3. Greatest Common Divisor - Least Common Multiple

Definition 3.1: The greatest common divisor of two natural numbers a and b is the largest natural number c which divides both a and b. We denote the greatest common divisor of a and b as gcd()ab , .

Example 3.2: Find gcd() 24,90 .

Solution: The natural numbers that divide 24 are 1, 2, 3, 4, 6, 12 and 24. The natural numbers that divide 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30 and 90. Therefore, gcd() 24,90= 6 .

One way to compute the greatest common divisor of two natural numbers a and b is to write their prime factorizations. For example, in the case of the numbers 24 and 90 above, we have 24= ()() 23 3 and 90= () 2( 3)2 ( 5) . Notice that we can include all of the prime factors from both numbers by using an exponent of 0. 24= ()()() 230 3 5 90= ( 2)( 3)2 ( 5) Now we take the smallest exponent that occurs for each prime factor and obtain gcd() 24,90= ( 2)min(3,1)( 3) min(1,2)( 5) min(0,1) 110 = ()()()235 = ()()23 = 6 We generalize this idea in the theorem below.

Theorem 3.3: Let ab, ∈ ` and suppose p12,pp ,..., n are prime numbers so that

α12ααn ap= ( 12) ( p) ...( pn ) and

β12ββn bp= ( 12) ( p) ...( pn ) where the values α12,ααβββ ,...,nn , 12 , ,..., are either natural numbers or 0. Then

min(αβ11 , ) min(α 2 ,β 2 ) min(αnn ,β ) gcd()()ab ,= p12 () p ... () pn

Proof: Let cab= gcd( , ) . Since p12,pp ,..., n are prime numbers, p j divides both a and b implies that p j divides c. α j β j α j β j If p j divides a and p j divides b, then the smaller of p j and p j divides both a and b. ( 22 divides 12 and 2 divides 30 implies 2 divides gcd(12,30)= 6 ). So,

min(α jj ,β ) p j divides both a and b; and hence c. If p j divides a but does not divide b,

min(α j ,0) 0 then p j does not divide c. But when we say ppjj= =1 divides c, we mean that p j does not divide c.

min(α11 ,βαβαβ ) min( 2 , 2 ) min(nn , ) Hence, gcd()()ab ,= p12 () p ...( pn ) . „

Example 3.4: Find the greatest common divisor of 1260 and 600.

Solution: The prime factorization of these numbers is given by 1260= ( 2)22( 3) ( 5)( 7) and 600= ()()() 232 3 5 So the combined prime factors are 2, 3, 5 and 7. Notice that these are all factors of 1260. Also, 600 can be written using these prime factors as 600= ( 2)320( 3)( 5) ( 7) Then Theorem 3.3 implies min() 2,3 min() 2,1 min() 2,1 min() 1,0 gcd()() 1260,600= 2() 3() 5() 7 = ()()()()23572110 = 60

Remark 3.5: The TI-83 function gcd can be used to find the greatest common divisor of two natural numbers. This function can be found by pressing MATH, selecting NUM, and then scrolling down to item 9. This will place gcd( on the calculator screen. Typing 1260,600) and pressing ENTER results in the screen shot below.

Gcd(1260,600) 60

How Do Calculators Compute The Greatest Common Divisor?

Calculators DO NOT use prime factorizations to compute the greatest common divisor of two natural numbers. Instead, they use a very efficient process called the Euclidean algorithm. We give this process after the definition and theorem below.

Definition 3.6: Let ab, ∈ ` . The integer part of ab/ is the largest value c chosen from {0,1,2,3,...} so that cab≤ / . We denote the integer part of ab/ by iPart(a/b).

Remark 3.7: iPart is a built-in function on the TI-83. You can access this function by pressing MATH and selecting NUM. The function iPart is item 3.

