Download Number Theory

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Addition wikipedia , lookup

List of prime numbers wikipedia , lookup

Mathematics of radio engineering wikipedia , lookup

Factorization of polynomials over finite fields wikipedia , lookup

Proofs of Fermat's little theorem wikipedia , lookup

Elementary mathematics wikipedia , lookup

Transcript
Park Forest Math Team
Meet #2
Number
Theory
Self-study Packet
Problem Categories for this Meet:
1. Mystery: Problem solving
2. Geometry: Angle measures in plane figures including supplements and
complements
3. Number Theory: Divisibility rules, factors, primes, composites
4. Arithmetic: Order of operations; mean, median, mode; rounding; statistics
5. Algebra: Simplifying and evaluating expressions; solving equations with 1 unknown
including identities
Meet #2 – Number Theory
Ideas you should know:
Prime Factorization: Every number 2,3,4,… can be written as a product of
prime number factors. For example, 12 = 2x2x3, or 22!3. What is the prime
factorization of 60? Answer: 22!3!5, or 22!31!51.
12
=
22!31
24
=
23!31
100
=
22!52
210
=
21!31!51!71
To the Zero Power: Any non-zero number to the zero power is one. 50=1.
This makes sense if you think that 53÷5=52, and 52÷5=51, so 51÷5=50,
which is the same as 5÷5=50=1. (00 is undefined, not 1.)
50=1
51=1x5
=5
52=1x5x5=25
Least Common Multiple (LCM): LCM(a,b) is the least (smallest) number that
is a multiple of both a and b. For example, LCM(6,10) = 30 because both 6
and 10 are factors of 30, and there is no smaller number than 30 that does
this. One way of computing LCM(a,b) is by prime factoring a and b. Then,
for each prime factor, take the larger of the two exponents. For example,
6=21!31!50, and 10=21!30!51. The larger exponents: 21!31!51 =30.
LCM(2, 3) = 6
LCM(2,4) = 4
LCM(6,10) = 30
LCM(8,12) = 24 LCM(6, ?) = 18
Meet #2, Number Theory
LCM(10,15) = 30
(?=9 or 18)
Greatest Common Factor (GCF): GCF(a,b) is the greatest (largest) number
that is a factor of both a and b. For example, GCF(6,10) = 2 because 2 is a
factor of both 6 and 10, and there is no larger number that does this. One
way of computing GCF(a,b) is by prime factoring a and b, and then, for each
prime factor, take the smaller of the two exponents. For example,
6=21!31!50, and 10=21!30!51. The smaller exponents: 21!30!50 =2.
Useful trick: GCF(a,b) x LCM(a,b) = a x b.
GCF(2,3) = 1
GCF(4,6) = 2
GCF(12, 16) = 4
GCF(12,18)=6 GCF(17,24)=1
GCF(10,?) x LCM(10,?) = 30
GCF(11, 22) = 11
(? = 3, because GCFxLCM=3x10)
Proper Factor: A factor of a number other than the number itself or one.
The proper factors of 6 are 2 and 3, but not 1 or 6. There are no proper
factors of prime numbers.
Factors of 6: 1, 2, 3, 6
Proper Factors: 2, 3
Factors of 7: 1, 7
Proper Factors: None!
Relatively Prime: Two numbers are relatively prime if they have no common
factors other than 1. Their GCF is 1. For example, 7 and 8 are relatively
prime as they share no factors greater than 1. If the difference between
two numbers is prime, then the numbers are relatively prime.
If GCF(A,B)=1, then A and B are relatively prime.
If A – B = prime, then GCF(A,B)=1
Are 21 and 91 relatively prime?
No, GCF(21,91)=7
Are 10 and 21 relatively prime?
Yes, 2x5 and 3x7 share no factors.
Meet #2, Number Theory
