Sent you a PM. Check it.
x MUST be 4, 7, 10, 13... etc or any other number that when divided by 3 the remainder is 1 (x = 3k+1)
y MUST be 17, 26, 35, 44... etc or any other number that when divided by 9 the remainder is 8 (y = 9m+8)
xy+1 is divisible by 3 because:
(3k+1)(9m+8) +1
27km+24k+9m+8+1
27km+24k+9m+9 is Divisible by 3
Note that (9km + 8k + 3m + 3) * 3 = 27km+24k+9m+9
If you want to see the inputs:
(4*17)+1 = 69 is Divisible by 3
(7*26)+1 = 183 is Divisible by 3
(4*26)+1 = 105 is Divisible by 3
(13*17)+1 = 222 is Divisible by 3
ccassel wrote:The question requires us to determine if it is sufficient but to take it a step further, how can we determine that the equation "xy + 1" IS certainly divisible by 3 with both 1 and 2?
Does anyone have material on remainder questions that you can direct me to? I need some work in this area.
Thanks in advance.
Chris
"There's a difference between interest and commitment. When you're interested in doing something, you do it only when circumstance permit. When you're committed to something, you accept no excuses, only results."