Arithmetic Flashcards

Question

Sum of Multiples in an Interval ## Footnote What is the sum of multiples of 5 from 105~355?

Answer 1

1. Calculate # of multiples in the interval 2. Calculate sum of each pair: (First multiple)+(Last multiple) 3. Calculate # of pairs of multiples: (# of multiples)÷2 →If the result is a decimal, it means there is 1 odd pair (9÷2=4.5, means there are 4 pairs and 1 odd) 4. Calculate sum of multiples in the interval: (sum of each pair) x (# of pairs) = total sum of multiples 4. In the case of an odd pair, add the odd pair (Sum of pair÷2) ## Footnote What is the sum of multiples of 5 from 105~355? 1. (355-105)÷5 + 1 = 50 +1 = 51 multiples 2. 105+355=460 = sum of each pair 3. 51 multiples÷2 = 25.5 = 25 pairs + 1 odd 4. 25 pairs x 460 (sum of each pair) = 11,750 (total sum of pairs) Odd pair = 460÷2 = 230 (sum of odd pair) Add Odd pair = 11,750 + 230 = 11,730 →Total sum of multiples of 5 = 11,730

Answer 2

1→All integers are divisible 2→If the number is even (ends in 0,2,4,6,8) 3→If the sum of digits can be divisible by 3: 816 = 8 + 1 + 6 = 15→15÷3=5 4→If the last two digits in that integer are also divisible by 4: 34,864→64÷4=16, 5→If the number ends in 0 or 5 6→If the number is even and divisible by 3: 750→7+5+0=12→12÷3=4 7→Use a calculator 8→If the last 3 digits is divisible by 8: 34,120→120÷8=15 9→If the sum of digits is divisible by 9: 5,841→5+8+4+1=18→18÷9=2 10→If the number ends in 0 | 80,000→00 is also divisible by 4 and 8

Answer 3

Integers that only have factors of 1 and itself. ## Footnote 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53

Answer 4

Take square root of number and test if it is divisible by all prime numbers less than the square root. ## Footnote Is 101 prime? 1. Take square root: √101 = 10.04 →Prime numbers<10.04: 2,3,5,7 2. Test if 101 is divisible by 2,3,5,7 →Not even, doesn't end in 0=not divisible by 2, 5 →1+0+1=2, not divisible by 3 →101÷7=14.4, not divisible by 7 Therefore, it is prime.

Answer 5

Every integer>1 can be written as a product of primes in one way ## Footnote 15→3¹ x 5¹ 100→2² x 5²

Answer 6

List up factorial integers, any multiple of these integers are factors of the factorial ## Footnote 5!=5 x 4 x 3 x 2 x 1 →5x4x3=60 is a factor of 10! →4x3x2=24 is a factor of 10!

Answer 7

* To find non-factors, simply find a prime number>factorial integer. * To find non-factors that aren't prime, calculate multiples of the prime number>factorial integer. ## Footnote 20!=23, 29 are prime numbers>20→non-factors of 20! →23x2=46, 23x3=69 are non-prime factors of 20! →29x2=58, 29x3=87 are non-prime factors of 20!

Answer 8

Simplify the number to an exponent and see if there are enough counts of number. ## Footnote Is 32 a factor of 15!? 1. Simplify 32=2^5 2. List out 15!: 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 3. 14=2x7, 12=2x6, 10=2x5, 8=2^3, 4=2^2 →Since there are more than five 2s, we can say that 32 is a factor of 15!

Answer 9

When you add/subtract a common factor of two integers, the result will also be a multiple of that common factor. ## Footnote 13+26=39 Notice that 13 and 26 are both multiples of 13. Also notice that the sum 39 is ALSO a multiple of 13. 1. Multiples of 30=1x30, 2x15, 3x10, 5x6 If you add one of the factors such as 3→30+3 2. Multiples of 30+3→33=1x33, 3x11 →We can see that 30 and 33 both share 3 as a factor

Answer 10

Use the bow tie method a/b>c/d = if ad>bc Is ¾>⁵⁄₇? 3x7=21 4x5=20 21>20, so yes it is larger.

