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Playing with Numbers

Playing with Numbers
science Square
Science Square

  • Numbers are four types: Natural Number, Whole Number, Integer, and Rational Numbers.

Numbers in general form

  • Any two digit number xy made of digits x and y can be written as = 10 X x + 1X y

  • Any three digit number xyz made of digits x, y, and z can be written as = 100 X x + 10Xy + 10Xz

EXAMPLE 1: Write the number 12 in generalized form.

SOLUTION: 12 = 10X1 + 1X2 = 10+2

EXAMPLE 2: Write the number 10X3 + 4 in usual form.

SOLUTION: 10X3 + 4 = 30 + 4 = 34

Games with numbers

  • Reversing a two-digit number:
    STEP 1: Suppose we choose a number xy, which is a short form for the 2-digit number 10x + y.
    STEP 2: On reversing the digits, we get the number yx = 10y + x.
    STEP 3: If we add the two numbers we get: (10x + y) + (10y + x) = 11x + 11y = 11 (x + y).
    So, the sum is always a multiple of 11.
    Observe here that if we divide the sum by 11, the quotient is x + y, which is exactly the sum of the digits of chosen number xy.

  • Reversing a 3-digit number:
    STEP 1: Let the 3-digit number be xyz = 100x + 10y + z.
    STEP 2: After reversing the order of the digits, we get the number zyx= 100z + 10y + x.
    STEP 3: On subtraction:
    If x > z, then the difference between the numbers is (100x + 10y + z) – (100z + 10y + x) = 100x + 10y + z – 100z – 10y – x = 99x – 99z = 99(x – z).
    • If z>a, then the difference between the numbers is (100z + 10y + x) – (100x + 10y + z) = 99z – 99x = 99(z – x).
    • And, of course, if x = z, the difference is 0. In each case, the resulting number is divisible by 99. So the remainder is 0.
    Observe that quotient is x – z or z – x.

  • Forming three digit numbers with given 3-digits:

    xyz = 100x + 10y + z
    zxy = 100z + 10x + y
    yzx = 100y + 10z + x
    xyz + zxy + yzx = 111(x + y + z)
    = 37 × 3(x + y + z), which is divisible by 37

EXAMPLE 1: Prove that the addition of a 2-digit no. and its reversed number is equal to the product of 11 and sum of the digits.

SOLUTION: 12= 10X1 + 2
21= 10X2 + 1
12+21= 11X3= 11X(1+2) [PROVED]

Hence, the addition of a 2-digit no. and its reversed number is equal to the product of 11 and sum of the digits.

EXAMPLE 2: Form 3-digit numbers with given 3 digit no. 123

SOLUTION: Combinations of the number 123 are 231,312 [Keeping in mind the no.s are in order]

123+231+312= 666= (1+2+3)X111= (1+2+3)X3X37

Hence, a 3 digit number formed by another 3-digit number can always be divided by 37.

Letter for digits

  • Here letters take the place of digits in an arithmetic ‘addition’, and the we have to find out which letter represents which digit.

  • Each letter must stand for just one digit. Each digit must be represented by just one letter.

  • The first digit of a number can’t be zero. Hence, we write the no. ‘ninety-five’ as 95, and not as 095, or 0095.

EXAMPLE 1: Find X and Y in the addition. X + X + X
Y X

SOLUTION: This has two letters X and Y whose values are to be found.
Study the addition in the one's column: the sum of three X’s is a number whose ones digit is X.
Therefore, the sum of two X’s must be a number whose ones digit is 0.
This happens only for X = 0 and X = 5.
If X = 0, then the sum is 0 + 0 + 0 = 0, which makes Y = 0 too.
We do not want this (as it makes X = Y, and then the tens digit of YX too becomes 0), so we reject this possibility.
So, X = 5.
Therefore, the puzzle is solved as shown below.

5
+ 5
+ 5
15 That is, X = 5 and Y = 1.

EXAMPLE 2: Find the digits X and Y.
Y X
× Y 3
5 7 X

SOLUTION: This also has two letters X and Y whose values are to be found.
Since the ones digit of 3 × X is X, it must be that X = 0 or X = 5.
Now, look at Y.
If Y = 1, then YX × Y3 would at most be equal to 19 × 19; that is, it would at most be equal to 361.
But the product here is 57X, which is more than 500.
So we cannot have Y = 1.
If Y = 3, then YX × Y3 would be more than 30 × 30; that is, more than 900.
But 57X is less than 600.
So, Y cannot be equal to 3.
Hence, Y = 2 .
Therefore, the multiplication is either 20 × 23 or 25 × 23.
But, 20 × 23 = 460. Thus, the second one is the correct multiplication.
The second one works out correctly, 25 × 23 = 575. The answer is X = 5, Y = 2.
2 5
×2 3
575

Tests of Divisibility

  • Divisibility by 10: If the ones digit of a number is 0, then the number is a multiple of 10; and if the ones digit is not 0, then the number is not a multiple of 10. So, we get a test of divisibility by 10.

