Q3) Number properties Flashcards

Question

If you know the LCM and GCF of TWO positive integers, what do you know?

Answer 1

You know the PRODUCT of the two positive integers This is because GCF and LCM are like two sides of the same coin and together, you cover ALL the prime factors Recall: > LCM consists of the largest prime factors + non-repeated prime factors > GCF consists of the smallest common prime factors

Answer 2

x = numerator y = divisor Factor = divisor Dividend = multiple Also can express as: x is divisible by y MATHEMATICALLY: x / y = INT (remainder = 0) INT often assigned k

Answer 3

ALWAYS prime factorize both the dividend and divisor > you can see if there is a remainder AND turn into PRODUCT of numbers

Answer 4

If x and y are positive integers and x / y is an integer (or y is a factor of x), then x / any factor of y is also an integer > A positive integer is divisible by all factors of a factor DEALING WITH POSITIVE NUMBERS

Answer 5

Is x = 2*3*m, where m is a non-negative integer?\ Or 2 and 3 are both factors of x

Answer 6

If z is divisible by BOTH of the two numbers, then z must also be divisible by the LCM of the two numbers (which is NOT always the product of the two numbers because of overlapping factors) Alternatively: z is divisible by the LCM of its factors Please note: It is NOT always true that z is divisible by the PRODUCT of its factors because of overlapping factors e.g., z is divisible by 15 and 20 The smallest value of z = LCM of 15 and 20 = 60, which does NOT equal the product 300

Answer 7

1) Factor of factors: A positive integer is divisible by all factors of a FACTOR 2) A positive integer is divisible by the LCM of ITS FACTORS (but not necessarily the product of its factors) ---> single integer 3) The PRODUCT of two or more integers is divisible by the PRODUCT of their factors ---> product of multiple integers

Answer 8

1) Must be given information about POSITIVE INTEGERS or WHOLE NUMBERS / NON-NEGATIVE numbers 2) You can determine what "could" be the value using divisibility rules or whether an unknown is "divisible by x" (in order to make one of the unknowns a whole number) Strategy: Set up the equation and rearrange so that one side is one unknown integer value, and the other side has the actual unknown integer value you are trying to solve for > then simplify > Then compare the actual unknown integer value with the denominator e.g., At a certain department store, a rug was originally priced at W dollars, where W is a whole number. During a liquidation sale, the rug was sold for 6 percent of its original price. Which of the following could be the sale price of the rug? > Let P = sale price > What value of P would make W a whole number? > W = 50P/3 --> P must be a multiple of 3 and W must be a multiple of 50

Answer 9

If the last two digits are divisible by 4, then the entire number is divisible by 4 Includes -00 (all multiples of 100 are divisible by 4, because will have 4*25 as factors)

Answer 10

If the number is EVEN and the last 3 digits are divisible by 8 (including -000, all multiples of 1000 are divisible by 8 because 8*125 = 1000)

Answer 11

Sum of all digits is divisible by 9

Answer 12

Sum of the odd-numbered place digits - sum of the even-numbered place digits is divisible by 11 Odd-numbered place digits --> ones, hundreds, ten-thousands etc. Even-numbered place digits --> tens, thousands, hundred-thousands etc.

Answer 13

Number is divisible by both 3 and 4

Answer 14

Number is an EVEN number whose digits SUM to a multiple of 3 (so divisible by both 2 and 3, the factors of 6)

Answer 15

Rule: The product of N consecutive integers will always be divisible by N! AND all the factors of N! Alternatively, the product of N consecutive integers is a MULTIPLE of N! (does not have to be equal to N!) > Also the product is divisible by all the FACTORS of N! / factor combinations (factors of factors rule) > so (N-1))! is a factor, (N-2)! is a factor etc. Rationale: > every N numbers, the pattern repeats for N and for N-1, N-2 etc. e.g., the product of any three consecutive integers will always be divisible by 3!, but also 1 and 2 e.g., the product of any four consecutive integers will always be divisible by 4!, but also factors of 4! (24) --> 1, 2, 3, 4, 6, 8, 12, 24 Many algebraic forms: (n-1)*n*(n+1); (n+10)*(n+11); (n-100)*(n-99)

