Divisibility Flashcards
Divisibility rules 2-12
/2 when the units digit is 0,2,4,6,8
/3 when the sum of the numbers is divisible by 3
/4 if the last two digits of a number are divisible by 4 then the whole numb is
/5 units digit 0,5
/6 if the numb is even and its digit sum up to a multiple of 3
/8 divide the last three digits by 8, if no remainder the whole numb is divisible by 8
/9 if the sum of all digits are divisible by 9
/11 if the sum of the odd numbered place digit - the sum of the even numbered place digit is divisible by 11 (0,11,22…)
2915/11= (5+9)-(1+2)=11
/12 if a number is divisible by both 3 and 4, it is also divisible by 12
remainder
when numerator is not a multiple of the denominator
when 2 numbers cannot divide evenly
formula for division
x/y= Q+ r/y
converting a reminder from fraction to decimal form
i.e 32/5 = 6 2/5= 6,40 (2/5=0,40)
convert only the fractional part and add to the whole number
converting a remainder from decimal to fraction
we can’t be sure what is the actual remanider as it can be multiple remainders
we need to know the value of x,y or both to determine this
by determine the most reduced fractional value (0,48 can be reduced to 12/25 and this means the remanider is a multiple of 12)
converting decimal remainder to an integer
9/5 =1.8 multiply 0,8 (the decimal part) with the denominator 0,8*5=4
remainders could be any integers ranging
n-1
determining the number of trailing zeros in a number
if a number after prime factorization contains 52 that number will have 0 as a units digit
trailing zero is a number that contains zeros (520, 52000, 520000,..)
are created by powers of 10 (5210=520 and has one trailing zero)
any factorial >= 5! will always have 0 as a units digit
because it will contain at least one pair of 5*2
using trailing zeros to determine the number of digits in an integer
25^10 * 8^6= (5^2)^10* (2^3)^6= 5^20 * 2^18= 18 trailing zeros and 55=25
18+2=20 digits
1. prime factorization
2. count the number of the pair (52)
3. multiply the remaining numbers together
4. sum the number of digits from step 2 and 3
leading zeros in decimals
zeros that occur to the right of the decimal point before the first non-zero number
2/10=0,2 has no leading zeros
2/100=0,02 has 1 leading zero
determine the numb of leading zeros in a form of 1/x and if x is an integer and x is not a perfect power of 10, then 1/x will have (k-1) leading zeros
1/5000= x=5000 k=4(digits) 4-1=3 leading zeros
if x is a single digit number there will be NO leading zeros
1/8=0,125
1/4=0,25
1/2=0,5
when x is 2 digit number: one leading zero
1/80=0,0125
when x is 3 digit numb: two leading zeros
1/800=0,00125
determine the numb of leading zeros in a form of 1/x when x is a perfect power of 10
hence, 1/x will have (k-2) leading zeros
- when the denominator’s perfect power of 10 is a two digit number (10) no leading zeros
- 10^2= 100 -three digit numb there is one leading zero
- 10^3= 1000 -four digit numb there are two leading zero
the product of any set of consecutive integer is always divisible by
any of the integers in the set and by any of the actor combinations of the number
the product of any set of n consecutive integers is divisible by n!
i.e what is the largest numb that must be a factor of the product of any 4 consecutive integers
4!= 432=24
shortcut for determining the number of primes in a factorial
- divide the given factorial i.e 21 with 3^1, 3^2, 3^3, 3^k and ignore any remainders until we get to 0
21/3=7
21/3^2=21/9=2 (ignore the remainder)
21/3^3= 21/27=0 - there are 7+2= 9 threes in 21!
hence the largest possible integer value of n is 9
determining the numb of primes in a factorial when the base of the divisor is not a prime number
- break the number into primes i.e 6^n =2^n * 3^n
repeat the process the same way (only the higher numb) in this case numb 3
determining the numb of primes in a factorial when the base of the divisor is a power of a prime number
- break into primes (4^n= 2^2n)
- repeat the process of division
- sum the numbers
- create an inequality i.e 2n<=26 n<=13
first 9 perfect squares
0,1,4,9,16,25,36,49,64
prime factorization of a perfect square will contain only even exponents
except 0,1
perfect cubs
if a cube root of some integer x is an integer then x is a perfect cube
0,1,8,27,64,125
prime factorization of a perfect cube will contain only exponents that are multiples of 3
64= 4^3
125=5^3
terminating decimals
a number that has a finite number of non-zero digits to the right of decimal points
1/4= 0,25, 1/10= 0,10, 1/25 =0,04,
decimal equivalent of a fraction will terminate only if the denominator of the reduced fraction has a p.f that contains
only 2s or 5s, or both, otherwise the decimal equivalent won’t terminate
- reduce the fraction
- prime factorization of the denominator
what is the remainder when 3^123/4
- find the pattern by dividing 3/4, 3^2/4, 3^3/4…
2. positive even exponents yields a remainder of 1 always, positive odd exponents yields a remainder of 3
patterns of units digit when raised by a power
number 2= will follow the pattern of 2-4-6-8
number 3= will follow the pattern of 3-9-7-1
number 4 will follow the pattern of 4-6
number 5=will follow the pattern of 5
number 6=will follow the pattern of 6
number 7= will follow the pattern of 7-9-3-1
number 8=will follow the pattern of 8-4-2-6
number 9=will follow the pattern of 9-1
integers greater than 9 will follow the same pattern as its units digit (12=2, 345=5, 6789=9)
88^599^677^3=
- cycle of patterns, 88^5 = 8^5= unit digit of 8, 99^6= unit digit 1, 77^3= unit digit 3
- multiply the units digit= 813=24
- the answer is the unis digit hence 4
remainders after division by 10^n
a whole number/10- the remainder will be the units digit of the numerator (153/10=15.3= 15 3/10 r=3)
a number/100- r= to last two digits (153/100 r=53)
a number/1000 r= to last theree digits (153/1000 r=153)
remainders after division by 5
when the integers with the same units digit is divided by 5 the remainder is constant
7/5 r=2, 17/5 r=2, 57/5 r=2