  • Divisibility by 5: If the ones digit of a number is 0 or 5, then it is divisible by 5.

  • Divisibility by 2: If the one’s digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.

  • Divisibility by 9 and 3:
    (a) A number is divisible by 9 if the sum of its digits is divisible by 9.
    (b) A number X is divisible by 3 if the sum of its digits is divisible by 3. Otherwise, it is not divisible by 3.

EXAMPLE 1: Check the divisibility of 15284 by 3.

SOLUTION: The sum of the digits of 15284 is 1 + 5 + 2 + 8 + 4 = 20. This number is not divisible by 3. We conclude that 15284 too is not divisible by 3.

EXAMPLE 2: If the three-digit number 12a is divisible by 9, what is the value of a?

SOLUTION: Since 12a is divisible by 9, sum of its digits, i.e., 1 + 2 + a should be divisible by 9, i.e., 3 + a should be divisible by 9.
This is possible when 3 + a = 9 or 18, .... But, since a is a digit, therefore, 3 + a = 9, i.e., x = 6.

Practice these questions

Q1) If 12a3 is a multiple of 3, where a is a digit, what might be the values of a?

Q2) If 34a is a multiple of 3, where a is a digit, what is the value of a?

Q3) If 56a7 is a multiple of 9, where a is a digit, what is the value of a?

Q4) Find the value of the letters 1 2 X
+ 6 A Y
X 0 9

Q5) Find the value of the letters 2 A B
+A B 1
B 1 8

Q6) If 417 is added with its other combinations then check if the sum is divided by 37 or not.

Q7) Write the number 100 × x + 10 × y + z in its usual form.

Q8) Write the number 302 in its generalised form.

Recap

  • Numbers are four types: Natural Number, Whole Number, Integer, and Rational Numbers.
  • Any three digit number xyz made of digits x, y, and z can be written as = 100 X x + 10Xy + 10Xz
  • xyz = 100x + 10y + z
    zxy = 100z + 10x + y
    yzx = 100y + 10z + x
    xyz + zxy + yzx = 111(x + y + z) = 37 × 3(x + y + z), which is divisible by 37
  • If the ones digit of a number is 0, then the number is a multiple of 10
  • If the digit at one’s place of a number is 0 or 5, then the number is divisible by 5.
  • If the ones digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.

Quiz for Playing with Numbers

Q.1

If 27y1 is a multiple of 9, where y is a digit, what is the value of y?

a) 8
b) 3
c) 2
d) 1

Q.2

Identify the pattern and find the next three terms.

3, 9, 15, 21, _, _, _

a) 28, 34, 40
b) 25, 31, 37
c) 27, 33, 39
d) 26, 32, 38

Q.3

Identify the pattern and find the next three terms.

1, 4, 9, 16, ___, ___, ___.

a) 23, 34, 51
b) 26, 37, 50
c) 24, 35, 48
d) 25, 36, 49

Q.4

Identify the pattern and find the next three terms.

1, 1, 2, 3, 5, 8, 13, 21, ___, ___, ___.

a) 33, 56, 87
b) 37, 58, 83
c) 34, 58, 89
d) 34, 55, 89

Q.5

Identify the pattern and find the next three terms.

2, 5, 10, 17, 26, __, __, __,

a) 34, 48, 63
b) 37, 50, 65
c) 36, 49, 67
d) none of the above

Q.6

Identify the pattern and find the next three terms.

2, 11, 26, 107, __, __, __.

a) 137, 866, 911
b) 136, 865, 910
c) 136, 864, 911
d) none of the above

Q.7

What is the value of 13+6/2+5x4 ?

a) 36
b) 45
c) 25
d) 28

Q.8

What is the number in ______ , if 8x6=2X_____X8

a) 32
b) 4
c) 3
d) 2

Q.9

Identify the missing number

163x38=(160+3)x_______

a) 35-3
b) 30+3
c) 35+3
d) 30+5

Q.10

Identify the product of 684 and 43 rounded off to the nearest 100

a) 30600
b) 29400
c) 29500
d) 29000

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