Answer 11

* Roots Can be Written as Exponents. * Remove the root and set a fraction exponent, move the root to the denominator. √5 = 5¹ᐟ² ∛5 = 5¹ᐟ³ * If the integer inside the root has an exponent, that exponent will be the numerator ∛5² = 5²ᐟ³

Answer 12

(ⁿ⁄ₘ)² = ⁿ²⁄ₘ² √ⁿ⁄ₘ = √ⁿ⁄√ₘ

Answer 13

1. An exponent can "cancel" a root out: (√a)² = a 2. A root can "cancel" an exponent out: √a² = |a| 3. If you multiply two roots together (of the same order root), you can combine what's under the radicals together: : √a√b = √axb 4. If you divide one root by another (of the same order root), you can combine what's under the radicals: √12/√3 = √¹²⁄₃ = √4 = 2

Answer 14

If there is a single radical in the denominator of a fraction, multiply both the numerator and denominator by that radical. This will have the effect of removing the radical from the denominator. 2/√3 Multiply both denominator & numerator by √3 to get rid of √3 denominator: 2/√3 x √3/√3 = 2√3/3

Answer 15

* Integers * Fractions * Decimals * Rational Numbers (also known as fractions) * Irrational Numbers

Answer 16

* Casual Definition: The absolute value is a machine that "makes whatever is inside it positive." * The "Correct" Definition: The absolute value of something is its "distance from 0" on the number line |4| = What values of n are a distance of 4 away from 0? 4 and −4 * Don't forget that the absolue value of something can equal zero but never a negative value. * For any negative value of x, the sign would have to be changed, so for x < 0, |x| = −x |-4| = -(-4) = 4 = opposite of x which is -x ## Footnote x = 4, then |x| = 4 |x|=4, x = 4 and -4 x = -4, then |x|=-4 b/c |-4|=4 which is the opposite of -4 a.k.a x, so we make the x opposite→-x

Answer 17

1. Brackets have no impact when we are just multiplying or adding: 3 + (4 + 5) = 3 + 4 + 5 or 3 (4 x 5) = 3 x 4 x 5 2. The absolute value of the sum of two integers is always ≤ the sum of the absolute values of the two integers: |a+b|≤|a|+|b| 3. The absolute value of the product of two integers is = the product of the absolute values of the two integers: |ab| = |a x b| 4. If a = 1→a = a² = √a 5. If a > 1→a² > a > √a 6. If 0 < a < 1→√a > a > a²

Answer 18

An equation involving two fractions, or two ratios. Basically, it's just two fractions equal to each other. ⅗ = ²¹⁄₃₅

Answer 19

* Increase: (Larger Number - Smaller Number) ÷ Smaller Number % increase can never be negative * Decrease: (Larger Number - Smaller Number) ÷ Larger Number

Answer 20

Fractions, decimals, percents, and ratios can all be used to represent the same value. Fraction → Decimal → Ratio → Percent ½ → 0.5 → 1:2 → 50% ¾ → 0.75 → 3:4 → 75% ⁷⁄₅ → 1.4 → 7:5 → 140%

Answer 21

1. List out asking number into prime number exponents that total less than the Factorial number 2. Calculate each exponent to its total 3. Calculate counts of that number that is in the Factorial number by Factorial number÷exponent total 4. Add up all the counts→# of asking number ## Footnote How many 3s are there in 30!? 1. List out 3¹, 3², 3³ 2. 3¹=3, 3²=9, 3³=27 3. 30÷3=10, 30÷9=3.33, 30÷27=1.1 4. 10+3+1=14 counts of 3s in 3!

