Strategies for Numbers

(√ ------ SQ root sign)

1) If the prime numbers p and t are the only prime factors of the integer m. Any possible product combination of the m's prime factors are also factors of m, or m is also a multiple of these product combinations (m would also be a multiple of p*t and p*p*p*t.


2) Memorize the fact that all even perfect squares are divisible by 4 (and thus give remainder 0 upon division by 4), and that all odd perfect squares are one more than a multiple of 4 (and thus give remainder 1 upon division by 4). therefore, if one of the squares is even and the other odd (as stipulated by statement 2), the overall remainder will be 0 + 1 = 1


3) For any 2 numbers x & Y.......

X * Y = HCF * LCM

HCF = Highest common factor
LCM = Lowest common factor

4) Remainder Problems....
general rule: if N '<' M (both positive integers) and you divide N by M, then the remainder is N. e.g x/6 and the remainder is 3 ----- Number can be (x = 3,9,15,21........)[Very useful rule in remainder problems]


5) Consider consecutive numbers :
n-1,n,n+1
a) If n is odd, then n-1 & n+1 will be even. Since every other even number is a multiple of 4, we can find the factor of (n+1)(n-1) as one multiple of 2 and the other a multiple of 4.
b) Exactly one of the 3 intergers will be divisible by 3.
c) if n is not divisible by 2, then n is odd, so both (n - 1) and (n + 1) are even. moreover, since every other even number is a multiple of 4, one of those two factors is a multiple of 4. so the product (n - 1)(n + 1) contains one multiple of 2 and one multiple of 4, so it contains at least 2 x 2 x 2 = three 2's in its prime factorization.
d)if n is not divisible by 3, then exactly one of (n - 1) and (n + 1) is divisible by 3, because every third integer is divisible by 3. therefore, the product (n - 1)(n + 1) contains a 3 in its prime factorization.


6) POWERS of a number have EXACTLY THE SAME PRIME FACTORS as does the ORIGINAL NUMBER.


7) only perfect squares have 3 factors : e.g 2^3 = 1,2,8 are the factors.

8) if you want a terminating decimal from a fraction, you have to have only 2's and/or 5's in the denominator .  

9) Whenever you do number plugging in a number properties problem, you have to figure out what types of numbers will have different effects upon the problem. in this case, you're considering both even and odd powers, and you're comparing those powers. therefore, you must consider signs (because there are both even and odd powers) AND fractions (because power comparisons are reversed with fractions). plug numbers accordingly