Determine all values of n for which the following statement is true: There exists integers x and y such that 63x + 147y = n.
Give a convincing argument to justify your answer.
63x+ 147y can be written as 21(3x+7y)
Now, 3x + 7y can give all integers values depending on the values of x and y.
For x=0 and y=0 , 3x+7y=0
For x=-2 and y=1 , 3x+7y=1
For x=3 and y=-1 , 3x+7y=2
So, 3x+7y can generate all values from 0 to 2. We can add a multiple of 3 to each of them to get all other integer values.
So, n = 63x+ 147y = 21(3x+7y)
=> n = 21z (where z is an integer)
So, n is a multiple of 21
Possible values of n are ......., -42, -21, 0, 21, 42, 63,.......... and so on
Please upvote. Thanks.
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