Pearls
Seven friends lost each other in a jungle. After some time one of them found a bag of pearls. He was very happy. But suddenly one of his friend arrived. They both agreed to divide the pearls between themselves. After division, one pearl resulted in extra. At this moment their third friend arrived, and they again distributed all the pearls among themselves. But again one pearl left. Similarly when the fourth, fifth and sixth friend arrived, the result was still one pearl left. At last when their seventh friend met them, this time all the pearls were distributed equally. Find the minimum numbers of pearls in the bag.
The answer is 301.
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Excellent problem! I solved it by finding first gcd ( 2 , 3 , 4 , 5 , 6 ) , which is simply 6 0 . From there, for our required to leave a remainder of 1 , our number must be 6 0 + 1 = 6 1 , or any other number in the form 6 0 x + 1 . Now, examine our second bound... it's divisible by seven. Thus we're looking for a solution to the Diophantine equation 6 0 x + 1 = 7 y . Solving, we find that the minimum possible amount of pearls is 3 0 1 .