Answer 22

Consecutive 0s at the end of an integer. 10 = 1 trailing zero 100 = 2 trailing zero It can also be simplified into exponents of 10. 10¹ = 10 = 1 trailing zero 10² = 100 = 2 trailing zero

Answer 23

Use the finding # of numbers in a Factorial! formula by finding how many 10s there are in the factorial! ※Remember to simplify 10 = 5¹ 2¹ Because there will be less counts of 5s than 2s in an integer, we will always calculate to find how many 5s there are in the factorial! ## Footnote 1. Simplify 10 = 5¹ 2¹ 2. List out exponents of 5s within 100: 5¹, 5² 3. 5¹ = 5, 5² = 25 4. Calculate counts of exponents in 100! 100÷5 = 20, 100÷25 = 4 5. 20 + 4 = 24 6. Therefore, there are 24 counts of 10s in 100! = 24 trailing zeros

Answer 24

・It is neither pos nor neg ・It is not prime ・It is even b/c it is divisible by 2 ・It is a multiple of every integer ・Zero is a factor of Zero (only factor of itself) ・Any integer÷0 = undefined ・0⁰ = undefined ・Any interger⁰ = 1 ・0! = 1

Answer 25

The number of primes within a fixed interval tend to decrease as the numbers get larger. * 1~100: 25 * 101~200: 21 * 201~300: 16

Answer 26

3 consecutive numbers going up on a number line. Alternates between even and odd. ・n, n+1, n+2 ・n-1, n, n+1

Answer 27

* (n) + (n+1) + (n+2) = 3n + 3 * Always results in a multiple of 3 (consecutive even/odd integers or consecutive multiples) * Consec Even: (n) + (n+2) + (n+4) * Consec Odd: (n-2) + (n) + (n+2) * Consec Multiples: For multiples of 5, (n) + (n+5) + (n+10) = 3n + 15 (Since 3n is a multiple of 3) * When the middle number is even, it will result in even (odd + even + odd) * When the middle number is odd, it will result in odd (even + odd + even)

Answer 28

* (n) x (n+1) x (n+2) * Always results in even * Always results in a multiple of 2, 3, 6 * When the middle number is odd, the product is always a multiple of 2, 3, 4, 6, 8

Answer 29

1. Simplify asking number into prime factors 2. Find the limiting prime factor. 3. Find # of limiting prime factor in the factorial! 4. After finding the total counts of the limiting prime factor, we need to revert it back to the original asking number ## Footnote How many 24s are there in 517!? 1. Simplify 24 into 24=2³ x 3¹ 2. For 24=2³ x 3¹, there will be less 2³ than 3¹, so 2³ is the limiting factor 3. Find # of 2s in 517! (counts of 2s = 514) 4. Revert back to 24. 　Since 24=2³ x 3¹, we need three 2s to create 24s, therefore we calculate # of three 2s in 514, 514÷3 = 171 →Therefore there are 171 counts of 24 in 517!

Answer 30

PF both numbers and if there are prime factors that match within the bigger number then it is a factor. ## Footnote 1. PF 648 = 2³ 3⁴ 2. PF 54 = 2¹ 3³ 3. Since there are one 2s and three 3s in 648 (2³ 3⁴) we can say that 54 is a factor of 648

Answer 31

PF the number and by creating combinations of the prime factors, you can calculate factors of that number. ## Footnote 1. PF 100 = 2² 5² 2. Factors: 2¹ 5² = 50, 2² 5¹ = 20, 2⁰ 5¹ = 5 etc...

Answer 32

PF the number and by multiplying any number to the combinations of prime factors, you can calculate multiples. ## Footnote 1. PF 200 = 2³ 5² 2. Multiple any number to create multiples: (2³ 5²) x 2 = 2⁴ 5², (2³ 5²) x 7² = 2³ 5² 7² etc...

Answer 33

1. PF both numbers 2. Pick out prime factors (smallest exponents) that they share in common. 3. Multiply the selected prime factors ## Footnote 1. PF of 720 = 2⁴ 3¹ 5¹ 　PF of 1500 = 2² 3¹ 5³ 2. Prime factors in common = 2² 3¹ 5¹ 3. Multiple common prime factors: 4 x 3 x 5 = 60→GCF

Answer 34

1. PF both numbers 2. Select prime factors that they don't share in common + the largest exponents of those that they share in common. 3. Multiply the selected prime factors ## Footnote 1. PF of 175 = 5² 7¹ 　PF of 245 = 5¹ 7² 2. Prime factors = 5² 7² 3. Multiple prime factors: 25 x 49 = 1225→LCM

