QUANT ERROR
√.5>.5?
YES. √.5=.7 (since # gets smaller if you multiply decimal x decimal) BUT ANY FRACTION RAISED TO FRACTION WILL BE < 1.
If integer is perfect square...
has odd # of factors (16-> not perfect square of prime, but has odd # of factors: 1, 2, 4, 8, 16 (2⁴ = 16, ⁴+¹ = 5 factors)
If integer has odd # of factors...
it's a perfect square! (e.g. 81-> not perfect square of prime, but has odd# of factors: 1, 3, 9, 27, 81 (3⁴=81, ⁴+¹=5 factors).
Nth root of any positive # >1
will be >1, since any root of 1 = 1
Root√2√3√4√5√6√7√8√9√10
√2=1.4, √3=1.7, √4=2, √5=2.2, √6=2.4, √7=2.6, √8=2.8, √9=3, √10=3.16
EVENLY SPACED SET MEAN/MEDIAN PROPERTIES:
1) Mean ALWAYS = MEDIAN 2) AVG OF FIRST + LAST TERMS = MEAN = MEDIAN 3) MEAN OF A SET w/ ODD NUMBER OF TERMS ALWAYS AN INTEGER, MEAN OF OF A SET w/ EVEN NUMBER OF TERMS NEVER AN INTEGER
X is same sign as X^3, so if -X>X^3, then x must be negative... so that -X will be positive #>negative...if X is positive, it would be -X> positive^3 which is impossible ***BLUF: PLUG IN or THINK THROUGH INEQUAL.
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PATTERNS OF POWERS (Units Digit)
2¹=2, 2²=4, 2³≅8, 2⁴=16, 2⁵=32, 2⁶=64 (rpt: 2,4,8,6 every 4th) 3¹=3, 3²=9, 3³≅27, 3⁴=81, 3⁵=243, 3⁶=729 (rpt: 3,9,7,1 every 4th) 4¹=4, 4²=16, 4³≅64, 4⁴=256 (rpt: 4,6,4,6 every 2nd) 5¹=5, 5²=25 (rpt 5) 6¹=6, 6²=36 (rpt 6) 7¹=7, 7²=49, 7³≅343, 7⁴=2401, 7⁵=16807 (rpt: 7,9,3,1 every 4th) 8¹=8, 8²=64, 8³≅512, 8⁴=4096, 8⁵=32768 (rpt: 8,4,2,6 every 4th) 9: (9,1: every 2nd)
Special Right Triangles: 3-x-5
3-4-5, 6-8-10, 9-12-15, 12-16-20, 15-20-25
BEWARE DS problems where Statement info is SAME as GIVEN info, just in different form!
4x=5y SAME AS y=.80x, since 4/5 = 80%
Special Right Triangles: 5-12-x
5-12-13, 10-24-26
Special Right Triangles:7-x-25
7-24-25
Special Right Triangles: x-15-17
8-15-17, 16-30-34
4⁰ or 2987⁰ or any positive #⁰
=1
If integer K is equal to sum of ALL EVEN MULTIPLES OF 15 between 295 and 615, what is the greatest prime factor of K?
ALL EVEN MULTIPLE OF 15 IS: MULTIPLES OF 15 x 2=30, SO FIND ALL MULTIPLES OF 30!!!
Sum of Odd # of consecutive integers...1+2+3
ALWAYS multiple of # of integers (1+2+3=6/3 integers)
Product of N evenly spaced terms...
ALWAYS multiple of N!
Remainders upon Division by 10
Always UNITS DIGIT (e.g. 252 divided by 10, remainder=2)
The PF's of Perfect Squares
Contain only EVEN POWERS
PF's of Perfect Cubes
Contain only POWERS WITH MULTIPLES OF 3
Shortest distance from vertex of cube to surface of sphere...
Diagonal - Diameter of Sphere... or 1/2 Diagonal - radius of sphere
Repeating Decimal to Fraction=.4444
Divide by 9 or 99....4/9 or 44/99
If M=√4+³√4+⁴√4, then M is: a) less than 3 b) equal to 3 c) between 3 and 4 d) equal to 4 e) greater than 4
E) Greater than 4.
To determine factors of N evenly spaced terms, substitute the smallest possible values for those N terms. The factors of resulting product will all be factors of the product in question!
IF N is an even integer >0, is N³+3N²+2N div. by 24? Yes, this can be re-written as N(N+1)(N+2). If N must be an even int. the smallest possible values we can substitue for N(N+1)(N+2) are 2,3,and4. Since product of 2x3x4=24 and contains factors 1,2,3,4,6,8,12,24, N³+3N²+2N div. by 24.
COUNT CONSEC. #'s
INCLUSIVE: Difference between end numbers of set and ADD 1 EXCLUSIVE: Difference between end numbers of set and SUBTRACT 1 (IF MULTIPLES, then divide by given multiple while ensuring end #'s are multiples, and add or subtract 1)
IF integer has exactly three factors...
It's a perfect square of a prime #: Factors: 1, itself, and square root of itself.... 3² = 9, factors are 1, 3, 9. 5²=35, factors: 1, 5, 25
Sum of Even # of consecutive integers...1+2
NEVER multiple of # of integers (1+2=3/2 integers) (Rule applies ONLY to CONSECUTIVE INTEGERS. If Set N contains 4 consecutive multiples of 4, the sum of set N is ALWAYS divisible by 4, since they are evenly spaced MULTIPLES NOT CONSECUTIVE INTEGERS. If N is an integer, the sum and difference of any multiples of N are ALWAYS multiples of N as well! (e.g. N = 4, so difference of multiples of 4 such as 16-8 will equal a multiple of 4, 16-8=12...sums as well: 8+16=24)
If 60! is written out as an integer, with how many consecutive 0's will that integer end?
Number of 0's at end of integer is number of times integer divisble by 10. Since 10=2x5, this number (# of times integer div. by 10) is smaller of the powers to which 2 and 5 are raised in integer's PF. Integer whose PF has 2⁸x5⁷ and 2⁷x5⁸ will both end with SEVEN 0's since it will have the # of 0's of the LESSER POWER. Since many more 2's than 5's in PF of 60!, REPHRASE: HOW MANY 5's are in PF of 60!? 60! is product of all integers from 1 to 60 inclusive. Each of twelve multiples of 5 contributes a 5 to the PF. ALSO, two multiples of 25 (25 and 50) also contribute a 5 each. so 60! will end in fourteen 0's.
Diagonal of Cube
Side√3
Exponents/powers of 10...10³, 10⁶
THE POWER = # OF ZEROS AFTER 1!!!! 10³=1000 10²=100 10⁶=1000000
OVERLAP Problems
1) Find Overlap 2) Test Extremes (1: Best case, Where all games are won at ends/minimum overlap, 2: Worst case, Where all games are won at overlap/max overlap)
PERFECT CUBES: Contain only prime factors whose exponents are multiples of 3
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PERFECT SQUARES: Contain only prime factors whose exponents are even
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