# Worksheet on L.C.M of Monomials | Least Common Multiple of Monomials Worksheet

Practice will improve your knowledge and gives you a grip on a subject. To become perfect in maths, you must have perfect practice. So, to overcome all troubles to learn math concepts like the least common multiple of two or more monomials, we are providing the Worksheet on L.C.M. of Monomials here.

To find the L.C.M of Monomials, you have to find the factors of monomials and get the L.C.M of numerical coefficients and literal coefficients. Multiply the common factors and extra common factors of numerical and literal coefficients to get the result. Find the Worksheets on Lowest Common Multiple of Monomials and practice them for effective learning.

1. Find the lowest common multiple (L.C.M.) of the two monomials:

(a) 12a³bc, 22ab³

(b) 11u²v, 121u³v

(c) 81p⁵qr²s³t, 12pq⁴s⁵

(d) 117mn, 65m³no

(e) x²y⁴z, x³y²z

Solution:

(a) 12a³bc, 22ab³

The L.C.M. of numerical coefficients = The L.C.M. of 12 and 22.

Since, 12=2x2x3, 22=2×11

Therefore, the L.C.M. of 12 and 22 is 2x2x3x11 = 132

The L.C.M. of literal coefficients = The L.C.M. of a³bc, ab³ = a³b³c

Since, in a³bc, ab³,

The highest power of a is a³.

The highest power of b is b³.

The highest power of c is c.

Thus, the L.C.M. of 12a³bc, 22ab³

= The L.C.M. of numerical coefficients × The L.C.M. of literal coefficients

= 132 × (a³b³c) = 132a³b³c

(b) 11u²v, 121u³v

The L.C.M of numerical coefficients = The lowest common multiple of 11, 121

Since, 11=1×11, 121=11×11

The lowest common multiple of 11, 121 = 11²x1 = 121

The L.C.M. of literal coefficients = The L.C.M. of u²v, u³v = u³v

Since, in u²v, u³v

The highest power of u is u³.

The highest power of v is v.

Thus, L.C.M of 11u²v, 121u³v = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 121 x u³v = 121u³v.

(c) 81p⁵qr²s³t, 12pq⁴s⁵

The Least Common Multiple of numerical coefficients = L.C.M of 81, 12

Since, 81=3x3x3x3, 12=3×4

Therefore, the L.C.M. of 81, 12 = 3⁴ x 4 = 3x3x3x3x4 = 324

The lowest common multiple of literal coefficients = L.C.M of p⁵qr²s³t, pq⁴s⁵ = p⁵q⁴r²s⁵t

Since, in p⁵qr²s³t, pq⁴s⁵

The highest power of p is p⁵.

The highest power of q is q⁴.

The highest power of r is r².

The highest power of s is s⁵.

The highest power of t is t.

Thus, L.C.M of 81p⁵qr²s³t, 12pq⁴s⁵ = L.C.M. of numerical coefficients × L.C.M. of literal coefficients

= 324 x p⁵q⁴r²s⁵t = 324p⁵q⁴r²s⁵t.

(d) 117mn, 65m³no

The least common multiple of numerical coefficients = l.c.m of 117, 65

Since, 117=13x3x3, 65=13×5

The l.C.M of 117, 65 = 13×3²x5 = 13 x 3 x 3x 5 = 585

The Lowest common multiple of literal coefficients = L.C.M of mn, m³no = m³no

Since, in mn, m³no

The highest power of m is m³.

The highest power of n is n.

The highest power of o is o.

Thus, the L.C.M of 117mn, 65m³no = L.C.M. of numerical coefficients × L.C.M. of literal coefficients

= 585 x m³no = 585m³no.

(e) x²y⁴z, x³y²z

The L.C.M of literal coefficients = Lowest common multiple of x²y⁴z, x³y²z = x³y⁴z

Since, in x²y⁴z, x³y²z

The highest power of x is x³.

The highest power of y is y⁴.

The highest power of z is z.

The least common multiple of numerical coefficients = 1.

Therefore, L.C.M of x²y⁴z, x³y²z = L.C.M. of numerical coefficients × L.C.M. of literal coefficients

= 1 x x³y⁴z = x³y⁴z.

2. Find the least common multiple of three monomials:

(a) 14xy²z, 10xy, 2xyz³

(b) 65a²b⁵c, 10ab, 5ab²

(c) 18pqr, 30pqr, 5pqr

(d) mn²o, m²n, op³

(e) 9u²v, 5uv³, 4u³v

Solution:

(a) 14xy²z, 10xy, 2xyz³

The L.C.M of numerical coefficients = L.C.M of 14, 10, 2

Since, 14=2×7, 10=2×5, 2=2×1

Thus, Least common multiple of 14, 10, 2 = 2x7x5x1 = 60

The L.C.M of literal coefficients = L.C.M of xy²z, xy, xyz³ = xy²z³

Since, in xy²z, xy, xyz³

The highest power of x is x.

The highest power of y is y².

The highest power of z is z³.

Therefore, the Lowest common multiple of 14xy²z, 10xy, 2xyz³ = L.C.M. of numerical coefficients × L.C.M. of literal coefficients

= 60 x xy²z³ = 60xy²z³.

