Stiv wrote:
If k, m, and t are positive integers and \frac {k}{6} + \frac {m}{4} = \frac {t}{12} , do t and 12 have a common factor greater than 1?
(1) k is a multiple of 3.
(2) m is a multiple of 3.
In the explanation of this question they say that the sum of two multiples of 3 give the number that is also a multiple of 3.
Is that a general rule for any number? If someone can elaborate I would be grateful!
If k, m, and t are positive integers and \frac{k}{6} + \frac{m}{4} = \frac{t}{12}, do t and 12 have a common factor greater than 1 ?\frac{k}{6} + \frac{m}{4} = \frac{t}{12} -->
2k+3m=t.
(1) k is a multiple of 3 -->
k=3x, where
x is a positive integer -->
2k+3m=6x+3m=3(2x+m)=t -->
t is multiple of 3, hence
t and 12 have a common factor of 3>1. Sufficient.
(2) m is a multiple of 3 -->
m=3y, where
y is a positive integer -->
2k+3m=2k+9y=t -->
t and 12 may or may not have a common factor greater than 1. Not sufficient.
Answer: A.
As for your question:
If integers a and b are both multiples of some integer k>1 (divisible by k), then their sum and difference will also be a multiple of k (divisible by k):Example:
a=6 and
b=9, both divisible by 3 --->
a+b=15 and
a-b=-3, again both divisible by 3.
If out of integers a and b one is a multiple of some integer k>1 and another is not, then their sum and difference will NOT be a multiple of k (divisible by k):Example:
a=6, divisible by 3 and
b=5, not divisible by 3 --->
a+b=11 and
a-b=1, neither is divisible by 3.
If integers a and b both are NOT multiples of some integer k>1 (divisible by k), then their sum and difference may or may not be a multiple of k (divisible by k):Example:
a=5 and
b=4, neither is divisible by 3 --->
a+b=9, is divisible by 3 and
a-b=1, is not divisible by 3;
OR:
a=6 and
b=3, neither is divisible by 5 --->
a+b=9 and
a-b=3, neither is divisible by 5;
OR:
a=2 and
b=2, neither is divisible by 4 --->
a+b=4 and
a-b=0, both are divisible by 4.
Hope it's clear.
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