what is the biggest number to fit evenly into 42 and 60

Smallest positive number divisible by two integers

A Venn diagram showing the least common multiples of combinations of 2, three, 4, 5 and 7 (half dozen is skipped as it is ii × 3, both of which are already represented).
For example, a card game which requires its cards to be divided as among up to 5 players requires at least sixty cards, the number at the intersection of the 2, 3, four, and v sets, just not the 7 gear up.

In arithmetic and number theory, the to the lowest degree common multiple, lowest common multiple, or smallest mutual multiple of 2 integers a and b, commonly denoted by lcm(a,b), is the smallest positive integer that is divisible by both a and b.^{[1] [ii]} Since division of integers past goose egg is undefined, this definition has meaning only if a and b are both different from zero.^[3] However, some authors define lcm(a,0) every bit 0 for all a, since 0 is the but common multiple of a and 0.

The lcm is the "lowest common denominator" (lcd) that can be used before fractions can be added, subtracted or compared.

The least common multiple of more than two integers a, b, c, . . . , usually denoted by lcm(a,b,c, . . .), is likewise well defined: Information technology is the smallest positive integer that is divisible past each of a, b, c, . . .^[1]

Overview [edit]

A multiple of a number is the product of that number and an integer. For case, 10 is a multiple of v considering 5 × two = x, so ten is divisible by 5 and 2. Because 10 is the smallest positive integer that is divisible by both v and 2, it is the to the lowest degree common multiple of v and 2. Past the same principle, 10 is the least common multiple of −5 and −two as well.

Notation [edit]

The least common multiple of 2 integers a and b is denoted as lcm(a, b).^[1] Some older textbooks employ [a, b].^{[3] [4]}

Instance [edit]

${\displaystyle \operatorname {lcm} (4,vi)}$

Multiples of four are:

${\displaystyle iv,8,12,16,20,24,28,32,36,forty,44,48,52,56,60,64,68,72,76,...}$

Multiples of half dozen are:

${\displaystyle 6,12,eighteen,24,30,36,42,48,54,60,66,72,...}$

Common multiples of 4 and 6 are the numbers that are in both lists:

${\displaystyle 12,24,36,48,60,72,...}$

In this list, the smallest number is 12. Hence, the least common multiple is 12.

Applications [edit]

When adding, subtracting, or comparison simple fractions, the to the lowest degree common multiple of the denominators (frequently called the lowest mutual denominator) is used, because each of the fractions can be expressed as a fraction with this denominator. For case,

${\displaystyle {2 \over 21}+{1 \over half dozen}={4 \over 42}+{7 \over 42}={xi \over 42}}$

where the denominator 42 was used, because information technology is the to the lowest degree mutual multiple of 21 and 6.

Gears trouble [edit]

Suppose there are 2 meshing gears in a machine, having 1000 and northward teeth, respectively, and the gears are marked by a line segment drawn from the middle of the kickoff gear to the centre of the 2nd gear. When the gears begin rotating, the number of rotations the kickoff gear must complete to realign the line segment can be calculated by using ${\displaystyle \operatorname {lcm} (m,n)}$ . The first gear must complete ${\displaystyle \operatorname {lcm} (thou,due north) \over 1000}$ rotations for the realignment. By that fourth dimension, the second gear will have made ${\displaystyle \operatorname {lcm} (g,n) \over n}$ rotations.

Planetary alignment [edit]

Suppose there are three planets revolving around a star which have l, grand and n units of time, respectively, to complete their orbits. Assume that fifty, m and n are integers. Assuming the planets started moving effectually the star after an initial linear alignment, all the planets reach a linear alignment over again later ${\displaystyle \operatorname {lcm} (l,m,n)}$ units of time. At this time, the offset, second and third planet will have completed ${\displaystyle \operatorname {lcm} (fifty,m,due north) \over fifty}$ , ${\displaystyle \operatorname {lcm} (l,g,due north) \over m}$ and ${\displaystyle \operatorname {lcm} (l,one thousand,n) \over north}$ orbits, respectively, around the star.^[5]