Answer 16

Remember: > Order of the terms in a product don't matter n^3 - n = n*(n^2 - 1) = n*(n + 1)*(n - 1) still a product of 3 consecutive integers if rearranged as so Another example: n^5 - 5n^3 + 4n = n*(n^4 - 5n^2 + 4) = n*(n^2 - 1)*(n^2 - 4) = n*(n + 1)*(n - 1)*(n - 2)*(n+2) = (n-2)*(n-1)*n*(n+1)*(n+2)

Answer 17

Rule: Product of N consecutive EVEN integers is divisible by 2^N * N! Rationale: > each even term will have a "2" factor (2^N) > then multiply by N! to account by consecutive nature Many algebraic forms (gap must be two AND you know whether even/odd): (n-1)*(n+1) NOTE: there is NO special divisibility rule for the product of consecutive ODD integers

Answer 18

1. at least one even integer, so the product is EVEN ---> rule to memorize (N! --> also divisible by factors of N!) 2. product is divisible by 3! --> 2, 3, and 6 ---> rule to memorize (N!) 3. If the middle term is ODD, the product is divisible by 8 (two consecutive even integers)

Answer 19

x / y = Q + r / y Rearranged: x = Qy + r r = x - Qy Keep in mind: > The remainder MUST be NONNEGATIVE and be LESS than DIVISOR y ------> the only possible values of r can be 0 to one less the divisor y = [0, 1, ... up to y - 1] > r/y is always LESS THAN 1 > start off every remainder question by writing out "y > r" > In other words, the DIVISOR (not x) determines the range of possible remainders > DO NOT SIMPLIFY THE FRACTION r / y ---> remainders only make sense in the context of a DIVISOR > Q = quotient = must be an integer (but could be in other forms like n^2 + 1) Any integer can be expressed as the product of some integer quotient Q and some divisor y, plus some remainder

Answer 20

If you know the remainders when X is divided by TWO DIFFERENT DIVISORS, you can come up with a COMBINED EQUATION FOR X Approach #1) Combine into one equation for X x = LCM of divisors * Integer + Smallest Value of X e.g., n = 15Q + 2 and n = 17Z + 2 LCM of 15 and 17 is 15*17 = 255 Smallest possible value of n that satisfies both equations = 2 So combined equation for n = 255*int + 2 NOTE: If the form has a NEGATIVE (e.g., n = 15Q - 2), just convert negative number into positive form by ADDING the divisor e.g., 15Q - 2 = 15Q + 13 Approach #2) Set the two equations equal to each other to understand relationship between QUOTIENTS

Answer 21

x = Qy + r (x - r) = Qy ---> Q and y are factors of x - r ---> list them out! > combined with y > r, you should be able to figure out properties of y (divisor)

Answer 22

YES HOWEVER, there are WATCHOUTS (1) Converting FRACTION REMAINDER to DECIMAL REMAINDER > It is easier to convert fraction remainder to decimal remainder, than from decimal remainder to fraction remainder > with a fraction remainder, you already know the divisor (2) Converting DECIMAL REMAINDER to FRACTION REMAINDER > When converting from decimal remainder to fraction remainder, we need MORE INFO to determine the remainder BECAUSE there is an infinite number of remainders possible (basically any fraction that simplifies into decimal is possible) REMEMBER: The "remainder" is really a non negative INTEGER (numerator part of fraction, all DEPENDENT ON THE DIVISOR) e.g., x / y = 9.48 ---> cannot say remainder is 48 (only true if divisor y = 100) x / y = 948/100 = 9 + 48/100 ---> remainder is 48 x / y = 474/50 = 9 + 24/50 ---> remainder is 24 Without further information on x or y, what we can do is determine what the MOST REDUCED FRACTION COULD BE ---> remainder must be a MULTIPLE of this number e.g., 48/100 = 24/50 = 12/25 ---> actual remainder must be a multiple of 12 (3) Converting DECIMAL REMAINDER to INTEGER (ONLY possible if you know the DIVISOR) Integer remainder = Decimal component of the division * divisor y