Answer 35

1. PF the integer 2. Add 1 to each exponent 3. Multiply exponents = # of pos factors 4. To find both pos and neg factors, just multiple # of pos factors x 2 ## Footnote 1. PF 300 = 2² 3¹ 5² 2. Add 1 to each exponent 　2²⁺¹ 3¹⁺¹ 5²⁺¹ = 2³ 3² 5³ 3. Multiply the exponents 　3 x 2 x 3 = 18→300 has 18 pos factors 4. 18 x 2 = 36→300 has 36 total (pos & neg) factors

Answer 36

1. PF the integer 2. Disregard the even prime (2) and add 1 to each exponent of the odd prime factors 3. Multiply exponents = # of odd factors ## Footnote 1. PF 6750 = 2¹ 3³ 5³ 2. Add 1 to each odd factor exponent 　3³⁺¹ 5³⁺¹ = 3⁴ 5⁴ 3. Multiply the exponents 　4 x 4 = 16→6750 has 16 odd factors

Answer 37

1. PF the integer 2. Leave the even prime (2) as is and add 1 to each exponent of the odd prime factors 3. Multiply all exponents = # of even factors If the PF has a 2¹, then the number of even and odd factors are the same. ## Footnote 1. PF 6750 = 2¹ 3³ 5³ 2. Keep even prime factor exponent as is and add 1 to each odd factor exponent 2¹ 3³⁺¹ 5³⁺¹ = 2¹ 3⁴ 5⁴ 3. Multiply the exponents 　1 x 4 x 4 = 16→6750 has 16 even factors

Answer 38

1. PF both numbers 2. Pick out factors in common 3. Add 1 to each exponent 4. Multiply the exponents ## Footnote 1. PF 264 = 2³ 3¹ 11¹ 　PF 1980 = 2² 3² 5¹ 11¹ 1. Pick out common factors: 2² 3¹ 11¹ 2. Add 1 to each factor exponent 2²⁺¹ 3¹⁺¹ 11¹⁺¹ = 2³ 3² 11² 3. Multiply the exponents 　3 x 2 x 2 = 12→264 and 980 share 12 common factors

Answer 39

Multiplication: Add exponents Division: Subtract exponents ## Footnote 1. 3²⁺² = 3⁴ 2. 5³⁻² = 5¹

Answer 40

A problem that presents three or four numbers that occur in regular intervals. The problem usually states that all of these instances occur at the same time at some point. They then ask you how much time will elapse before they all occur simultaneously again. Solution: Find the LCM. ## Footnote 1. PF each Integer: 　Bell A: 10 secs = 2¹ 5¹ 　Bell B: 15 secs = 3¹ 5¹ 　Bell C: 20 secs = 2² 5¹ 2. LCM = 2² 3¹ 5¹ = 4 x 3 x 5 = 60 secs 3. It will take 60 secs/1 min for all 3 bells to ring again simultaneously

Answer 41

Use exponent unit digit patterns to find the unit digit. Find the closest multiple to 4 of the exponent then refer to the pattern. ## Footnote Unit Digit of 2³⁶ 1. List up exponents of 2 patterns 　2¹ = 2 　2² = 4 　2³ = 8 　2⁴ = 16→UD is 6 　...There is a re-occurring pattern of 2, 4, 8, 6. This pattern is only in blocks of four. 2. 36 is a multiple of 4, so we refer to the UD of exponent 4 = 6 3. Therefore the unit digit of 2³⁶ = 6

Answer 42

* 0ⁿ = Always 0 * 1ⁿ = Always 1 * 2ⁿ = 2, 4, 8, 6...repeats * 3ⁿ = 3, 9, 7, 1...repeats * 4ⁿ = 4, 6...repeats * 5ⁿ = Always 5 * 6ⁿ = Always 6 * 7ⁿ = 7, 9, 3, 1...repeats * 8ⁿ = 8, 4, 2, 6...repeats * 9ⁿ = 9, 1...repeats

