find the greatest number which can divide 284,698 & 1618 leaving the same remainder 8 in each case
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10
Since,
284, 698 and 1618 leave remainder 8 therefore, (284-8), (698-8) and (1618-8) will be completely divisible to that number, that is,
276, 690 and 1610
the required number is their HCF.
and their HCF = 46
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284, 698 and 1618 leave remainder 8 therefore, (284-8), (698-8) and (1618-8) will be completely divisible to that number, that is,
276, 690 and 1610
the required number is their HCF.
and their HCF = 46
Plz mark it as brainliest
Answered by
0
Answer: (284-8),(698-8),(1618-8)
Hcf of 276,690,1610 = 23
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