Question

If HCF of 65 and 117 is of the form
65m -117, then the value of m is​

Answer 1

$\large\underline{\sf{Solution-}}$

Given numbers are 65 and 117

Let we first find the prime factors of 65 and 117

So,

$\rm \: Prime\:Factorization \: of \: 65$

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{13}}}&{\underline{\sf{\:\:65 \:\:}}} \\ {\underline{\sf{5}}}& \underline{\sf{\:\:5\:\:}}\\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

Now,

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{13}}}&{\underline{\sf{\:\:117 \:\:}}}\\ {\underline{\sf{3}}}& \underline{\sf{\:\:9 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:3\:\:}}\\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

So, we have

$\rm \: Prime\:Factorization \: of \: 65 \: = \: 13 \times 5$

and

$\rm \: Prime\:Factorization \: of \: 117 \: = \: 13 \times 3 \times 3$

$\rm\implies \:HCF \: (65, \: 117) \: = \: 13$

Now, it is given that

$\rm \:HCF \: (65, \: 117) \: = \: 65m - 117$

So, from above two, we concluded that

$\rm \: 65m - 117 = 13$

$\rm \: 65m = 13 + 117$

$\rm \: 65m = 130$

$\bf\implies \:m \: = \: 2$

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ADDITIONAL INFORMATION

If a and b are two natural number having HCF, H and LCM, L then

1. H × L = a × b

2. H always divides L, a as well as b

3. L is always divisible by a, b as well as H

4. HCF of two co - prime numbers is 1

5. LCM of two co - prime numbers a and b is ab.

6. Smallest composite number is 4 and Smallest prime number is 2.

9. Smallest odd composite number is 9.

Answer 2

Solution-

Given number are 65 and 117

To find the prime factors of 65 and 117

Prime factorisation of 65

13 | 65

5 | 5

| 1

Now,

13 | 117

3 | 9

3 | 3

| 1

Prime factorisation of 65 = 13×5

Prime factorisation of 117 = 13×3×3

= HCF (65,117) = 13

HCF (65, 117) = 65m -117

Now,

65m - 117 = 13

65m = 13+117

65m = 130

m = 130/65

m = 2

Hence, the value of m is 2