Answer 43

* Quotient: A number of times an integer can go into another integer without exceeding it. 23÷5, 5 can go into 23 four times w/out exceeding it. 5x4=20, 5x5=25←This exceeds 23! So 4 is the quotient. * Remainder: The amount left over needed to reach the integer. 23÷5, 4 is the quotient. 5x4=20, there is still 3 needed to reach 23. So 3 is the remainder. * The remainder is always pos, can never be neg

Answer 44

1. Divide 300÷9 = 33.33... 2. Take the integer and multiply by the divisor to get the Quotient. 　Quotient: 33 x 9 = 297 (9 can go in 300, 297 times) 3. To find the remainder, just subtract the quotient from the integer 　Remainder = 300 - 297 = 3→3 is the remainder

Answer 45

In the case of a negative number, we need to find how many times 4 can go in -14 w/out exceeding it. In this case, it has to be less than -14 (-15~) 1. The closest multiple of 4 that is less than -14 is 4 x -4 = -16 　Quotient: -4 3. To find the remainder, we need to calculate how much we need to get -16 to -14. -16 - (-14) = 2→In order to reach -14, we need to add 2 　Remainder = 2 is the remainder

Answer 46

* When Denominator>Numerator, Remainder = Numerator * Whenever the Number Being Divided is Less than the Divisor, and both integers are positive, the remainder is simply the number being divided. 5÷6=r 5 b/c 0 (quotient) x 6 +5 (remainder) = 5

Answer 47

* **÷1**= Remainder is Always 0 * **÷2** = If the integer being divided is even = always 0 　　　If the integer being divided is odd = always 1 * **÷3** = List up first 1~5 remainder to find a pattern.The pattern depends on the integer being divided. 　　7¹ ÷ 3 = 2 r1 　　7² ÷ 3 = 16 r1 　　7³ ÷ 3 = 114 r1 　　→We can see that there is a pattern of remainder = 1 * **÷4** = Remainder of last two digits of the integer ÷ 4 　　Remainder of 422 when divided by 4. 　　1.Take the last two digits 22. 　　2. Divide 22 by 4: 22 ÷ 4 = 5 with a remainder of 2. 　　3. 422/4=105 r 2→Remainder is also 2 * **÷5**= If the unit digit = 0, 1, 2, 3, 4→That is the remainder 　　　If the unit digit = 5, 6, 7, 8, 9→Unit Digit-5 = remainder 　　　Numbers ending in 0 or 5: have a remainder of 0 (e.g., 10 ÷ 5 = 2 R 0; 25 ÷ 5 = 5 R 0). 　　　Numbers ending in 1 or 6: have a remainder of 1 (e.g., 11 ÷ 5 = 2 R 1; 36 ÷ 5 = 7 R 1). 　　　Numbers ending in 2 or 7: have a remainder of 2 (e.g., 12 ÷ 5 = 2 R 2; 47 ÷ 5 = 9 R 2). 　　　Numbers ending in 3 or 8: have a remainder of 3 (e.g., 13 ÷ 5 = 2 R 3; 58 ÷ 5 = 11 R 3). 　　　Numbers ending in 4 or 9: have a remainder of 4 (e.g., 14 ÷ 5 = 2 R 4; 69 ÷ 5 = 13 R 4). * **÷6** = Try to find pattern like ÷3 * **÷7** = Try to find pattern like ÷3 * **÷8** = Remainder of last three digits of the integer ÷ 8. 　　　The only possible remainders are 0, 1, 2, 3, 4, 5, 6, and 7. * **÷9**= Try to find pattern like ÷3 * **÷10**= Unit Digit = Remainder

Answer 48

Calculate the remainder of each term, then add them together. 1. (17 + 12)/10 = 17/10 + 12/10 2. 17/10 = 1 r7, 12/10 = 1 r2 3. Add remainders = 7 + 2 = 9→Final remainder. If the resulting sum is > than the divisor, then we divide the resulting sum/divisor. The remainder from the above formula will be the Final remainder.