Calculation [edit]

Using the greatest mutual divisor [edit]

The following formula reduces the problem of computing the least common multiple to the problem of computing the greatest common divisor (gcd), too known as the greatest mutual factor:

${\displaystyle \operatorname {lcm} (a,b)={\frac {|ab|}{\gcd(a,b)}}.}$

This formula is also valid when exactly i of a and b is 0, since gcd(a, 0) = |a|. However, if both a and b are 0, this formula would cause division by cipher; lcm(0, 0) = 0 is a special case.

There are fast algorithms for calculating the gcd that practise not require the numbers to exist factored, such as the Euclidean algorithm. To return to the example above,

${\displaystyle \operatorname {lcm} (21,half-dozen)={21\cdot 6 \over \gcd(21,6)}={21\cdot 6 \over \gcd(3,half-dozen)}={21\cdot 6 \over three}={\frac {126}{3}}=42.}$

Because gcd(a, b) is a divisor of both a and b, information technology is more than efficient to compute the lcm by dividing before multiplying:

${\displaystyle \operatorname {lcm} (a,b)=\left({|a| \over \gcd(a,b)}\right)\cdot |b|=\left({|b| \over \gcd(a,b)}\correct)\cdot |a|.}$

This reduces the size of ane input for both the division and the multiplication, and reduces the required storage needed for intermediate results (that is, overflow in the a×b computation). Because gcd(a, b) is a divisor of both a and b, the partition is guaranteed to yield an integer, so the intermediate issue tin can exist stored in an integer. Implemented this way, the previous instance becomes:

${\displaystyle \operatorname {lcm} (21,vi)={21 \over \gcd(21,6)}\cdot half-dozen={21 \over \gcd(3,vi)}\cdot half-dozen={21 \over 3}\cdot half dozen=7\cdot 6=42.}$

Using prime factorization [edit]

The unique factorization theorem indicates that every positive integer greater than 1 can exist written in merely one manner equally a production of prime number numbers. The prime numbers tin can exist considered as the atomic elements which, when combined, brand upward a composite number.

For example:

${\displaystyle xc=2^{1}\cdot 3^{ii}\cdot 5^{ane}=2\cdot 3\cdot 3\cdot 5.}$

Hither, the composite number 90 is made up of ane atom of the prime number 2, two atoms of the prime number iii, and one atom of the prime number v.

This fact tin exist used to find the lcm of a set of numbers.

Case: lcm(8,9,21)

Gene each number and limited it every bit a product of prime number number powers.

${\displaystyle {\begin{aligned}8&=two^{three}\\9&=iii^{2}\\21&=three^{ane}\cdot seven^{one}\end{aligned}}}$

The lcm volition be the production of multiplying the highest power of each prime number together. The highest power of the three prime numbers 2, 3, and 7 is 2³, iii², and 7^ane, respectively. Thus,

${\displaystyle \operatorname {lcm} (8,9,21)=ii^{three}\cdot 3^{2}\cdot 7^{1}=8\cdot ix\cdot seven=504.}$

This method is not as efficient equally reducing to the greatest common divisor, since there is no known general efficient algorithm for integer factorization.

The aforementioned method can too be illustrated with a Venn diagram as follows, with the prime factorization of each of the two numbers demonstrated in each circumvolve and all factors they share in common in the intersection. The lcm so can be found past multiplying all of the prime numbers in the diagram.

Here is an case:

48 = two × 2 × 2 × 2 × 3,

180 = 2 × 2 × three × 3 × 5,

sharing 2 "two"s and a "3" in common:

Least mutual multiple = 2 × 2 × ii × two × 3 × 3 × 5 = 720

Greatest mutual divisor = 2 × ii × 3 = 12

This besides works for the greatest mutual divisor (gcd), except that instead of multiplying all of the numbers in the Venn diagram, i multiplies only the prime factors that are in the intersection. Thus the gcd of 48 and 180 is ii × 2 × three = 12.

