osirus0830 wrote:These are the only problems that I can't solve at least some of the time. Would you have been able to figure these problems out without MGMAT numbers properties book? Is there any web resource that would help or should I just break down and purchase that book?
on REMAINDER problems, such as these,
you should almost always be able to solve the problem by SYSTEMATIC PLUGGING-IN OF NUMBERS.
since remainders are fundamentally based on repetition - i.e., remainders themselves repeat in cycles, over and over again, predictably and systematically - you should be able to
discover PATTERNS in them, simply by trying enough examples.
just GENERATE EXHAUSTIVE LISTS, try them, and, there you go.
the only catch with this method is that this is not fast, so you have to DECIDE RIGHT AWAY if you want to use this method.
the idea is to try the lists until one of two things happens:
(a) you see an obvious PATTERN
(b) you get "INSUFFICIENT"
as soon as either of these things happens, you are finished.
--
here's a taste of this method, applied here:
statement (1)
this means odd numbers. so, try odd numbers, and see what happens.
n = 1 --> (n + 1)(n - 1) = 0 --> remainder when 0 is divided by 24 = 0
n = 3 --> (n + 1)(n - 1) = 8 --> remainder when 8 is divided by 24 = 8
if you don't like dividing numbers smaller than 24 by 24, then try the next two or three:
n = 5 --> (n + 1)(n - 1) = 24 --> remainder when 24 is divided by 24 = 0
n = 7 --> (n + 1)(n - 1) = 48 --> remainder when 48 is divided by 24 = 0
n = 9 --> (n + 1)(n - 1) = 80 --> remainder when 80 is divided by 24 = not 0 (we don't care what it is, as long as it's different from the previous answers, since that's "insufficient")
insufficient.
statement (2)
n = 1 --> (n + 1)(n - 1) = 0 --> remainder when 0 is divided by 24 = 0
n = 2 --> (n + 1)(n - 1) = 3 --> remainder when 3 is divided by 24 = 3
if you don't like dividing numbers smaller than 24 by 24, then try the next two or three:
n = 5 --> (n + 1)(n - 1) = 24 --> remainder when 24 is divided by 24 = 0
n = 7 --> (n + 1)(n - 1) = 48 --> remainder when 48 is divided by 24 = 0
n = 8 --> (n + 1)(n - 1) = 63 --> remainder when 63 is divided by 24 = not 0 (we don't care what it is, as long as it's different from the previous answers, since that's "insufficient")
insufficient.
together:
n = 1 --> (n + 1)(n - 1) = 0 --> remainder when 0 is divided by 24 = 0
n = 5 --> (n + 1)(n - 1) = 24 --> remainder when this is divided by 24 = 0
n = 7 --> (n + 1)(n - 1) = 48 --> remainder when this is divided by 24 = 0
n = 11 --> (n + 1)(n - 1) = 120 --> remainder when this is divided by 24 = 0
n = 13 --> (n + 1)(n - 1) = 168 --> remainder when 48 is divided by 24 = 0
ok, the pattern here is pretty obvious.
it's always 0.
sufficient.