Theorem 3.8: Let ab, ∈ ` with ab> , and suppose cab= iPart( / ) . Then racb= − and 0 ≤

a Proof: cab= iPart() / implies that = ck+ where 01≤ k < (otherwise c will not be the b largest integer which is less than or equal to ab/ ). a =+c k ⇒ a = b ( c + k ) = bc + bk ⇒ a − bc = bk . b Since 01≤

Remark 3.9: The value r in Theorem 3.8 is sometimes referred to as the remainder associated with the division ab/ , and the integer part c of ab/ is sometimes referred to as the quotient associated with the division ab/ .

Example 3.10: The integer part of 2/3 is 0, the integer part of 9/4 is 2, and the integer part of 67/13 is 5. That is, iPart(2/3) = 0, iPart(9/4) = 2 and iPart(67/13) = 5. Theorem 14 can be realized with the equations 19=−( 24)( ) and 267513=−( )( )

The Euclidean algorithm uses the Theorem 14 to find the greatest common divisor of two natural numbers. This algorithm was originally proposed in Euclid’s Elements around 300 B.C. It is amazing that this process is still widely used in practice by high speed computing devices. The process is given below.

Euclidean algorithm Let ab, ∈` with ab> . ra=−ipart( abb / ) 1 Continue until the first time rb=−ipart( brr / ) 211 ri = 0

rr31=−ipart( rrr 122 / ) If i = 1 then gcd(ab ,) = 1.

rr42=−ipart( rrr 233 / ) Otherwise, gcd(ab , ) = ri−1 . #

Theorem 3.11: Let ab, ∈` with ab> . The Euclidean algorithm computes gcd()ab , .

Proof: Let ab , ∈ ` with ab> . We are looking for gcd(ab , ) . Suppose the remainder of the division of a by b is c. Then aqbc= + , where q is the quotient of the division. Any common divisor of a and b also divides c (since c can be written as caqb= − ); similarly any common divisor of b and c will also divide a. Thus, the greatest common divisor of a and b is the same as the greatest common divisor of b and c. Therefore, it is enough if we continue the process with numbers b and c. Since c is smaller in absolute value than b, we will reach c = 0 after finitely many steps. „

Example 3.12: Use the Euclidean algorithm to compute gcd( 1260,600) .

Solution: You can verify that r =−1260 ipart() 1260 / 600( 600) =− 1260( 2)( 600) = 60 1 r2 =−600 ipart( 600 / 60 )() 60 =− 600 ()() 10 60 = 0 So, gcd() 1260,600= 60.

Example 3.13: Use the Euclidean algorithm to compute gcd( 860,1375) .

Solution: The algorithm results in

r1 =−1375( 1)( 860) = 515

r2 =−860()( 1 515 ) = 345

r3 =−515() 1 345 = 170

r4 =−345()( 2 170 ) = 5

r5 =−170()() 34 5 = 0 Consequently, gcd() 860,1375= 5.

Example 3.14: Use the Euclidean algorithm to compute gcd( 1326,741) .

Solution: You can verify that

r1 =−1326( 1)( 741) = 585

r2 =−741()( 1 585 ) = 156

r3 =−585()( 3 156 ) = 117

r4 =−156()( 1 117 ) = 39

r5 =−117()() 3 39 = 0 Consequently, gcd() 1326,741= 39 .

A quantity which is closely related to the greatest common divisor is the least common multiple.

Definition 3.15: Let ab, ∈` . The least common multiple of a and b is the smallest natural number which is a multiple of both a and b. The least common multiple of a and b is denoted by lcm()ab , .

Remark 3.16: lcm(ab , ) ≤ ab whenever ab , ∈ ` .

Since the lcm()ab , is the smallest number that is a multiple of both a and b, it is also the smallest number for which both a and b are divisors. As a result, all of the prime factors of a must divide lcm()ab , , and all of the prime factors of b must divide lcm()ab , . This leads to the following theorem.

