​YESTERDAY I HAD ASKED A QUESTION IN MATHS, and THE EXPERTS PROVIDED ME WITH A CERTIFIED ANSWER. BUT,THE SOLUTION WAS NOT CLEAR... HERE IS THAT QUESTION:
QUESTION:

A number"X" leaves the same reminder while dividing 5814, 5430, 5958. What is the largest possible value of "X" . 
{ Please find the solution with proper explanation and please don't give any weblink or any certified answer}

Let r be the same remainder left in each case when X divides 5814, 5430 and 5958.According to division algorithm,5814 = aX + r     where a is the quotient5430 = bX + r     where b is the quotient5958 = cX + r     where c is the quotientNow, 5814-5430 = a-bX  384 = a-bX Now, 5958 - 5430 = c-bX 528 =c-bX Now, 5958 - 5814 = c-aX  144 = c-aXNow, 384, 528 and 144 each of the given number is divisible by X.The prime factorisation of 384, 528 and 144 is,384 = 2×2×2×2×2×2×2×3528 = 2×2×2×2×3×11144 = 2×2×2×2×3×3HCF of 384, 528 and 144 = 2×2×2×2×3 = 48So, largest possible value of X is 48. 

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