find the greatest number of 258, 321 and 602 leaving remainder of 6, 7 and 8 respectively
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Answer:
Step-by-step explanation:To find the Greatest number which divides 258 and 323 leaving remainder 2 and 3 in each case i.e. HCF.
Consider HCF as x.
In order to make 258 and 323 completely divisible by x, we need to deduct the remainder 2 and 3 from both the cases.
256 = 4×64
320= 5×64
Greatest number which divides 256 and 320 leaving remainder 2 and 3 in each case is 64.
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2
Answer:
258-6=252
321-7=314
602-8=594
hcf of 252,314,594=2
the greatest number is 2
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