Answer 23

1) Divide each of the integers by the divisor 2) Focus on the remainders --> MULTIPLY the remainders 3) Correct for any "excess remainders" at the end of the multiplication (if remainder is greater than the original divisor) OR see if R=0 first based on divisibility rules Algebraically: (ab)/c --> Remainder = (Ra*Rb)/c (product of the remainders) Alternatively, you may be asked about the product of two unknown integers and given expressions --> find the combined equation and divide by divisor to figure out remainder e.g., n = 9Q + 6; m = 12k + 4 So nm = (9Q + 6)*(12k + 4) = 108QK + 36Q + 72k + 24 nm/18 = int + int + int + 1 + 6/18 ---> remainder = 6

Answer 24

1) Divide each integer by the divisor 2) Focus on the remainders --> add or subtract (based on the sign) 3) Correct for any "excess remainders" at the end of addition (by subtracting multiples of the divisor) // any "negative remainders" at the end of subtraction (by adding multiples of the divisor) e.g., 12/5 --> R = 2 13/5 --> R = 3 17/5 --> R = 2 x/5 has a remainder = 2 + 3 + 2 = 7 --> after adjusting for excess remainder, we get R = 2 Algebraically: (a + b)/c --> Remainder = (Ra + Rb)/c (sum of the remainders) (a - b)/c --> Remainder = (Ra - Rb)/c (subtraction of the remainders)

Answer 25

Each factor of 10 or (2x5) PAIR in the prime factorization of that number > so the limiting factor is the number of 2s and 5s present, whichever is fewer Application Qs: > Trailing zero rule can be used to determine the TOTAL # OF DIGITS in a number **Other things to keep in mind: > Any factorial >= 5! will have a zero in its units digit > if we have even ONE 2x5 pair, the UNITS DIGIT will always be zero

Answer 26

You can leverage the understanding that (2x5) pairs create trailing zeros 1) prime factorize the number 2) determine the number of trailing zeros after the non-zero digit (how many 2x5 pairs are there) 3) find the PRODUCT of the remaining factors (unpaired 2s or 5s along with other nonzero prime factors) 4) Number of digits = # of trailing zeros + # of digits in the product of the remaining factors

Answer 27

MUST BE 1 / # When dealing with decimals, it refers to the zeros that occur to the RIGHT of the decimal point, but before the first non-zero number e.g., 0.02 has 1 leading zero 0.2 has no leading zeros 0.002 has 2 leading zeros Basically do not count the 0 before the decimal Approach: #1) X (denom integer) is NOT a perfect power of 10 (e.g., unequal number of 2s and 5s, or non 2 and 5 integers) > If X is an integer with k digits, and if X is not a perfect power of 10, then 1/X will have (k-1) leading zeros e.g., 1/8 --> 1 digit, so 0 leading zeros (0.125) e.g., 1/40 --> 2 digits, so 1 leading zero (0.025) #2) X (denom integer) IS a perfect power of 10 (e.g., equal number of 2s and 5s) > If X is an integer with k digits, and if X is a perfect power of 10, then 1/X will have (k-2) leading zeros e.g., 1/100 --> 3 digits and X is a perfect power of 10, so 1/X will have 1 leading zero (0.01)

Answer 28

1) Divide N by each power of the prime (ignoring remainders) until the quotient equals zero e.g., 21/3^1, 21/3^2 ... 2) Add the quotients from the previous divisions = sum represents the number of prime number x in the prime factorization of N! Rationale: > trying to COUNT the number of individual instances / double instances / triple instances etc. of a prime number e.g., 21/3 --> 7 instances where 3 shows up as a factor 21/9 --> 2 instances where 3 shows up AGAIN (as an extra factor)

Answer 29

Similar to the steps involved to determine the largest number of times a PRIME divides into a factorial, except you want to figure out the max number of PRIME NUMBER PAIRS that divide into the factorial e.g., max number of (2x3) pairs that divide into 40 > The number of pairs is CONSTRAINED by the larger of the pair (in this case, the number of 3s) --> short cut is to just solve for # of the larger of the pair > ** Make sure to adjust for EVERY NUMBER IN THE NUMERATOR e.g., there are three 500!, so multiply the total number of 5s by 3 (you must do this BEFORE setting up the inequality) > If there is an EXPONENT on the prime, remember to just do an extra step of solving for the unknown in the exponent after e.g., 50!/25 --> 50!/(5^2n) We set 2n <= # of 5s and solve for n Keep in mind: > you cannot just divide 40 by 6 because you are missing lone numbers that can form a 2x3 pair

