question_answer

  Find the largest positive integer that will divide 398, 436 and 542 leaving remainders 7, 11 and 15 respectively. OR Prove that n3 - n is divisible for 6, for every positive integer n.

Answer:

  Clearly, the required number is the HCF of the numbers 398 - 7 = 391, 436 - 11 = 425, and 542 - 15 = 527. First we find the HCF of 391 and 425 by Euclid's algorithm as given below.

11	391	425 391	1
	(HCF)	34	2
		0 (Remainder)

Clearly, H.C.P. of 391 and 425 is 17. Let us now find the HCF of 17 and the third number 527 by Euclid's algorithm.

17 (HCF)
	0 (Remainder)

The HCF of 17 and 527 is 17. Hence, HCF of 391, 4250 and 527 is 17. Hence, the required number is 17. OR n3 - n = n (n2 - 1) = n (n - 1) (n + 1) = Product of three consecutive positive integers and hence divisible by 3! =6