DonQuixote wrote:
That's pretty interesting. Does the book also say how exactly this works? What are the properties of the numbers that allow this system to be applied?
Separate the last digit from the rest of the number.
Let x be the units digit, and let y be the rest of the number.
So for any number N, N = 10y+x.
Using rahulsn84's method, we say that N should be divisible by prime p if y+Mx is divisible by p, where M is some 'check multiplier'.
10y + x = y + Mx
10y + x = 10y + 10Mx
x = 10Mx
0 = 10Mx - x
0 = x(10M - 1)
Now the question is, is x(10M - 1) divisible by p? We have defined the check multiplier such that 10M - 1 is divisible by p. So yes, p|(10M - 1).
Check multiplier for 7 is 5. 10*5 - 1 = 49, which is 7*7.
Check multiplier for 13 is 4. 10*4 - 1 = 39, which is 13*3.
Check multiplier for 17 is 12. 10*12 - 1 = 119, which is 17*7.
Check multiplier for 19 is 2. 10*2 - 1 = 19, which is 19*1.
Check multiplier for 23 is 7. 23*4 - 1 = 69, which is 23*3.
etc.
This works for all primes.
You can simplify (or complicate, depending on how you see things) this method by also realizing that you don't things to end in 9, but ending it with 1 works as well.
For example, for p = 17, instead of using a check multiplier of 12, we can use a check multiplier of 5. 17*3 = 51 = 10*5 + 1. In this case, instead of adding Mx, we just need to subtract Mx. That is, 10y + x = y - Mx. Doing the algebra, we get x(10M + 1) = 0.
But yes, as so many others pointed out, this will not be on the GMAT.