Can you find the divisor??
Find the largest prime number such that when each of the three numbers 29787, 31339, 33861 are divided by , we get the same remainder.
The answer is 97.
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1 solution
This problem has a very easy solution. Since we have all three numbers leaving same remainder say k when divided by the number p each of the differences of any two numbers is divisible by p . Thus we get 1 5 5 2 ( 3 1 3 3 9 − 2 9 7 8 7 ) , 2 5 2 2 ( 3 3 8 6 1 − 3 1 3 3 9 ) , 4 0 7 4 ( 3 3 8 6 1 − 2 9 7 8 7 ) all are divisble by p . On factorizing, we find that the largest "COMMON" prime factor of theirs is 9 7 Kindly upvote if you liked this!