Theorem 3.17: Let ab, ∈ ` , and suppose the prime factorizations of a and b are given by

α12ααn ap= ()()12 p... () pn and

β12ββn bp= ( 12) ( p) ...( pn ) where the values α12,ααβββ ,...,nn , 12 , ,..., are either natural numbers or 0. Then

max(α11 ,βαβαβ ) max( 2 , 2 ) max(nn , ) lcm()()ab ,= p12( p) ...( pn )

Proof: Let cab= lcm( , ) . If a and b are both multiples of p j , then c is a multiple of p j

α j β j too. If a is a multiple of p j and b is a multiple of p j , then c is a multiple of the

α j β j bigger of p j and p j ( since c is a multiple of both a and b).Hence, c is a multiple of

max(α jj ,β ) p j . Note that, if a is not a multiple of p j but b is; then c should be a multiple

max(0,β jj ) β of p j . Using p j = p j solves this problem. Thus,

max(α11 ,βαβαβ ) max( 2 , 2 ) max(nn , ) lcm()()ab ,= p12( p) ...( pn ) „

Example 3.18: Use Theorem 17 to compute lcm( 24,15) .

Solution: We can see that 24= () 23 ( 3) and 15= ( 3)( 5) . These can be written in common form as 24= ( 2)30( 3)( 5) and 15= ( 2)0 ( 3)( 5) As a result, max() 3,0 max() 1,1 max() 0,1 lcm()() 24,15= 2() 3() 5 311 = ()()()235 =120

Example 3.19: We show the value of lcm(ab , ) , gcd(ab , ) and ab for several choices of ab, ∈ ` .

a, b lcm(a,b) gcd(a,b) ab 16, 28 112 4 448 18, 24 72 6 432 14, 21 42 7 294 15, 24 120 3 360

Although it might not be readily apparent, the table above gives some insight to the relationship between lcm()ab , , gcd(ab , ) and ab . The table above has been reproduced below, along with an additional column containing the product of lcm()ab , and gcd()ab , .

a, b lcm(a,b) gcd(a,b) ab lcm(a,b)gcd(a,b) 16, 28 112 4 448 448 18, 24 72 6 432 432 14, 21 42 7 294 294 15, 24 120 3 360 360

For the choices of a and b above, it appears that lcm(ab ,) gcd( ab , ) = ab This result is always true.

Theorem 3.20: Let ab, ∈` . Then lcm(ab ,) gcd( ab , ) = ab.

α12ααn ββ12 βn Proof: Let ap= ()()12 p... () pn and bp= ()()12 p... () pn where the values

α12,ααβββ ,...,nn , 12 , ,..., are either natural numbers or 0. Then

max(α11 ,βαβαβ ) max( 2 , 2 ) max(nn , ) lcm()()ab ,= p12( p) ...( pn ) and min(α11 ,βαβαβ ) min( 2 , 2 ) min(nn , ) gcd()()ab ,= p12( p) ...( pn ) .

max(αβ , ) max( αβ , )max(αβ , ) min( αβ , ) min( αβ , ) min( αβ , ) lcmab , gcd ab , = ⎡⎤⎡⎤ p11 p 2 2... pnn p 11 p 2 2 ... p nn () ()()⎣⎦⎣⎦12 () ()nn() 12 () ()

max(αβ11 , )++ min( αβ 11 , ) max( αβ 2 , 2 ) min( αβ 2 , 2 ) max(αβnn , )+ min( αβ nn , ) = ()pp12 () ...() pn

If max(α11 ,βα ) = 1 then min(α11 ,ββ ) = 1. And if min(α11 ,βα ) = 1, then max(α11 ,ββ ) = 1.

