Math, asked by Advered8434, 1 year ago

The greatest value of k such that each one of the numbers 4/3k, 20 /42k, 8 /6k and 36/63k is an integer is

Answers

Answered by CarlynBronk
0

Solution:

The value of k such that, \frac{4 }{3k}, \frac{20}{42k}, \frac{8}{6k}, {\text{and}, \frac{36}{63k} will be an integer if,

Lowest common multiple(LCM) of  \frac{4 }{3}, \frac{20}{42}, \frac{8}{6}, {\text{and}, \frac{36}{63} , that is

LCM of numerator(4, 20,8,36)

4= 2 × 2

20=2×2×5

8=2×2×2

36=2×2×3×3

LCM(4, 20,8,36)=2× 2× 2 ×3 ×3 ×5

                        = 360

LCM of denominator(3, 42,6,63)

3= 3 ×1

42= 2×3×7

6=2×3

63=3×3×7

LCM(3, 42,6,63)=126

\frac{\text{LCM of numerator(4, 20,8,36)}}{\text{LCM of denominator(3, 42,6,63)}}=\frac{360}{126}

The value of k such that, \frac{4 }{3k}, \frac{20}{42k}, \frac{8}{6k}, {\text{and}, \frac{36}{63k} will be an integer if, \frac{360}{126k} will be an integer, that is ,

\frac{360}{126k}=Any integer {........-1,0,1,2,.......} suppose M

If , k= \frac{360}{126 M},

Where, M = an integer


Answered by josimagic
0

Answer:

The greatest value of k such that each one of the numbers 4/3k, 20 /42k, 8 /6k and 36/63k is an integer is  = 1/126

Step-by-step explanation:

The given numbers are 4/3k, 20 /42k, 8 /6k and 36/63k

let the number be  4/3k become integer when k = 1/3

if the number is 20/42k become integer when k = 1/42

Therefore to find the greatest value of k  such that all the given numbers become an integers, we have to calculate the, LCM of denominators.

LCM(3,42,6,63) = 126

Therefore the value of k = 1/126

Check all the numbers with k

4/3k = 4/(3*1/126) = 4/(1/42) = 42*4 = 168

20/42k = 20/(42*1/126) = 20/(1/4) = 80

8/6k = 8/(6*1/126) = 8/(1/21) = 8*21 = 168

36/63k = 36/(63*1/126) = 36/(1/2) = 36*2 = 72

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