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GCD

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by yellowho » Mon Mar 07, 2011 12:19 am
If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?

1) x = 12u, where u is an integer.

2) y = 12z, where z is an integer.
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Source: — Problem Solving |

by sanju09 » Mon Mar 07, 2011 12:58 am
yellowho wrote:If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?

1) x = 12u, where u is an integer.

2) y = 12z, where z is an integer.

(1) If x = 12 u, where u is an integer, then the equation in the stem x = 8 y + 12 can be read as

A multiple of 12 = 8 y + 12

or simply, A multiple of 12 = 8 y.

Now the big question-Is y a multiple of 12? Don't know, it may or may not be so. Insufficient

(2) If y = 12 z, where z is an integer, then the equation in the stem x = 8 y + 12 can be read as

x = 8 × A multiple of 12 + 12 = Another multiple of 12.

Now since y is a given multiple of 12 and x is hence proved to be another multiple of 12 and in the present case, x and 8 y are both separable by 12, and the two numbers are also spaced out by 12 digits on the real number line, so [spoiler]12 is the GCD of x and y.


B
[/spoiler]
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by yellowho » Mon Mar 07, 2011 2:04 am
Here's my approach: (please poke holes thru it if I am wrong)

S1:

12u=8y+12
12(u-1)+12=8y+12
12(u-1)=8y
divide by 4: 3(u-1)=2y
Y is divisible by 3 and u divisible by 2, x divisible by 12.

No singular result. Insuf.

S2:

(same method as above)
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