Answer 30

** 1) First express the divisor as a power of a PRIME number e.g., 4^n = 2^2n e.g., 500^n = 2^2n * 5^3n 2) Determine the number of the base prime that divide into the factorial (Divide N! by each power of the prime (IGNORING REMAINDERS **) until the quotient equals zero) —> constrained by the largest prime e.g., 30/2 = 15 30/4 = 7 30/8 = 3 30/16 = 1 Total number of 2s in 30! = 26 ***** Make sure to adjust the answer for EVERY NUMBER IN THE NUMERATOR e.g., there are three 500!, so multiply the total number of 5s by 3 (you must do this BEFORE setting up the inequality) 4) Set up an inequality using just the exponents 2n <=26 (trying to determine the max value of n) n <= 13 Keep in mind: > you cannot just divide 30 by powers of 4 because you are missing single values of 2 that can combine into 4

Answer 31

Perfect square --> *** square root of a perfect square must be an integer => sqrt(perfect square) = integer Note: 0 and 1 are the ONLY unique among perfect squares because they don't have prime factors > does not need to be represented as having an even exponent May also be expressed as 230x = y^2

Answer 32

0 1 8 27 64 125 216 343 512

Answer 33

Perfect cube > has prime factorization form with exponents that are MULTIPLES OF 3 / divisible by 3 May also be expressed as: 450p = q^3 Note: 0 and 1 are unique among perfect cubes because they don't have prime factors

Answer 34

Strategy: Utilize the knowledge that REMAINDERS follow PATTERNS > so we can determine the remainder when a very large number is divided by another number Tactically, this means finding (1) what the pattern of remainers is (pattern repeats every xth power) (2) take exponent and RE-EXPRESS so it matches the pattern (3) determine remainder > similar approach is followed for UNITS DIGITS (but still different approaches) e.g., What is the remainder when 3^200 / 5? Step 1: Find the pattern in the remainders when powers of 3 are divided by 5 3^1 --> R = 3 3^2 --> R = 4 3^3 --> R = 2 3^4 --> R = 1 3^5 --> R = 3 (pattern repeats itself every 4th power). In other words, all powers that are a multiple of 4 has a remainder = 1 Step 2: Re-express exponent to match pattern 3^200 (already a multiple of 4) Step 3: Determine remainder > we know every exponent that is a multiple of 4 will have a remainder = 1 > BUT a multiple of 4 is also a multiple of 2 --> that's okay, it is still a multiple of 4

Answer 35

Units digit will follow a pattern: > stay the same (regardless of power) > two-number pattern (tied to whether you have an even vs odd exponent) > four-number pattern (every power that is a multiple of 4 will have units digit = x) Stay the same regardless of power: 0, 1, 5, 6 Two-number pattern: 4 and 9 > For 4 ---> 4 for odd exponents and 6 for even exponents > For 9 --> 9 for odd exponents and 1 for even exponents Four-number pattern: 2, 3, 7, 8 Note: For base integers greater than 9 (not single digits), the same units-digit pattern holds true --> just ignore all other digits and focus on the units digit and the exponent e.g., 123384^33 --> read as 4^33 Applications of this pattern: > Short cut to PS (scan answer choices and pick answer based on matching units digit) > Product of powers (units digit of the product = product of each term's units digit) e.g., 7^(8a + 9b + 10c) has what units digit? Can re-express as 7^8a * 7^9b * 7^10 c (if one of the units digit = 0, all 0) > Sum of powers (e.g., 3^15 + 2^15 --> choose PS option set that matches unit digit) > for DIVISION of powers --> usually you can re-express as a common base and simplify into a single term (e.g., 27^324 / 3^134 = 3^838)

