Find the greatest number that will divide 186, 321 and 402 so as to leave the same remainder in each case.
27
The greatest number leaving the same remainder divides the differences between the numbers exactly.
Differences: \(321-186=135\), \(402-321=81\), \(402-186=216\).
HCF of 135, 81 and 216: \(\text{HCF}(135,81)=27\), and \(\text{HCF}(27,216)=27\).
Hence, the greatest such number is 27.
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A and B are two prime numbers such that A > B and their LCM is 209. The value of A 2 - B is:
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