Hence, in any case max(α11 ,βαβαβ )+=+ min( 11 , ) 1 1. So we have;

αβ11++ α 2 β 2 αnn + β lcm()ab , gcd ()()() ab ,= p12 p ... () pn = ⎡ ppαα12 ... p αnn⎤⎡ pp ββ 12 ... p β ⎤ ⎣()()12 ()nn⎦⎣ ()() 12 () ⎦ = ab „

Exercises 1. Use the prime factorizations of 860 and 1375 to compute gcd( 860,1375) . 2. Use the prime factorizations of 6300 and 1584 to compute gcd() 6300,1584 . 3. Use the prime factorizations of 1260 and 2640 to compute gcd() 1260,2640 . 4. Use the prime factorizations of 2373 and 1374 to compute gcd() 2373,1374 . 5. Use the division algorithm to compute gcd( 6300,1584) . 6. Use the division algorithm to compute gcd( 1260,2640) . 7. Use the division algorithm to compute gcd( 2373,1374). 8. Use the prime factorizations of 75 and 124 to find lcm( 75,124) . 9. Use the prime factorizations of 236 and 125 to find lcm(236,125) . 10. Use the prime factorizations of 84 and 118 to find lcm( 84,118). 11. Suppose ab = 900 and lcm(ab ,) = 300 . Give gcd(ab , ) . 12. Let mab,,∈` . Show that gcd(ma , mb) = m gcd( a , b) ; i.e. the greatest common divisor satisfies a distributive property. 13. Let abc, , ∈ ` . Show that gcd( gcd(ab , ), c) = gcd( a ,gcd( bc , )) ; i.e. the greatest common divisor satisfies an associative property. 14. Let a ∈ ` . Show that gcd(aa , ) = a; i.e. the greatest common divisor is idempotent. 15. Let ab, ∈ ` . Show that gcd(ab , )= gcd( ba , ) ; i.e. the greatest common divisor satisfies a commutative property. 16. Let ab, ∈` . Show that lcm(aaba ,gcd( , )) = and gcd(aaba ,lcm( , )) = . 17. Let a ∈ ` . Show that lcm(aa , ) = a; i.e. the least common multiple is idempotent. 18. Let ab, ∈` . Show that lcm(ab , )= lcm( ba , ); i.e. the least common multiple satisfies a commutative property. 19. Let abc,,∈` . Show that lcm( lcm(ab , ), c) = lcm( a ,lcm( bc , )) ; i.e. the least common multiple satisfies an associative property. 20. Let mab, , ∈ ` . Show that lcm(ma , mb) = m lcm( a , b) ; i.e. the least common multiple satisfies a distributive property.

Solutions:

1. 860=⋅⋅ 223 5 43 and 1375 =⋅ 5 11. Therefore;

min(0,2) min(1,3) min(0,1) min(0,1) gcd(860,1375)= () 2 () 5 ( 11 ) ( 43 ) = ()()()( 2000 5 11 43 ) = 5

2. 6300=⋅⋅⋅ 222 5 7 9 and 1584 =⋅⋅ 2 4 9 11. Therefore;

gcd 6300,1584=⋅⋅⋅⋅ 2min(2,4) 5 min(0,2) 7 min(0,1) 9 11 min(0,1) ()()( ) ( ) ( ) ( ) =⋅= 22 9 36

3. 1260=⋅⋅⋅ 224 5 7 9 and 2640 =⋅⋅⋅ 2 3 5 11. Therefore;

gcd()() 1260,2640= 2min(2,4)( 3) min(0,1)( 5)( 7) min(0,1)( 9) min(0,1)( 11) min(0,1)

==()() 22 5 20

4. 2373=⋅⋅ 3 7 113 and 1374 =⋅⋅ 2 3 229. Therefore;

gcd()()() 2373,1374== 20000 3( 7) ( 113) ( 229) 3 .

5. gcd() 6300,1584

r1 =−6300( 3)( 1584) = 1548

r2 =−1584()( 1 1548 ) = 36

r3 =−1548() 43 36 = 0 Thus; gcd() 6300,1584= 36 .

6.

r1 =−2640( 2)( 1260) = 120

r2 =−1260()() 10 120 = 60

r3 =−120() 2 60 = 0 Consequently, gcd() 1260,2640= 60 .

7.

r1 =−2373( 1)( 1374) = 999

r2 =−1374()( 1 999 ) = 375

r3 =−999() 2 375 = 249

r4 =−375()( 1 249 ) = 126

r5 =−249()( 1 126 ) = 123

r6 =−126()( 1 123 ) = 3

r7 =−123()() 41 3 = 0 Hence, gcd() 2373,1374= 3 .