Using a uncomplicated algorithm [edit]

This method works easily for finding the lcm of several integers.^{[ citation needed ]}

Let there be a finite sequence of positive integers X = (x ₁, ten _ii, ..., 10 _n ), n > 1. The algorithm proceeds in steps equally follows: on each stride m it examines and updates the sequence Ten ^(m) = (10 ₁ ^(m), x _ii ^(m), ..., x _n ⁽¹⁰⁰⁰⁾), X ⁽¹⁾ = X, where 10 ^(m) is the thousandth iteration of X, that is, X at step m of the algorithm, etc. The purpose of the test is to pick the least (possibly, one of many) element of the sequence X ^(m). Assuming x _{k 0} ^(m) is the selected element, the sequence X ^(m+1) is defined as

x _k ^(m+i) = x _k ^(yard), k ≠ k ₀

10 _{k 0} ^(m+1) = x _{m 0} ^(thou) + x _{k 0} ⁽¹⁾.

In other words, the to the lowest degree element is increased by the respective x whereas the rest of the elements pass from Ten ^(m) to Ten ^(thou+one) unchanged.

The algorithm stops when all elements in sequence X ^(m) are equal. Their mutual value L is exactly lcm(X).

For example, if X = X ⁽ⁱ⁾ = (three, 4, 6), the steps in the algorithm produce:

X ⁽²⁾ = (6, 4, 6)

X ⁽ⁱⁱⁱ⁾ = (six, viii, 6)

Ten ^(four) = (6, eight, 12) - by choosing the second 6

X ^(five) = (nine, 8, 12)

X ⁽⁶⁾ = (9, 12, 12)

X ⁽⁷⁾ = (12, 12, 12) so lcm = 12.

Using the table-method [edit]

This method works for whatsoever number of numbers. One begins by list all of the numbers vertically in a table (in this example 4, 7, 12, 21, and 42):

The process begins by dividing all of the numbers past 2. If 2 divides any of them evenly, write 2 in a new column at the summit of the tabular array, and the event of division past ii of each number in the infinite to the right in this new column. If a number is non evenly divisible, only rewrite the number again. If ii does not carve up evenly into any of the numbers, repeat this procedure with the next largest prime number, 3 (see below).

×	2
4	2
7	7
12	half-dozen
21	21
42	21

Now, bold that 2 did divide at least 1 number (every bit in this example), check if 2 divides again:

×	2	2
4	2	one
7	vii	7
12	vi	3
21	21	21
42	21	21

Once 2 no longer divides any number in the current column, echo the procedure by dividing past the next larger prime, iii. One time 3 no longer divides, try the next larger primes, 5 then 7, etc. The process ends when all of the numbers have been reduced to 1 (the column nether the last prime number divisor consists only of 1's).

×	ii	2	3	7
iv	ii	1	one	1
7	seven	7	vii	1
12	6	three	i	1
21	21	21	7	i
42	21	21	7	ane

Now, multiply the numbers in the summit row to obtain the lcm. In this case, information technology is 2 × 2 × 3 × 7 = 84.

As a full general computational algorithm, the above is quite inefficient. One would never desire to implement it in software: information technology takes too many steps and requires too much storage space. A far more efficient numerical algorithm can be obtained by using Euclid'southward algorithm to compute the gcd beginning, and then obtaining the lcm by division.

Formulas [edit]

Fundamental theorem of arithmetic [edit]

Co-ordinate to the cardinal theorem of arithmetic, every integer greater than 1 can be represented uniquely as a product of prime numbers, up to the society of the factors:

${\displaystyle due north=two^{n_{2}}3^{n_{3}}5^{n_{5}}7^{n_{7}}\cdots =\prod _{p}p^{n_{p}},}$

where the exponents due north ₂, due north ₃, ... are not-negative integers; for example, 84 = 2ⁱⁱ 3¹ 5⁰ 7¹ xi⁰ 13⁰ ...

