Number Properties, 4th Edition, Hard Practice Question Set

[bal2155]

Question #16: we are given a set of three consecutive integers x, y, z and asked to answer which of a number of additive and multiplicative operations involving x, y, z are divisible by 3.

Why is it that "any group of three consecutive integers must include a multiple of 3", as the solution states? What about, for example, the set [-1, 0, 1]? I feel like I'm missing something there.

I'm using the Kindle eBook version so here are screenshots of the question and solution:

https://www.dropbox.com/s/ar5c6ug0yxqkl56/NP%20Hard%20Set%20Question%2016.png?dl=0

https://www.dropbox.com/s/nb1ae8assj3c03b/NP%20Hard%20Set%20Solution%2016.png?dl=0

Thanks!

[n00bpron00bpron00b]

A set of 3 consecutive integers will always have a value that is divisible by 3 (or a multiple of 3)

In general, the product of N consecutive integers is always divisible by N!

1 2 3
4 5 6 (6 is 3x2)
7 8 9 (9 is 3x3)
10 11 12 (12 is 4x3) and so on.


Regarding the set of values,

(-1,0,1) - the product of this is "0" and "0" is a multiple of all positive integers. So 0/3 = Quotient of 0 and Remainder of 0

[bal2155]

Thanks for the reply - still having trouble understanding how 0 is a multiple of 3, though.

I always thought the multiples of 3 would be 3, 6, 9, 12 etc...

[n00bpron00bpron00b]

x - numerator (dividend)
y - denominator (divisor)

whenever you come across a case where "x" < "y" (where x and y are both positive integers) - the result of x/y will be

0 quotient (y goes into x zero times)
and dividend as the remainder

This is true for all positive values.

[bal2155]

I think the proof I was looking for was:

For any 3 integers x, y, z

If xy=z then z is a multiple of x and y

Let x=3, y=0; xy = 0 = z
Therefore 0 is a multiple of 3.

Thanks!

[tommywallach]

Great convo here. Proof is unnecessary though. Easier just to have a rule: 0 is a multiple of everything.

-t