8. 75=⋅ 3 522 and 124 = 2 ⋅ 31. Thus, max() 2,0 max() 1,0 max() 2,0 max(1,0) lcm()() 75,124= 2() 3() 5() 31 = ()()()(23531212 ) 1 = 9300

9. 236=⋅ 223 59 and 125 = 5 . Hence, max() 2,0 max() 3,0 max(1,0) lcm()() 236,125= 2() 5() 59 = ()()(255923 ) 1 = 29500

10. 84=⋅⋅ 22 3 7 and 118 =⋅ 2 59. Therefore, max() 2,1 max() 1,0 max() 1,0 max(1,0) lcm()() 84,118= 2() 3() 7() 59 = ()()()(23759211 ) 1 = 4956

11. Suppose ab = 900 and lcm(ab ,) = 300 . We know that lcm(ab ,)() gcd ab , = ab. ab 900 So, gcd()ab ,=== 3. The greatest common divisor is: lcm()ab , 300 gcd()ab ,= 3 .

12. Let mab, , ∈ ` .

α12ααn Suppose p12,pp ,..., n are prime numbers so that ap= ( 12) ( p)()... pn

ββ12 βn and bp= ()()12 p... () pn , where the values α12,ααβββ ,...,nn , 12 , ,..., are either natural numbers or 0. Then

min(α11 ,βαβαβ ) min( 2 , 2 ) min(nn , ) gcd()ab ,= ( p12)() p ...( pn )

α12ααn β12ββn We have: ma= ()() m p12( p) ...( pn ) and mb= ( m)( p12)( p )... ( pn ). So, the greatest common divisor is:

min(1,1) min(α11 ,βαβαβ ) min( 2 , 2 ) min(nn , ) gcd()()()()()ma , mb= m p12 p ... pn

min(αβ11 , ) min( α 2 , β 2 ) min( αnn , β ) = mp()12 () p ... () pn = mab gcd( , )

Hence, the greatest common divisor satisfies a distributive property.

13. Let abc, , ∈ ` .

α12ααn Suppose p12,pp ,..., n are prime numbers so that ap= ()()12 p... () pn ,

β12ββn γ12γγn bp= ()()12 p... () pn and cp= ( 12) ( p) ...( pn ) , where the values

α12,ααβββγγγ ,...,nnn , 12 , ,..., , 12 , ,..., are either natural numbers or 0. Then

min(α11 ,βαβαβ ) min( 2 , 2 ) min(nn , ) gcd()ab ,= ( p12)() p ...( pn ) and

min(βγ11 , ) min( β 2 , γ 2 ) min( βnn , γ ) gcd()()bc ,= p12 () p ... () pn .

min(min(α11 ,βγ ), 1 ) min(min( αβγ 2 , 2 ), 2 ) min(min( αβγnn , ), n ) gcd(gcd()ab , , c )= () p12( p) ...( pn ) and

min(αβγ111 ,min( , )) min( α 2 ,min( βγ 22 , )) min( αnnn ,min( βγ , )) gcd()abcp ,gcd() ,= ()12( p) ...( pn ) .

Here, it is important to observe that

min(min(αnn ,βγ ), n )== min( α n , min( βγ nn , )) min( αβγ nnn , , ) . The reason behind this is; while comparing 3 natural numbers, you can start comparing from whichever you want. If you want to find the smallest of 5, 9 and 6, then arrange them in order; 5 < 6 < 9. The smallest is 5, this does not change. So, it is not important if you compare 5 & 9 first and then the result with 6; or if you compare 6 & 9 first and then the result with 5.

Therefore, gcd() gcd(ab , ), c= gcd( a ,gcd( bc , )) ; i.e. the greatest common divisor satisfies an associative property.

