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The HCF of 2472, 1284 and a third number 'N' is 12. If their LCM is 23 × 32 × 5 × 103 × 107, then the number 'N' is:
Answer & Solution
Correct Answer:
Option
D
We have,
HCF of the numbers × LCM of the numbers = Multiplication of the numbers.
(12) × (23 × 32 × 5 × 103 × 107) = 2472 × 1284 × N
Hence,
N = $$\frac{{\{ \left( {12} \right) \times \left( {{2^3} \times {3^2} \times 5 \times 103 \times 107} \right)\} }}{{2472 \times 1284}}$$
Or, N = 3 × 5
HCF of the numbers × LCM of the numbers = Multiplication of the numbers.
(12) × (23 × 32 × 5 × 103 × 107) = 2472 × 1284 × N
Hence,
N = $$\frac{{\{ \left( {12} \right) \times \left( {{2^3} \times {3^2} \times 5 \times 103 \times 107} \right)\} }}{{2472 \times 1284}}$$
Or, N = 3 × 5
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Logintaken the lcm of given three numbers is
lcm=12*206*107*y (N=12*y)
and given lcm is 2^3*3^2*5*103*107
and equating both lcms and get the value of y is 15
so the third number is 2^2*3^2*5
Solution for this question = 2³×3²×5×103×107×12²=2472×1284×b
Find from here the value of b.... Here, value of b is our answer. It is also option C
Right all the factors for all the values. Figure out the GCD from that.
Then the remaining values that are not present in LCM has to be present in the value N.
can you give your explanation for this question? @Sakthivel