Answer 36

Ex #1) Finding the xth digit to the right of the decimal place in a fraction > We can use our understanding of digit patterns because there is a consistent pattern in DIGITS to the right of the decimal point > To solve, first do division to get a sense of the pattern e.g., 2/11 = 0.181818... each digit in the ODD numbered place to the right is 1 and each even numbered place digit to the right of the decimal point is 8. So because 79 is an odd number, the 79th digit to the right of the decimal place must be 1 Ex #2) Other common pattern applications > To solve, first right out the CORE PATTERN that repeats (it is best to write out each component separately, even for repeated elements after each other) > e.g., Wheat, Wheat, White, White, White, Rye ---> Pattern repeats every 6 items > Use your knowledge of MULTIPLES to figure out how many COMPLETE ITERATIONS can occur within the total number of items > e.g., 2000/6 --> 333 complete iterations, Remainder 2 > Figure out the pattern at the end of the total number of items

Answer 37

Just move the decimal point to the LEFT. The digits to the right of the decimal point represent the remainder e.g., divide by 10 --> remainder will be the UNITS digit of the dividend e.g., divide by 100 --> remainder will be the last two digits of the dividend So if you KNOW the units digit of a number, you KNOW the remainder after division by power of 10 Application of this concept: > typically combined with finding patterns in large powers (units digit) Falls into category of special remainder rule (similar to division by power of 10)

Answer 38

For a given UNITS digit, the remainder after division by 5 is ALWAYS THE SAME So if you KNOW the units digit of a number, you KNOW the remainder after division by 5 (sufficient) Falls into category of special remainder rule (similar to division by power of 10) Applications: > Often combined with patterns in units digit of large powers

Answer 39

1) consecutive integers (constant difference = 1) 2) consecutive MULTIPLES of a number (constant difference >1) 3) consecutive numbers with the same remainder when divided by some integer (constant difference > 1) e.g., [1, 7, 13, 19, 25] --> d = 6, remainder when divided by 6 all equal 1

Answer 40

(1) Product is even // divisible by 2! (2) Two integers do NOT have common prime factors // GCF = 1

Answer 41

Concept: Divisibility (multiples/factors) OPTION 1 Algebraic way (factors/multiples) 1) Translate the prompt Let x be the number we are trying to solve for? (148 + x)/12 = int 2) To make the testing easier, first SIMPLIFY the expression so you take out the WHOLE number 148 = 144 + 4 So re-express as (144 + 4 + x)/12 = 12 + (4+x)/12 3) Then sub in values of x that produce an integer result Ans E (8) OPTION 2: Divisibility rules > to be divisible by 12, a number must be divisible by both 3 and 4 > so try each number to see if the sum with 148 is divisible by 12

Answer 42

Concept: Factors / multiples / divisibility > We are trying to see what number must be a FACTOR of FACTORS > it is true that x or y must be divisible by EVERY PRIME FACTOR of 660 > BUT x or y may not always be divisible by the PRODUCT OF A COMBINATION of prime factors (especially powers of primes can be split) 1) prime factorize 660 = 2^2 * 3 * 5 * 11 2) Process of elimination a. 3 --> x or y must always be divisible by 3 e.g., 2^2 * 5 * 11 x 3 ---> ONLY PRIME FACTOR in the option set b. 4 --> x or y do NOT need to be divisible by 4 e.g., 2^2 * 5 * 11 x 3 (one number is divisible by 4) e.g., 2 * 3 x 2 * 5 * 11 (neither number is divisible by 4)

Answer 43

Concept: Factorials and units digit

Answer 44

Concept: Factorials and units digit > Any factorial >= 5! will have a zero in its units digit (1) a = 4! --> NS because 4! = 24 but b could be 10 or any other number (2) b = 7! --> S because any factorial >= 5! has a zero in its unit digit B

Answer 45

RULE for remainders (addition, subtraction and multiplication): (ab)/c --> Remainder = (Ra*Rb)/c (product of the remainders) 5/7 ---> remainder is 5/7 y/7 --> remainder is 1/7 SO remainder of 5y/7 = (5*1)/7 = 5/7 (5 is the remainder)

