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
Solution:
The value of k such that, will be an integer if,
Lowest common multiple(LCM) of , 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
The value of k such that, will be an integer if, will be an integer, that is ,
=Any integer {........-1,0,1,2,.......} suppose M
If , k= ,
Where, M = an integer
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