14. Let a ∈ ` . Show that gcd(aa , ) = a; i.e. the greatest common divisor is idempotent.

α12ααn Suppose p12,pp ,..., n are prime numbers so that ap= ()()12 p... () pn

where the values α12,αα ,..., n are either natural numbers or 0. Then

min(α11 ,ααααα ) min( 2 , 2 ) min(nn , ) gcd()()aa ,= p12 () p ...( pn ) where, of course,

min(αnn ,αα ) = n. Hence,

gcd()()aa ,= pmin(α11 ,ααααα )( p) min( 2 , 2 ) ...( p) min(nn , ) 12 n αα12 αn ==()()pp12 ... () pn a

It is easier to see this result if we restate the question as “what is the greatest number that divides both a and a?”. The answer is, of course, a itself.

15. Let ab, ∈ ` . Show that gcd(ab , )= gcd( ba , ) ; i.e. the greatest common divisor satisfies a commutative property.

α12ααn Suppose p12,pp ,..., n are prime numbers so that ap= ()()12 p... () pn

β12ββn and bp= ()()12 p... () pn , where the values α12,ααβββ ,...,nn , 12 , ,..., are either natural numbers or 0. Then

min(α11 ,βαβαβ ) min( 2 , 2 ) min(nn , ) gcd()ab ,= ( p12) ( p) ...( pn ) and

min(β11 ,α ) min(β 2 ,α 2 ) min(βnn ,α ) gcd()()ba ,= p12 () p ... () pn .

The conclusion follows from the observation that min(αnn ,ββα )= min( nn , ) . Hence, gcd(ab , )= gcd() ba , .

16. Let ab, ∈ ` .

α12ααn Suppose p12,pp ,..., n are prime numbers so that ap= ( 12) ( p)()... pn

ββ12 βn and bp= ()()12 p... () pn , where the values α12,ααβββ ,...,nn , 12 , ,..., are either natural numbers or 0. Then

max(α11 ,βαβαβ ) max( 2 , 2 ) max(nn , ) lcm()()ab ,= p12 () p ...( pn ) and

min(α11 ,βαβαβ ) min( 2 , 2 ) min(nn , ) gcd()ab ,= ( p12)() p ...( pn ) .

max(ααβ111 ,min( , )) max( ααβ 2 ,min( 22 , )) max( αnnn ,min( αβ , )) lcm(aabp ,gcd( , ))= ()12 () p ... () pn

Let’s find max(α111 , min(αβ , )) . If α11< β , then min(α11 ,βα ) = 1 and

max(α111111 , min(αβ , ))== max( αα, ) α . If β11< α , then min(α11 ,ββ ) = 1

and max(α111111 , min(αβ , ))= max( αβ, ) = α. Similarly, for any n,

max(αnnnn , min(αβ , )) = α. Therefore;

lcm(aabp ,gcd( , ))= ()max(ααβ111 ,min( , ))( p) max( α 2 ,min( αβ 22 , )) ...( p) max( αnnn ,min( αβ , )) 12 n αα12 αn ==()()pp12 ... () pn a

min(ααβ111 ,max( , )) min( ααβ 2 ,max( 22 , )) min( αnnn ,max( αβ , )) gcd(aabp ,lcm( , ))= ()12 () p ... () pn

With the reasoning we used above we can conclude that

min(αnnnn , max(αβ , )) = α. Thus,

gcd(aabp ,lcm( , ))= ()min(ααβ111 ,max( , ))( p) min( α 2 ,max( αβ 22 , )) ...( p) min( αnnn ,max( αβ , )) 12 n αα12 αn ==()()pp12 ... () pn a

Hence, lcm(aaba ,gcd( , )) = and gcd(aaba ,lcm( , )) = .

17. Let a ∈ ` . Suppose p12,pp ,..., n are prime numbers so that

α12ααn ap= ()()12 p... () pn where the values α12,αα ,..., n are either natural numbers or 0. Then

lcm()()aa ,= pmax(α11 ,ααααα ) () p max( 2 , 2 ) ...( p) max(nn , ) 12 n αα12 αn ==()()pp12 ... () pn a

Therefore, lcm(aa , ) = a; i.e. the least common multiple is idempotent.

