Question

The H.C.f of 1284,2472 & N is 12. If their L.C.M is 2 3 *3 2 *5*103*107 then find the number "N".......? which "DAREDEVIL" can solve this itz my CHALLENGE...come on..!!

Answer 1

1284=2×2×3×107 and
2472=2×2×2×3×103
∴, the 3rd no.= n =2×2×3×a=12×a=H.C.F×a
Now, b=1284/12=107
c=2472/12=2×103
L.C.M/12= (2³×3²×5×103×107)/12=2×3×5×103×107
Then, a=L.C.M/(b×c)
or, a=(2×3×5×103×107)/(107×2×103)
or, a=3×5
or, a=15
∴, a=12×15=180 Ans.

Answer 2

Solution:-
Prime factorization of the numbers.
1284 = 2*2*3*107 = 2² × 3 × 107
2472 = 2*2*2*3*103 = 2³ × 3 × 103
Now, since the H.C.F. of three numbers is 12, the number N will also have 12(2²*3).
So, N = 2²*3
Now, the L.C.M. contains the highest power of three as 2(3²), but the prime factorization of the remaining numbers does not contain them, so the number N should contain it.
Therefore, N = 2²*3²
The L.C.M. contains a prime number 5, but the prime factorization of the other two numbers does not have 5. So, the number N s must contain it.
So, N = 2²*3²*5 
N = 180
So, the value of N is 180.
Answer.