Number of Factors: 12 has 6 factors: 1,2,3,4,6,12 (we count 1 and 12). To
count them, do prime factorization first. Take the exponents, add one to
each, and multiply. For example, 12=22!31, so the exponents are 2 and 1.
Add one to each: 3 and 2. Multiply: 2x3=6. So, 12 has 6 factors, same as
we listed above. How many factors does 1000 have? 1000=23!53, so it has
4x4=16 factors.
Another way to count the number of factors is to list them in pairs and be
sure you got them all! That’s the hard part.
3+1=4
4x3=12 factors
2+1=3
200=23!52
1 x200
2 x100
4 x 50
= 12 factors
5 x 40
8 x 25
10x20
Perfect squares have an odd number of factors. 36=1x36,
2x18,3x12,4x9,6x6, or 9 distinct factors. 36=22x32, (2+1)x(2+1)=9 factors.
Factors of 49: 1, 7, 49
= 3 factors – an odd number
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 = 9 factors (odd)
Euler’s GCF Method (Fast way to compute Greatest Common Factor):
If A>B, then GCF(A,B) = GCF(A-B,B). This works because if A and B have a
common factor, then A-B has the same factor. Repeat until obvious.
GCF(130,91) = GCF(130-91, 91) = GCF(39, 91) =
GCF(91, 39) = GCF(91-39, 39) = GCF(52,39) =
GCF(52-39, 39) = GCF(13,39) = 13
Meet #2, Number Theory
!""#$%&$$$$$$'"(")*"+$&,-,$
!
Category 3 ± Number Theory
!
"# $%&'!()!'%*!+*&)'!,-..-/!012'(32*!-4!Ͷͷ!&/5!͸͸6!!
!
!
!
7# 8%*!9:*&'*)'!,-..-/!;&<'-:!=9,;>!-4!'?-!/&'1:&2!/1.@*:)!‫ܣ‬ǡ ‫!)(!ܤ‬ͷA!&/5!'%*(:!
3:-51<'!=‫ͳ!)(!>ܤ כ ܣ‬ǡͲͲͲ#!
$%&'!()!'%*!).&22*)'!3-))(@2*!)1.!-4!‫ ܣ‬൅ ‫ܤ‬6!
!
!
B# C/!&!4&:!&?&D!E&2&FDA!B!<-.*')!&:*!G()(@2*!4:-.!32&/*'!H"I!
,-.*'!J23%&!()!G()(@2*!*G*:D!K!D*&:)A!&/5!?&)!2&)'!)**/!(/!7LLM#!
,-.*'!N*'&!()!G()(@2*!*G*:D!M!D*&:)A!&/5!?&)!2&)'!)**/!(/!7LLO#!
,-..*'!9&..&!()!G()(@2*!*G*:D!P!D*&:)A!&/5!?&)!2&)'!)**/!(/!7LLO#!
C/!?%&'!D*&:!?(22!&22!'%:**!<-.*')!@*!G()(@2*!'-E*'%*:!&E&(/6!!
!
!
!
!
Answers
1. _______________
2. _______________
3. _______________
"""#$%&'%#()*!
!
!""#$%&$$$$$$'"(")*"+$&,-,$
!
!"#$%&'#()!&#!*+&,-#./!0!±!1%23,.!45,#./!
67 Ͷͷ ൌ ͵ଶ ‫ כ‬ͷ!
͸͸ ൌ ʹ ‫!ͳͳ כ ͵ כ‬
Answers
1. 990
2. 65
3. 2,121
45,!8,+)&!*#22#(!9%$&':$,!;8*9<!')!&5,!:.#=%>&!#?!+$$!:.'2,!!
?+>&#.)!@'&5!&5,'.!5'-5,)&!:#@,.)A!‫ܯܥܮ‬ሺͶͷǡ͸͸ሻ ൌ ʹ ‫͵ כ‬ଶ ‫ כ‬ͷ ‫ ͳͳ כ‬ൌ ͻͻͲ!
!
B7 45,!:.#=%>&!#?!&@#!(%23,.)!')!,C%+$!&#!&5,!:.#=%>&!#?!&5,'.!‫!ܨܥܩ‬+(=!‫ܯܥܮ‬A!!
‫ ܤ כ ܣ‬ൌ ‫ܨܥܩ‬ሺ‫ܣ‬ǡ ‫ܤ‬ሻ ‫ܯܥܮ כ‬ሺ‫ܣ‬ǡ ‫ܤ‬ሻD!@5'>5!'(!#%.!>+),!')!ͳǡͲͲͲD!)#!@,!E(#@!&5+&!
‫ܯܥܮ‬ሺ‫ܣ‬ǡ ‫ܤ‬ሻ ൌ ͳͲͲͲ ൊ ͷ ൌ ʹͲͲ ൌ ʹଷ ȉ ͷଶ !7!!"'(>,!‫ܨܥܩ‬ሺ‫ܣ‬ǡ ‫ܤ‬ሻ ൌ ͷ!&5,(!#($/!
#(,!#?!&5,!(%23,.)!>+(!5+F,!ʹ!+)!+!:.'2,!?+>&#.D!(#&!3#&5!#?!&5,27!!G#&5!
(%23,.)!5+F,!ͷ!+)!+!?+>&#.D!3%&!#($/!#(,!#?!&5,2!5+)!ͷଶ !+)!+!?+>&#.!;#&5,.@'),D!
&5,!‫@!ܨܥܩ‬#%$=!5+F,!3,,(!ͷଶ <7!!"#!,'&5,.!‫ ܣ‬ൌ ʹଷ ȉ ͷଶ ൌ ʹͲͲܽ݊݀‫ ܤ‬ൌ ͷD!#.!
‫ ܣ‬ൌ ʹଷ ȉ ͷ ൌ ͶͲܽ݊݀‫ ܤ‬ൌ ͷଶ Ǥ!!!45,!),>#(=!:+'.!5+)!&5,!)2+$$,.!)%2D!͸ͷ7!
!
07 H#.!>#(F,(',(>,!@,!>+(!=,=%>&!ʹͲͲͲ!?.#2!+$$!/,+.)A!
I$:5+!')!F')'3$,!'(!/,+.)A!͹ǡ ͳ͵ǡ ͳͻǡ ʹͷǡ ǥ!;.,2+'(=,.!6!@5,(!='F'=,=!3/!J<!
G,&+!')!F')'3$,!'(!/,+.)A!ͻǡ ͳ͸ǡ ʹ͵ǡ ͵Ͳǡ ǥ!;.,2+'(=,.!B!@5,(!='F'=,=!3/!K<!
L+22+!')!F')'3$,!'(!/,+.)A!ͻǡ ͳ͹ǡ ʹͷǡ ͵͵ǡ ǥ!;.,2+'(=,.!6!@5,(!='F'=,=!3/!M<!
"#!&#!?'(=!&5,!/,+.!@5,(!+$$!+::,+.D!@,!(,,=!&#!?'(=!+!(%23,.!&5+&!?'&)!+$$!&5.,,!
>#(='&'#()7!"'(>,!;&5,!/,+.)!?#.<!I$:5+!+(=!L+22+!3#&5!)5#%$=!$,+F,!+!
.,2+'(=,.!#?!6!@5,(!='F'=,=!3/!J!#.!MD!#%.!/,+.!)5#%$=!$,+F,!+!.,2+'(=,.!#?!6!
@5,(!='F'=,=!3/!+(/!2%$&':$,!#?!J!+(=!MD!+(=!):,>'?'>+$$/!3/!&5,'.!8*9D!ʹͶ7!