Answer 49

When the numerator>denominator ⁸⁄₇

Answer 50

An integer that can be written as a fraction 8 = ⁸⁄₁

Answer 51

・An integer than cannot be written as a fraction (π = 3.14...) ・An integer that is non-repeating & non-terminating when written as a decimal (√2, √3) * Examples: π, Square roots of non-perfect squares (√2, √3)

Answer 52

An integer + fraction (5⅜). This happens when there is an improper fraction. To reverse a mixed number back into an improper fraction, we multiple the quotient x denominator. Then add the remainder to it. (quotient x denominator) + remainder. ## Footnote 5⅜ 5 x 8 = 40 40 + 3 = 43 →⁴³⁄₈ (Improper fraction)

Answer 53

Flip the second fraction and multiply. ⅔ ÷ ½ ÷ ⅗ ⅔ x ²⁄₁ x ⁵⁄₃ = (2 x 2 x 5) / (3 x 1 x 3) = ²⁰⁄₉ Whenever you divide by 1/n, it is the same as multiplying by n/1 6 ÷ ½ = 6 x ²⁄₁ = 6 x 2 = 12

Answer 54

* A positional number system, meaning that the location of a digit in a number influences its value. Hundreds→Tens→Unit→Tenths→Hundredths→Thousandth 10² → 10¹→ 10⁰ → 10⁻¹ → 10⁻² → 10⁻³ * Decimals can be written as powers of 10: 177.35 = 1(10²) + 7(10¹) + 7(10⁰) + 3(10⁻¹) + 5(10⁻²)

Answer 55

・Terminates means that there is an end to the decimal. 1. To figure out whether it terminates, simplify the fraction and PF the denominator. 1. If there are only powers of 2 or 5 or a combination of both→it terminates. ## Footnote 63/210 =(3² 7¹) / (2¹ 3¹ 5¹ 7¹) =3¹ and 7¹ cancels out = (3¹) / (2¹ 5¹) Since the denominator only consists of 2s and 5s, it terminates.

Answer 56

*** Terminating Decimal** : Note the number of digits and put it over 10s * If only one digit repeats, put it over 10: 0.4...= ⁴⁄₁₀ * If two digits repeat, put them over 100: 0.45...= ⁴⁵⁄₁₀₀ * If three digits repeat, put them over 100: 0.456...= ⁴⁵⁶⁄₁₀₀₀ *** Non-Terminating Decimal:** Note the number of digits and put it over 9s * If only one digit repeats, put it over 9: 0.4...= ⁴⁄₉ * If two digits repeat, put them over 99: 0.45...= ⁴⁵⁄₉₉ * If three digits repeat, put them over 999: 0.456...= ⁴⁵⁶⁄₉₉₉

Answer 57

1. If nᵐ = nᵃ⁺ᵇ, then m = a + b 2. If you have a neg exponent, you can make it pos by flipping into a fraction: n⁻ᵃ = ¹⁄ₙᵃ 3. If you multiply two numbers together with the same base, you can add the exponents: (nᵃ)(nᵇ) = nᵃ⁺ᵇ 4. If you divide a number by another number with the same base, you can subtract the exponents: (nᵃ)/(nᵇ) = nᵃ⁻ᵇ 5. a⁰ = 1 except for 0 6. If you multiply two different numbers together with the same exponent, you can bracket it and set it to the same exponent: (nᵃ)(bᵃ) = (n x b)ᵃ 7. If you divide two different numbers with the same exponent, you can set the same exponent to each numerator&denominator: (ⁿ⁄ₐ)ᵐ = nᵐ/aᵐ 8. If you raise a number with an exponent to another exponent with (parentheses), you simply multiply the exponents: (nᵃ)ᵇ = nᵃᵇ 9. An exponent raised to another exponent w/out () : 5³^³ ≠ 5⁹ but 5²⁷ 10. (5³)³ = 5⁹

Answer 58

* Any real number that can be expressed as a fraction, or ratio, of two integers, where the denominator is not zero. * 0 is a Rational number * Rational + Irrational = Irrational

Answer 59

* 2⁵= 32 (approx 30) * 2¹⁰= 1024 (approx 1000) * 3⁵= 243 (a little under 250) * 5⁴= 625 * 5⁵= 3125 (approx 3000)

Arithmetic Flashcards

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