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Question: Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is Solution: N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) = H.C.F. of 3360, 2240 and 5600 = 1120. Sum of digits in N = (1 + 1 + 2 + 0) = 4 Aren't we supposed to find the HCF (1305, 4665, 6905) rather than that of the HCF(4665 - 1305), (6905 - 4665) and (6905 - 1305)? |
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Answer» Question: Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is Solution: |
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