Answer 46

4! = 24 5! = 120 6! = 720

Answer 47

Concept: Remainder patterns in powers (slightly different approach to pattern in unit digit of powers) (1) Determine if 2^256 is a multiple of 3 // is 2^256 divisible by 3 > check if Remainder = 0 Remainders when powers of 2 are divided by 3: 2^1 / 3 --> R = 2 2^2 / 3 --> R = 1 2^3 / 3 --> R = 2 2^4 / 13 --> R = 1 Pattern: every odd exponent of 2 has R = 2, while every even exponent of 2 has R = 1 Since 256 is even, 2^256/3 has R = 1 (so 2^256 is NOT divisible by 3) (2) If 2^256 is NOT divisible by 3, figure out what number you need to subtract from it so get remainder = 0 1

Answer 48

3^1 = 3 3^2 = 9 3^3 = 27 3^4 = 81 3^5 = 243

Answer 49

Concept: LCM of two different pairs of integers, trying to figure out product of two integers (1) No value of s, so NS (2) No value of r, so NS (3) Testing reveals two different answers, so NS ** COME UP WITH TEST VALUES FASTER Statement 1: 48 = 2^4 * 3 Statement 2: 80 = 2^4 * 5 t is the SHARED integer between the two statements ---> t must be 2^1 <= 2^4 LCM means HIGHEST POWER of shared factors and all non-repeated factors Test #1: t = 2^4, r = 3, s = 5 rs = 15 Test #2: t = 2^4, r = 2^4*3, s = 2^4*5 rs = 2^8 * 15

Answer 50

A decimal is terminating if the REDUCED FRACTIONAL EQUIVALNET contains only powers of 2's and/or powers of 5s in the denominator Always test denom = 100 (yes terminating decimal) e.g., n could be /100 such as 22/100

Answer 51

NO, exponents do NOT influence even / odd properties of a number Powers are simply multiplying the number by itself, either an odd number or an even number

Answer 52

Concept: Factorials > best strategy is to TEST > Tip when testing: test small values AND large values, and weird value (0, 1, fractions if relevant) (1) x^y > z Test 1: x = 2, y = 3, z = 1 2!*3! > 1! is TRUE Test 2: x = 2, y = 3 (KEEP ONE SIDE CONSTANT TO MAKE IT EASIER TO TEST), z = 6 2!*3! > 6! is FALSE NS (2) x + y = z From this statement, we can see that z will ALWAYS be greater than x, and z will ALWAYS be greater than y Logically, (x!)*(y!) > z! will always be FALSE Test 1 (SMALL values): x = 1, y = 1, z = 2 1!*1! > 2! is FALSE Test 2 (LARGE values): x = 10, y = 5, z = 15 10!*5! > 15! is FALSE S

Answer 53

Concept: Application of trailing zero rule You need to figure out the number of (2x5) pairs in the factorial > since there are fewer 5s than 2s factors, focus on determining the NUMBER OF 5s in the factorial Question becomes - how many 5s are there in 21!? Answer: Use the factorial shortcut 21/5^1 = 4 5's 21/5^2 = 0 5's So there are 4 5's in 21!

Answer 54

Concept: Divisibility involving decimals > convert to INTEGERS (in this case, price in CENTS is always an integer) Let F = final price in CENTS P = original price in CENTS F = P*(0.84) F = P * (84/100) P = 25F/21 ---> F must be divisible by both 3 and 7 A gives a sum of 17, which is not divisible by 3, so A could not be the final price of the computer

Answer 55

WATCH OUT - this type of question is NOT the same as finding the number of primes in a factorial 1) express potential factor as a PRIME e.g., 9^n = 3^2n > you need to do this to account for lone 3's that can combine into 9 2) Prime factorize the dividend 45 = 3^2 * 5 3) Count the number of 3's --> 2 3's 4) Set up inequality of the powers to solve for n 2n <= 2 n <= 1

