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# What is the gcd of a, b, and 3a+23b 1. a = 4 2. gcd(a, b,

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What is the gcd of a, b, and 3a+23b 1. a = 4 2. gcd(a, b, [#permalink]

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12 Apr 2004, 11:41
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What is the gcd of a, b, and 3a+23b

1. a = 4
2. gcd(a, b, 23a+3b) = 4
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16 Apr 2004, 03:29
hallelujah1234 wrote:
What is the gcd of a, b, and 3a+23b

1. a = 4
2. gcd(a, b, 23a+3b) = 4

say i take a = b = 2
then 23a + 3b = 3a + 23b = 52

so B is sufficient
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15 May 2004, 17:37
abeautifulmind wrote:
hallelujah1234 wrote:
What is the gcd of a, b, and 3a+23b

1. a = 4
2. gcd(a, b, 23a+3b) = 4

say i take a = b = 2
then 23a + 3b = 3a + 23b = 52

so B is sufficient

if you take a = b = 2, then how does that satisfy given second statement? The gcd in that case will be 2.

I would solve the question by applying the halle..'s methodology. Halle can then approve wether this solution is correct or not.

gcd(a, b, 23a+3b) = 4

=> a = 4m, where m > 0 (integer)
=> b = ak + 4 = (4m)k + 4 = 4mk + 4, where k >=0; integer
=> 3a+23b = 3(4m) + 23(4mk+4) = 4 [3m + 23(mk+1)]

So the gcd (a, b, 3a+23b) = 4

So the answer is B.
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15 May 2004, 17:44
gcd(a, b, ax+by) = gcd(a, b) (obvious)

B says, gcd(a,b) = 4. So, gcd(a, b, 3a+23b) = 4.
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15 May 2004, 17:58
hallelujah1234 wrote:
gcd(a, b, ax+by) = gcd(a, b) (obvious)

let
a=3
b=2
x=5
y=7

ax+by=15+14=29
gcd of 2, 3, 29 isn't 6
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15 May 2004, 18:02
mirhaque wrote:
hallelujah1234 wrote:
gcd(a, b, ax+by) = gcd(a, b) (obvious)

let
a=3
b=2
x=5
y=7

ax+by=15+14=29
gcd of 2, 3, 29 isn't 6 :panel

gcd(2, 3) = 1; gcd(2, 3, 29) = 1 = gcd(2, 3)

The proof is based on the fact: that gcd(a, b, c, ...) is the least positive integer in the set {am1+bm2+cm3+ ... | mi is an integer}
[#permalink] 15 May 2004, 18:02
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# What is the gcd of a, b, and 3a+23b 1. a = 4 2. gcd(a, b,

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