Given two positive integers ${\textstyle a=\prod _{p}p^{a_{p}}}$ and ${\textstyle b=\prod _{p}p^{b_{p}}}$ , their to the lowest degree common multiple and greatest common divisor are given by the formulas

${\displaystyle \gcd(a,b)=\prod _{p}p^{\min(a_{p},b_{p})}}$

and

${\displaystyle \operatorname {lcm} (a,b)=\prod _{p}p^{\max(a_{p},b_{p})}.}$

Since

${\displaystyle \min(ten,y)+\max(x,y)=ten+y,}$

this gives

${\displaystyle \gcd(a,b)\operatorname {lcm} (a,b)=ab.}$

In fact, every rational number tin exist written uniquely as the product of primes, if negative exponents are allowed. When this is washed, the higher up formulas remain valid. For example:

${\displaystyle {\begin{aligned}4&=two^{ii}3^{0},&6&=ii^{1}three^{ane},&\gcd(4,6)&=2^{ane}3^{0}=2,&\operatorname {lcm} (four,half dozen)&=2^{two}3^{one}=12.\\[8pt]{\tfrac {1}{3}}&=2^{0}three^{-one}5^{0},&{\tfrac {2}{5}}&=2^{1}3^{0}5^{-1},&\gcd \left({\tfrac {1}{iii}},{\tfrac {2}{5}}\correct)&=2^{0}3^{-1}5^{-i}={\tfrac {ane}{xv}},&\operatorname {lcm} \left({\tfrac {one}{3}},{\tfrac {two}{5}}\correct)&=2^{1}3^{0}5^{0}=2,\\[8pt]{\tfrac {one}{6}}&=2^{-i}iii^{-1},&{\tfrac {3}{4}}&=2^{-two}3^{1},&\gcd \left({\tfrac {one}{6}},{\tfrac {3}{4}}\right)&=ii^{-ii}3^{-1}={\tfrac {i}{12}},&\operatorname {lcm} \left({\tfrac {1}{half-dozen}},{\tfrac {3}{4}}\right)&=ii^{-1}three^{1}={\tfrac {iii}{2}}.\terminate{aligned}}}$

Lattice-theoretic [edit]

The positive integers may be partially ordered by divisibility: if a divides b (that is, if b is an integer multiple of a) write a ≤ b (or equivalently, b ≥ a). (Note that the usual magnitude-based definition of ≤ is not used here.)

Under this ordering, the positive integers become a lattice, with come across given by the gcd and bring together given by the lcm. The proof is straightforward, if a bit tedious; it amounts to checking that lcm and gcd satisfy the axioms for meet and join. Putting the lcm and gcd into this more general context establishes a duality between them:

If a formula involving integer variables, gcd, lcm, ≤ and ≥ is true, and then the formula obtained past switching gcd with lcm and switching ≥ with ≤ is likewise true. (Remember ≤ is defined equally divides).

The following pairs of dual formulas are special cases of general lattice-theoretic identities.

Commutative laws ${\displaystyle \operatorname {lcm} (a,b)=\operatorname {lcm} (b,a),}$ ${\displaystyle \gcd(a,b)=\gcd(b,a).}$

Associative laws ${\displaystyle \operatorname {lcm} (a,\operatorname {lcm} (b,c))=\operatorname {lcm} (\operatorname {lcm} (a,b),c),}$ ${\displaystyle \gcd(a,\gcd(b,c))=\gcd(\gcd(a,b),c).}$

Absorption laws ${\displaystyle \operatorname {lcm} (a,\gcd(a,b))=a,}$ ${\displaystyle \gcd(a,\operatorname {lcm} (a,b))=a.}$

Idempotent laws ${\displaystyle \operatorname {lcm} (a,a)=a,}$ ${\displaystyle \gcd(a,a)=a.}$

Define divides in terms of lcm and gcd ${\displaystyle a\geq b\iff a=\operatorname {lcm} (a,b),}$ ${\displaystyle a\leq b\iff a=\gcd(a,b).}$