Answer 56

A) Algebraic method (leverages understanding of REMAINDERS) > create the equation for n + m using information about the remainders m = 18Q + 12 n = 24k + 14 So, n + m = 18Q + 24k + 26 > then, divide n + m by a number that divides into 18 and 24, but leaves a remainder when divided into 26 (e.g., 3) > The correct answer that matches the algebraic expression of n + m will have the same remainder when divided by 3 6Q + 8k + 8 + 2/3 I. 50/3 = 16 R2 (True) II. 70/3 = 23 R1 (False) III. 92/3 = 30 R2 (True) I and III ALSO you can set the combined equation EQUAL to each of the potential answer choices and see if you can find INTEGER VALUES for the variables that make the equation work, if yes, then it is a possible answer

Answer 57

RULE 1: Perfect squares are either divisible by 4 (R=0) or leave a R=1 when divided by 4 > Even perfect squares are always divisible by 4 (Proof: (2n)^2 / 4 = 4n^2 / 4 --> integer with 0 remainder) > Odd perfect squares always have remainder = 1 if divided by 4 (Proof: (2n+1)^2 / 4 = (4n^2 + 4n + 1)/4 ---> R=1 RULE 2: Perfect squares are also either divisible by 3 (R=3) or leave a R=X when divided by 3 > Perfect squares that are divisible of 3 (e.g., 9, 36, 81) leave R=0 > Perfect squares that are NOT divisible by 3 (e.g., 1, 4,16, 25) leave R=1

Answer 58

RULE 1: Perfect squares are either divisible by 4 (R=0) or leave a R=1 when divided by 4 > Even perfect squares are always divisible by 4 (Proof: (2n)^2 / 4 = 4n^2 / 4 --> integer with 0 remainder) > Odd perfect squares always have remainder = 1 if divided by 4 (Proof: (2n+1)^2 / 4 = (4n^2 + 4n + 1)/4 ---> R=1 RULE 2: Perfect squares are also either divisible by 3 (R=3) or leave a R=X when divided by 3 > Perfect squares that are divisible of 3 (e.g., 9, 36, 81) leave R=0 > Perfect squares that are NOT divisible by 3 (e.g., 1, 4,16, 25) leave R=1

Answer 59

Concept: Factors and patterns involving consecutive integers What we know: > n is between 10 and 99, inclusive > n^3 - n is equivalent to the product of 3 consecutive integers > So the question is asking how many different values of n produce a remainder of 0 when n^3 - n is divided by 12 > the product of 3 consecutive integers is divisible by n! or 6 (so we already know that the product of 3 consecutive integers is divisible by 3, now we just need to determine when n^3 - n is divisible by 4!! For these GROUPS of consecutive integer questions, it is easiest to ANCHOR TO THE MIDDLE NUMBER (every instance of the middle term will produce a grouping, just double check the ends) ALSO note that only n needs to be between 10 and 99 inclusive; n-1 and n + 1 don't follow this rule!! so no special adjustments are needed Case 1) n is odd --> n^3 - n is divisible by 8 and therefore also by 4 > how many numbers between 10 and 99 are odd? Total # of integers [10,99] = (99-10)+1 = 90 integers, half of which are odd 45 odd integers from 10 to 99 inclusive = 45 instances where n^3 - n is divisible by 12 Case 2) n is even, meaning n-1 and n+1 are odd and don't have any factors of 2 > need to determine the number of two digit MULTIPLES OF 4 that can be n Formula when the increment does not equal 1: (Highest number divisible by the given number - lowest number divisible by the given number)/increment + 1 = (96 - 12)/4 + 1 = 21 + 1 = 22 instances where n^3 - n is divisible by 12 Putting it all together, we have 45 odd n + 22 even n = 67 outcomes where n^3 - n will be divisible by 12

Answer 60

Perfect square of a prime will always have only 3 factors > 1, itself, and the prime it is raised to the power of 2 This is because in prime factorization form, the power is 2 and to find the total # of factors, 2+1 = 3 If an integer has only 3 positive factors (including 1) or 2 positive factors other than 1, what does this tell you?

