the greatest number which divides 106 241 and 286 leaving the same remainder in each case is
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If a number 'a' and a number 'b' are divisible by a number 'n' then, a+b and a-b is also divisible by n
=> let the number be n which divides 62, 132 and 237 leaving a reminder r
=> the required number then becomes H.C.F of 62-r, 132-r and 237-r
=> it could also be the H.C.F of
(132-r)-(62-r) and (237-r)-(132-r)
i.e. 70 and 105
=> H.C.F of 70 and 105 = 35
the required number n is 35.
=> let the number be n which divides 62, 132 and 237 leaving a reminder r
=> the required number then becomes H.C.F of 62-r, 132-r and 237-r
=> it could also be the H.C.F of
(132-r)-(62-r) and (237-r)-(132-r)
i.e. 70 and 105
=> H.C.F of 70 and 105 = 35
the required number n is 35.
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