"#!#%.!/,+.!>+(!3,!#(,!#?!&5,!(%23,.)!ʹͷǡ Ͷͻǡ ͹͵ǡ ͻ͹ǡͳʹͳ ǥ!3%&!+$)#!(,,=)!&#!
$,+F,!+!.,2+'(=,.!#?!B!@5,(!='F'=,=!3/!K7!45,!?'.)&!(%23,.!'(!&5')!),.',)!&#!?'&!
&5,!3'$$!')!ͳʹͳ7!"#!#%.!/,+.!')!ʹǡͳʹͳ7!
"""#$%&'%#()*!
!
Category 3
Number Theory
Meet #2, December 2008
1.
What is the Greatest Common Factor of 216 and 504?
2.
The prime factorization of X is !" # $ % . None of the values of !& '& $&()*(+
is equal to 2. Only one of the four values is 1. All four values of
!& '& $&(,-.(+ are different. What is the smallest possible value of X?
3.
The Least Common Multiple of positive integers ! and ' is 144. The
Greatest Common Factor of ! and ' is 12. What is the value of !' / 01 ?
Answers
1. _______________
2. _______________
3. _______________
Solutions to Category 3
Number Theory
Meet #2, December 2008
Answers
1.
72
1.
!"# $ !% & '% and ()* $ !% & '+ & ,The Greatest Common Factor of 216 and 504 is equal to
!% & '+ $ ,!.
2. 405
3.
64
2. Since none of the values can be equal to 2, . and / should be 3
and 5 if we want X to be as small as possible (they can’t be 1 since
1 isn’t prime). Since we can use 1 just once, using it as the
exponent of 5 will keep X smallest. That leaves the exponent for
3. The smallest value left to use is 4. So X =0'1 & (- $ 2" & ( $
*)(.
Editor note: We can’t use zero as no proper “prime factorization”
has a zero as either a base or an exponent. Otherwise, there would
be no unique prime factorization: 10 = 21 ·51 or 21 ·51·70, etc. See
the Fundamental Theorem of Arithmetic.
3. The LCM(.3 4) times the GCF(.3 4) will always be equal to the product of .
and 4. So .4 $ "! & "** $ ",!2 and .4 5 !, $ ",!2 5 !, $ #*. A faster
-+&-11
1&-11
1&-8
way to compute that would be
$
$
$ #*.
+6
7
-
Category 3
Number Theory
Meet #2, December 2006
1. Give the prime factorization of 792. You may use exponents or not, but you
must list the primes from least to greatest. For example, the prime factorization of
2
60 can be written as 2 ⋅ 3⋅ 5 or as 2 ⋅ 2 ⋅ 3⋅ 5 .
2. The product of two numbers is 396. If the greatest common factor (GCF) of
these two numbers is 3, what is the least common multiple (LCM) of these two
numbers?
3. At the manufacturing plant of Gadgets & Gizmos, Inc., it takes 1 hour and 24
minutes to assemble a gadget and 1 hour and 52 minutes to assemble a gizmo.
At 9:12 AM one morning, a new gadget and a new gizmo roll off the assembly
lines simultaneously. Give the time of day, using AM or PM, for the next time that
a new gadget and a new gizmo will roll off the assembly line simultaneously.
Answers
1. _______________
2. _______________
3. _______________
www.imlem.org
Solutions to Category 3
Number Theory
Meet #2, December 2006
Answers
1. The prime factorization of 792 can be written in exponential
form, with the primes listed from least to greatest as 2 3 ⋅ 3 2 ⋅ 11
1. 2 3 ⋅ 3 2 ⋅11
or 2 3 ⋅ 3 2 ⋅ 111 . Without exponents, we would write
or
2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 11. We will also accept the symbol × for
2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅11 multiplication.
2. 132
3. 2:48 PM
Students with correct
answer in a cluster of
6 schools:
1. 30/36
2. 31/36
2. In general, the product of the GCF and the LCM of two
numbers is equal to the product of the two numbers. We know
the product of the two numbers, so if we divide by the GCF we
