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| If a and b are positive integers divisible by 6 |
| Re: If a and b are positive integers divisible by 6 |
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UPA
Guest
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From i: a = 2b + 6 if we suppose b = 6k where k is an integer a = 2(6k) +6 a = 6 (2k+1) then the GCD of 6 (2k+1) and 6k is always 6. so suff.... From ii: a = 3b suppose b = 6k where k is an integer a = 3 (6k) a = 18k then the GCD of 18k and 6k is 6k. since we donot know about k, 6k could be 6 or any multiple of 6. so nsf.. So A makes sense... |
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| Re: If a and b are positive integers divisible by 6 |
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Ron Purewal
MGMAT STAFF
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i don't know about simpler words, but i can probably help by illustration. specifically, consider the case in which 'x' is 6. then this means: if you have two multiples of 6 that are only 6 apart, then their greatest common factor is 6. this ought to make sense to you; here's a sketch of the necessary reasoning: (step 1) the greatest common factor is AT LEAST 6: this must be true, because 6 goes into both numbers. therefore 6 is a common factor of the two numbers. we're not done yet, though, because we haven't established that it's the greatest common factor. (step 2) the greatest common factor is AT MOST 6: the numbers are only 6 apart. therefore it's impossible that any number more than 6 could go into both of them, because they aren't far enough apart. (if you have two multiples of 12, for instance, they have to be at least 12 apart, and so on) the same is true for all other integers as well; there's nothing special about 6. |
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| If a and b are positive integers divisible by 6 |
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