Chapter 1

Real Numbers

It is given by Carl Friedrich Gauss, it states that every composite number can be expressed as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

Example:

\( displaystyle begin{array}{l}30=2times 3times 5250=2times {{5}^{3}}1771=7times 11times 13end{array}\)

Note:   The prime factorisation of a natural number is unique, except for the order of its factors

HCF and LCM by prime factorization method

We can find HCF and LCM of number by the prime factorization method.

HCF (Highest Common factor) =    product of the smallest power of each common factor in the numbers

LCM (Lowest Common Multiple) = Product of the greatest power of each prime factor involved in the number

Example: Find the LCM and HCF  of 6 and 20 by the prime factorisation method.

Sol.

We have \(displaystyle 6={{2}^{1}}times {{3}^{1}},text{and }20=2times 2times 5={{2}^{2}}times {{5}^{1}}\)

HCF (6, 20) = \(displaystyle {{2}^{1}}\)(Product of the smallest power of each common prime factor in the numbers.)

  =2

LCM (6, 20) = \( displaystyle {{2}^{2}}times {{3}^{1}}times {{5}^{1}}\)  (Product of the greatest power of each prime factor, involved in the numbers.)

   = 60

Example: Find the HCF and LCM of 126 and 156 using prime factorisation method.

Sol.

Prime factorisation of 126 and 156 are as follows:

\( displaystyle begin{array}{l}126=2times {{3}^{2}}times 7156={{2}^{2}}times 3times 13end{array}\)

HCF (126, 156) = product of smallest power of each common factors

      \( displaystyle begin{array}{l}=left( {{{2}^{1}}times {{3}^{1}}} right)=6end{array}\)

Example: Two tanks contain 504 and 735 litres of milk respectively. Find the maximum capacity of a container which can measure the milk of either tank an exact number of times.

Sol.

Prime factorisation of 504 and 735 are as follows:

\( displaystyle begin{array}{l}504={{2}^{3}}times {{3}^{2}}~times {{7}^{1}}735=3times 5times {{7}^{2}}end{array}\)

HCF(504, 735) = (3 × 7) = 21.

Hence, capacity of the required container = 21 litres.

Example: Find the least number which when divided by 9, 10 and 15 leaves the  remainders 4, 5 and 10, respectively.

Sol.

Here, 9–4 = 10–5=15–10=5

Also, L.C.M. (9, 10, 15)= 90

the required least number = 90–5=85.

Example: What is the smallest number which is exactly divisible by 36, 45, 63 and 80?

Sol.

The required smallest number = L.C.M. of 36, 45, 63 and 80

prime factorisation of 36, 45, 63 and 80 are as follows:

\( displaystyle begin{array}{l}36={{2}^{2}}times {{3}^{2}}45={{3}^{2}}times 563={{3}^{2}}times 780={{2}^{4}}times 5end{array}\)

L.C.M. of 36, 45, 63 and 80 \( displaystyle begin{array}{l}={{2}^{4}}times {{3}^{2}}times 5times 7=5040end{array}\)

Example: Find the greatest number which will divide 772 and 2778 so as to leave the               remainder 5 in each case.

Sol.

The required greatest number = H.C.F. of (772–5) and (2778–5)

    = H.C.F. of 767 and 2773

prime factorisation of 767 and 2773 are as follows:

\( displaystyle begin{array}{l}767=13times 592773=47times 59end{array}\)

The required greatest number = 59

AN IMPORTANT PROPERTY

               Product of two given numbers = product of their HCF and LCM.

               Thus, (a × b) = HCF(a, b) × LCM(a, b).

Example: The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers  is 161, find the other.

Sol.  

For two numbers a and b, we know that

(a × b) = { HCF(a, b)} × {LCM(a, b)}.

Here a = 161, b=?   HCF = 23 and LCM = 1449

\( displaystyle begin{array}{l}left( {161times b} right)=left( {23times 1449} right)b=left( {frac{{23times 1449}}{{161}}} right)b=207end{array}\)

Hence, the other number is 207.