will get the LCM: 396 ÷ 3 = 132.
3. Gadgets take 1 hour and 24 minutes, or 60 + 24 = 84
minutes. Gizmos take 1 hour and 52 minutes, or 60 + 52 = 112
minutes. We are looking for the least common multiple of 84
and 112. The prime factorizations are 84 = 2 2 ⋅ 3 ⋅ 7 and
112 = 2 4 ⋅ 7 . Taking the greatest number of each prime factor,
we find that the LCM is 2 4 ⋅ 3 ⋅ 7 = 336 . This means that after
336 minutes, a new gadget and a new gizmo will again roll off
the assembly line simultaneously. This many minutes is 5
hours and 36 minutes, so we need to find the time of day that is
5 hours and 36 minutes after 9:12 AM. In military time, this
would be 14:48. In regular time, it will be 2:48 PM.
3. 21/36!
www.imlem.org
Category 3
Number Theory
Meet #2, November 2004
1. Find the greatest common factor (GCF) of 5,409,149,074,560 and 37,644,750,
given their prime factorizations below.
5,409,149,074,560 = 2 7 ⋅ 3 2 ⋅ 5 1 ⋅ 7 3 ⋅115 ⋅13 0 ⋅17 1
37,644,750 = 2 1 ⋅ 3 4 ⋅ 5 3 ⋅ 7 0 ⋅111 ⋅13 2 ⋅17 0
2. If LCM(a,b) = 90 and LCM(a,c) = 36, then what is the value of LCM(a,b,c)?
Note: “LCM(a,b)” means the least common multiple of a and b.
3. If m is a proper factor of 208 and n is a proper factor of 288 and
GCF(m,n) = 2, then what is the greatest possible value of m + n?
Note: A proper factor is a factor of a number other than the number itself.
Answers
1. _______________
2. _______________
3. _______________
www.Imlem.org
Solutions to Category 3 Average team got 10.34 points, or 0.9 questions correct
Number Theory
Meet #2, November 2004
Answers
1. 990
2. 180
3. 170
The greatest common factor of the two numbers is the
product of the primes that they have in common. For
each prime factor, we use the lower exponent to create
the GCF. Thus the GCF of the two numbers is
2 1 ⋅ 3 2 ⋅ 5 1 ⋅ 7 0 ⋅111 ⋅13 0 ⋅17 0 = 2 ⋅ 9 ⋅ 5 ⋅11 = 10 ⋅ 99 = 990 .
2. The LCM(a,b,c) is the least common multiple of the
LCM(a,b) and the LCM(a,c). In other words, we need to
find LCM(90,36). It is helpful to consider the prime
factors of each number: 90 = 2 ⋅ 3 2 ⋅ 5 and 36 = 2 2 ⋅ 3 2 .
The LCM of these two numbers must have all the prime
factors of each and no extras. If we start with the prime
factorization of 90, we only need one more factor of 2 to
create a multiple of 36. Thus the LCM of 90 and 36—
and also of a, b, and c—is 90 × 2 = 180.
3. The proper factors of 208 are 1, 2, 4, 8, 13, 16, 26, 52,
and 104. The proper factors of 288 are 1, 2, 3, 4, 6, 8, 9,
12, 16, 18, 24, 32, 36, 48, 72, 96, and 144.
Taking one number from each list and making sure their
GCF is 2, we get the greatest possible sum from 144 and
26, namely 170.
www.Imlem.org
Meet #2
December 2010
Category 3 – Number Theory
1. What is the Least Common Multiple of
and
?
2. The Greatest Common Factor (GCF) of two natural numbers
product (
) is
is , and their
.
What is the smallest possible sum of
?
3. In a far away galaxy, 3 comets are visible from planet 51:
Comet Alpha is visible every 6 years, and was last seen in 2007.
Comet Beta is visible every 7 years, and was last seen in 2009.
Commet Gamma is visible every 8 years, and was last seen in 2009.
In what year will all three comets be visible together again?
Answers
1. _______________
2. _______________
3. _______________
www.imlem.org
Meet #2
December 2010
Solutions to Category 3 – Number Theory
Answers