It can as well exist shown^[half-dozen] that this lattice is distributive; that is, lcm distributes over gcd and gcd distributes over lcm:

${\displaystyle \operatorname {lcm} (a,\gcd(b,c))=\gcd(\operatorname {lcm} (a,b),\operatorname {lcm} (a,c)),}$

${\displaystyle \gcd(a,\operatorname {lcm} (b,c))=\operatorname {lcm} (\gcd(a,b),\gcd(a,c)).}$

This identity is self-dual:

${\displaystyle \gcd(\operatorname {lcm} (a,b),\operatorname {lcm} (b,c),\operatorname {lcm} (a,c))=\operatorname {lcm} (\gcd(a,b),\gcd(b,c),\gcd(a,c)).}$

Other [edit]

• Let D be the production of ω(D) distinct prime number numbers (that is, D is squarefree).

Then^[7]

${\displaystyle |\{(x,y)\;:\;\operatorname {lcm} (x,y)=D\}|=3^{\omega (D)},}$

where the absolute bars || announce the cardinality of a set.

• If none of ${\displaystyle a_{one},a_{2},\ldots ,a_{r}}$ is zero, then

${\displaystyle \operatorname {lcm} (a_{1},a_{2},\ldots ,a_{r})=\operatorname {lcm} (\operatorname {lcm} (a_{1},a_{two},\ldots ,a_{r-1}),a_{r}).}$ ^{[8] [ix]}

In commutative rings [edit]

The to the lowest degree common multiple tin can be divers generally over commutative rings as follows: Let a and b be elements of a commutative ring R. A mutual multiple of a and b is an element 1000 of R such that both a and b split grand (that is, there be elements ten and y of R such that ax = one thousand and by = m). A least common multiple of a and b is a mutual multiple that is minimal, in the sense that for whatsoever other common multiple north of a and b, one thousand dividesdue north.

In general, ii elements in a commutative ring can take no least mutual multiple or more i. However, any 2 least mutual multiples of the aforementioned pair of elements are associates. In a unique factorization domain, any two elements have a least common multiple. In a main ideal domain, the least common multiple of a and b can be characterised every bit a generator of the intersection of the ethics generated by a and b (the intersection of a collection of ethics is ever an platonic).

See also [edit]

• Dissonant cancellation
• Coprime integers
• Chebyshev function

Notes [edit]

1. ^ ^{a b c} Weisstein, Eric Due west. "Least Common Multiple". mathworld.wolfram.com . Retrieved 2020-08-30 .
2. ^ Hardy & Wright, § five.1, p. 48
3. ^ ^{a b} Long (1972, p. 39)
4. ^ Pettofrezzo & Byrkit (1970, p. 56)
5. ^ "nasa spacemath" (PDF).
6. ^ The next three formulas are from Landau, Ex. III.3, p. 254
7. ^ Crandall & Pomerance, ex. two.4, p. 101.
8. ^ Long (1972, p. 41)
9. ^ Pettofrezzo & Byrkit (1970, p. 58)

References [edit]

• Crandall, Richard; Pomerance, Carl (2001), Prime Numbers: A Computational Perspective, New York: Springer, ISBN0-387-94777-9
• Hardy, G. H.; Wright, E. M. (1979), An Introduction to the Theory of Numbers (5th edition), Oxford: Oxford University Printing, ISBN978-0-19-853171-5
• Landau, Edmund (1966), Elementary Number Theory, New York: Chelsea
• Long, Calvin T. (1972), Elementary Introduction to Number Theory (2nd ed.), Lexington: D. C. Heath and Company, LCCN 77-171950
• Pettofrezzo, Anthony J.; Byrkit, Donald R. (1970), Elements of Number Theory, Englewood Cliffs: Prentice Hall, LCCN 77-81766

bayquileste.blogspot.com

Source: https://en.wikipedia.org/wiki/Least_common_multiple

Share this post