Answer 61

FIND THE PATTERN in the TENS DIGIT (ignore what's going on in the units digit) using smaller exponents (don't give up just because the number is getting very large) e.g., Tens digit 7^1 = 0 7^2 = 4 7^3 = 4 7^4 = 0 7^5 = 0 7^6 = 4 Pattern in the tens digit of powers of 7 = 0, 4, 4, 0 1415 = R 3 when divided by 4, so tens digit = 4

Answer 62

Express as a product of two x+y's: (x+y)*(x+y) = even or odd

Answer 63

How to solve: For each LCM clue provided ... > Prime factorize the integers > Remember that LCM = non-repeated prime factors * highest power repeated prime factors shared by at least two of the integers > for the unknown integer, determine what the EXPONENTS for each prime factor COULD BE (e.g., 2^0to3) Example: 6 = 2*3 15 = 3*5 N = ?? (1) 40 = 2^3*5 8 = 2^3 N = 2^0to3 * 5 (must have five) NS because N could be 5, 10, 20, 40, which changes the LCM of 6, 15 and N (2) 60 = 2^2*3*5 12 = 2^2 * 3 N = 2^0to2 * 3^0to1 * 5 NS because N could be 5, 10 etc. (3) NS because N = 2^0to2 * 5 = 5, 10, 20, whcih changes the LCM of 6, 15 and N

Answer 64

The ONLY number that is EQUAL TO ITS OPPOSITE is ZERO So this statement is asking for the expression that EQUALS zero

Answer 65

Factorials LONG WAY is to prime factorize 14! then compare to the prime factorized forms of each answer SHORTER WAY is to look at the OPTIONS FIRST and compare starting from LARGEST factor to smallest factor (A) 1690 is 2*5*13^2 ---> we know that 14! only has one 13, so this is WRONG (B) 1210 is 2*5*11^2 --> we know that 14! has only one 11, so this is also WRONG (C) 726 is 2*3*11^2 --> we know that 14! has only one 11, so this is also WRONG (d) 625 is 5^4 --> instead of prime factorizing 14!, figure out how many 5s are in 14! using the factorial division SHORTCUT 14/5 = 2 14/25 = 0 Max 2 5's in 14! So this is also wrong (keep going until you realize that ans E is right)

Answer 66

> Create test cases that produce Y and N option or produce multiple different values > When creating Y and N cases, try to keep most of the numbers the SAME and move only ONE number to compare differences e.g., W is a multiple of 5 and W and Z share only ONE common prime. What is their GCF? Test case 1) W = 5 * 7, Z = 7 ---> GCF is 7 Test case 2) W = 5 * 3, Z = 3 ---> GCF is 3 NS since different values e.g., W is a multiple of 5, Z is a multiple of 9, and W and Z share only ONE common prime. What is their GCF? Test case 1) W = 5 * 7, Z = 3^2 * 7 ---> GCF = 7 Test case 2) W = 5 * 13, Z = 3^2 * 13 ---> GCF = 13 NS since different values

Answer 67

Dividend / Divisor = Quotient + remainder / divisor Divisor = what you divide by = denominator Dividend = numerator Quotient = answer after division Remainder = what does not get evenly divided

Answer 68

Remainders THINK ABOUT TWO CASES of the relationship between a and b: Case 1: a = b > a/b = a/a = 1 R = 0 > b/a also has R = 0 > Case 1 fits the statement that a / b has the same remainder as b / a > therefore looking for a PERFECT SQUARE (a^2) = ab = b^2 > Only III = 36 is a perfect square Case 2: a > b > b / a has a R = b (because b / a < 1) > a / b has a R < b (remainder rule is 0 <= R < divisor b) > does not match the statement Case 3: a < b > a / b has a R = a > b / a has a R < a > does not match the statement Therefore only when a = b is the statement true

Answer 69

If a and b are both multiples of an integer > 1, then a + b and a - b will also be a multiple of the same integer Example: a=6 and b=9, both are divisible by 3 ---> a+b=15 and a−b=−3 If only one of the integers is a multiple of an integer > 1, then a + b and a - b will not be a multiple of that integer Example: a=6 and b=5 → a+b = 11 and a-b = 1, neither is divisible by 3 If a and b are both not multiples of an integer > 1, then a + b and a - b may or may not be a multiple

Q3) Number properties Flashcards

Integers, zero and one, even and odd numbers, positive and negative numbers, evenly spaced numbers, divisibility, remainders, prime numbers, factors and multiples