1. 990
2. 65
3. 2,121
1.
The Least Common Multiple (LCM) is the product of all prime
factors with their highest powers:
2. The product of two numbers is equal to the product of their
, which in our case is
. Since
one of the numbers can have
numbers have
the
:
, so we know that
then only
as a prime factor, not both of them. Both
as a factor, but only one of them has
would have been
and
as a factor (otherwise,
). So either
, or
The second pair has the smaller sum,
3. For convenience we can deduct
.
from all years:
Alpha is visible in years:
(remainder 1 when divided by 6)
Beta is visible in years:
(remainder 2 when divided by 7)
Gamma is visible in years:
(remainder 1 when divided by 8)
So to find the year when all appear, we need to find a number that fits all three
conditions. Since (the years for) Alpha and Gamma both should leave a
remainder of 1 when divided by 6 or 8, our year should leave a remainder of 1
when divided by any multiple of 6 and 8, and specifically by their LCM,
So our year can be one of the numbers
.
but also needs to
leave a remainder of 2 when divided by 7. The first number in this series to fit
the bill is
. So our year is
.
www.imlem.org
Category 3
Number Theory
Meet #2, November/December 2012
1. Some IMLEM Clusters will take Meet #2 on Thursday, 12/06. Find
the prime factorization of 1206. You can use exponents or not, but you
must write the prime factors in order from least to greatest.
2. Grace was supposed to find the greatest common factor (GCF) of 330
and 462, but she found the greatest common prime factor (GCPF)
instead. What is the ratio of the GCF of 330 and 462 and the GCPF that
Grace found?
3. At the Gadgets and Gizmos factory, workers complete a gadget every
588 minutes and a gizmo every 882 minutes. The factory is open 24
hours a day, seven days a week, and the workers always complete the
gadgets and gizmos on schedule. If it happens that a gadget and a gizmo
are completed at the same time at 3:00 PM on Thursday, November 29,
how many additional times will a gadget and a gizmo be completed at
the same time before 3:00 PM on Thursday, December 6?
Answers
1. ______________________
2. ______________________
3. _______________ times
Solutions to Category 3
Number Theory
Meet #2, November/December 2012
1. A systematic way to find the prime
factorization of a number is to divide out
primes in order of the primes as shown in
the so-called ladder at right. The prime
factorization of 1206 is 2  3  3  67 or,
using exponents, it’s 2  32  67.

21206
3603
3201
6767
1
Answers
1. 2  32  67
or 2  3  3  67
2. 6
3. 5 times
2. The prime factorization of 330 is 2 3 5 11 and that of 462 is
2 3 7 11. The GCF is 2  3  11 = 66 and the GCPF is 11, so the

desired ratio is 66/11 = 6.
Editor note: The original problem
asked “How many times greater is the

actual GCF of 330 and 462 than the GCPF that Grace found?” This was
changed here to remove any ambiguity.
3. The prime factorization of 588 is 22  3  72, and that of 882 is 2  32
 72. The LCM of these two numbers is 22  32  72 = 1764. This means
that a gadget and a gizmo are completed simultaneously every 1764
minutes. There are 7  24  60 = 10,080 minutes in a week. Five times
1764 = 8820 and 6  1764 = 10584, so gadgets and gizmos will be
completed simultaneously an additional 5 times in the next week.