find the greatest greatest number which divides which divides 121, 226 and 259 and leaves remainder 1, 2 and 3 respectively
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78
Answer:
8 is the largest number that divides 121, 226 and 259 and leaves remainder of 1,2 and 3 respectively.
Step-by-step explanation:
Subtract the given remainders from the numbers that are given :-
121 - 1 = 120 = 2 x 2 x 2 x 3 x 5 = 2^3 x 3 x 5
226 - 2 = 224 = 2 x 2 x 2 x 2 x 2 x 7 =2^5 x 7
259 - 3 = 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2^8
Common prime factors: 2
Common prime factors with the lowest exponent: 2^3
HCF (120, 224, 256 ) = 8
∴ 8 is the largest number that divides 121, 226 and 259 and leaves remainder of 1,2 and 3 respectively.
anushkaverma39:
thank u so much
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8 is the answer of this question
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