rearrangement of the piece of area  (7² – 4²) cm²

 [ step 2]

Thus (7² – 4²) = ( 7 + 4) x ( 7-4)cm²

x²- y² = (x + y) (x-y). [ x =7, y=4]

Do Yourself:

Verify the identity, x²-y² = (x + y) (x-y) using the following pairs of numbers:

X = 4, y=3

(ii) X = 6, y = 2

(iii) x= 8, y = 4

(iv) X = 7, y=2

(v) x=10, y=4

(vi) X =9, y =6