18. Let ab, ∈` . Suppose p12,pp ,..., n are prime numbers so that

α12ααn β12ββn ap= ()()12 p... () pn and bp= ( 12) ( p) ...( pn ) , where the values

α12,ααβββ ,...,nn , 12 , ,..., are either natural numbers or 0. Then

max(αβ11 , ) max(α 2 ,β 2 ) max(αnn ,β ) lcm()()ab ,= p12 () p ... () pn and

max(β11 ,αβαβα ) max( 2 , 2 ) max(nn , ) lcm()()ba ,= p12 () p ...( pn ) . Since max(α11 ,ββα )= max( 11 , ) , we can conclude that lcm(ab , )= lcm( ba , ).

19. Let abc, , ∈ ` .Suppose p12,pp ,..., n are prime numbers so that

α12ααn β12ββn ap= ()()12 p... () pn , bp= ( 12) ( p) ...( pn ) and

γ12γγn cp= ()()12 p... () pn , where the values α12,ααβββγγγ ,...,nnn , 12 , ,..., , 12 , ,..., are either natural numbers or 0. Then

max(αβ11 , ) max(α 2 ,β 2 ) max(αnn ,β ) lcm()()ab ,= p12 () p ... () pn and

max(β11 ,γβγβγ ) max( 2 , 2 ) max(nn , ) lcm()bc ,= ( p12)() p ...( pn ) .

max(max(α11 ,βγ ), 1 ) max(max( αβγ 2 , 2 ), 2 ) max(max( αβγnn , ), n ) lcm(lcm()ab , , c )= () p12( p) ...( pn ) and

max(αβγ111 ,max( , )) max( α 2 ,max( βγ 22 , )) max( αnnn ,max( βγ , )) lcm()abcp ,lcm() ,= ()12( p) ...( pn ) .

As we have pointed out before, the order at which you start to compare natural numbers does not matter;

max(max(αnn ,βγ ), n )= max( α n ,max( βγ nn , ))= max( αβγ nnn , , ) .

Hence, lcm() lcm(ab , ), c= lcm( a ,lcm( bc , )) ; i.e. the least common multiple satisfies an associative property.

20. Let mab,,∈` .

α12ααn Suppose p12,pp ,..., n are prime numbers so that ap= ( 12) ( p)()... pn

β12ββn and bp= ()()12 p... () pn , where the values α12,ααβββ ,...,nn , 12 , ,..., are either natural numbers or 0. Then

max(αβ11 , ) max(α 2 ,β 2 ) max(αnn ,β ) lcm()()ab ,= p12 () p ... () pn

α12ααn β12ββn We have: ma= ()() m p12( p) ...( pn ) and mb= ( m)( p12)( p )... ( pn ). So, the least common multiple is:

max(1,1) max(α11 ,βαβαβ ) max( 2 , 2 ) max(nn , ) lcm()()()()()ma , mb= m p12 p ... pn

max(αβ11 , ) max( α 2 , β 2 ) max( αnn , β ) = mp()12 () p ... () pn = mab lcm( , )

Thus, lcm()ma , mb= m lcm( a , b) ; i.e. the least common multiple satisfies a distributive property.

Graphical Representation Of gcd(a,b) = 1

Numbers ab, ∈ ` are said to be relatively prime if and only if gcd(ab , )= 1. It is possible to visualize all of the pairs (ab, ) for which a and b are relatively prime by plotting the points ()ab, for a and b between 1 and 100. We give this plot below, and the pattern is very interesting! Surprisingly, there are more pairs in this range that are relatively prime than those that are not. In fact, of the 10,000 pairs (ab, ) for a and b between 1 and 100, there are 6,087 pairs which are relatively prime.

()ab, for which gcd(ab , )= 1 for ab,1,...,100∈{ }