(b) 65a²b⁵c, 10ab, 5ab²

The L.C.M of numerical coefficients = L.C.M of 65, 10, 5

Since, 65=5×13, 10=2×5, 5=1×5.

Thus, lowest common multiple of 65, 10, 5 = 5x13x2 = 130

The least common multiple of literal coefficients =L.C.M of a²b⁵c, ab, ab² = a²b⁵c

The highest power of a is a².

The highest power of b is b⁵.

The highest power of c is c.

Therefore, L.C.M of 65a²b⁵c, 10ab, 5ab² = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 130 x a²b⁵c = 130a²b⁵c.

(c) 18pqr, 30pqr, 5pqr

The L.C.M of numerical coefficients = L.C.M of 18, 30, 5

Since, 18=2x3x3, 30=3x2x5, 5=1×5

L.C.M of 18, 30, 5 = 2x3x3x5 = 90

The L.C.M of literal coefficients = L.C.M of pqr, pqr, pqr = pqr.

Since in, pqr, pqr, pqr

The highest power of p is p.

The highest power of q is q.

The highest power of r is r.

Therefore, L.C.M of 18pqr, 30pqr, 5pqr = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 90 x pqr = 90pqr

(d) mn²o, m²n, op³

The L.C.M of literal coefficients = Lowest common multiple of mn²o, m²n, op³ = m²n²op³

Since in, mn²o, m²n, op³

The highest power of m is m².

The highest power of n is n².

The highest power of o is o.

The highest power of p is p³.

The L.C.M of numerical coefficients = 1.

Therefore, L.C.M of mn²o, m²n, op³ = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 1 x m²n²op³ = m²n²op³.

(e) 9u²v, 5uv³, 4u³v

The L.C.M of numerical coefficients = least common multiple of 9, 5, 4

Since, 9=3×3, 5=1×5, 4=2×2

Thus, L.C.M of 9, 5, 4 = 3²x5x2² = 180

The L.C.M of literal coefficients =L.C.M of u²v, uv³, u³v = u³v³

Since in, u²v, uv³, u³v,

The highest power of u is u³.

The highest power of v is v³.

Therefore, L.C.M of 9u²v, 5uv³, 4u³v = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 180 x u³v³ = 180u³v³.

3. Calculate the L.C.M of the following monomials:

(a) 32y⁴, 40xy², 20x²y²

(b) 27u², 18u

(c) 3x⁴y⁴z³, 5x²y³z⁴

(d) 2pq⁵r³s, 21p³qrs, 4pq²rs², 9p²s

(e) 36mn², 27m³no², 8mo

Solution:

(a) 32y⁴, 40xy², 20x²y²

The L.C.M of numerical coefficients = L.C.M of 32, 40, 20

Since, 32=2x2x2x4, 40=2x4x5, 20=4×5

Thus, L.C.M of 32, 40, 20 = 2x8x2x5x2 = 320

The L.C.M of literal coefficients = least common multiple of y⁴, xy², x²y² = x²y⁴

Since in, y⁴, xy², x²y²

The highest power of y is y⁴.

The highest power of x is x².

Therefore, L.C.M of 32y⁴, 40xy², 20x²y²= L.C.M of numerical coefficients x L.C.M of literal coefficients

= 320 x x²y⁴ = 320x²y⁴.

(b) 27u², 18u

The L.C.M of numerical coefficients = L.C.M of 27, 18

Since, 27=3x3x3, 18=3x3x2

Thus, L.C.M of 27, 18 = 3x3x3x2 = 56

The L.C.M of literal coefficients = lowest common multiple of u², u = u²

Since in, u², u

The highest power of u is u²

Therefore, L.C.M of 27u², 18u= L.C.M of numerical coefficients x L.C.M of literal coefficients

= 56 x u² = 56u²

(c) 3x⁴y⁴z³, 5x²y³z⁴

The L.C.M of numerical coefficients = L.C.M of 3, 5

Since, 3=1×3, 5=1×5

Thus, L.C.M of 3,5 = 3×5 = 15

The least common multiple of literal coefficients = L.C.M of x⁴y⁴z³, x²y³z⁴ = x⁴y⁴z⁴

Since in, x⁴y⁴z³, x²y³z⁴

The highest power of x is x⁴

The highest power of y is y⁴

The highest power of z is z⁴

Therefore, L.C.M of 3x⁴y⁴z³, 5x²y³z⁴= L.C.M of numerical coefficients x L.C.M of literal coefficients

= 15 x x⁴y⁴z⁴ = 15x⁴y⁴z⁴.

(d) 2pq⁵r³s, 21p³qrs, 4pq²rs², 9p²s

The L.C.M of numerical coefficients = L.C.M of 2, 21, 4, 9

Since, 2=2×1, 21=3×7, 4=2×2, 9=3×3

Thus, L.C.M of 2, 21, 4, 9 = 2x3x7x2x3 = 252

The L.C.M of literal coefficients = least common multiple of pq⁵r³s, p³qrs, pq²rs², p²s = p³q⁵r³s²

Since in, pq⁵r³s, p³qrs, pq²rs², p²s

The highest power of p is p³

The highest power of q is q⁵

The highest power of r is r³

The highest power of s is s²

Therefore, L.C.M of 2pq⁵r³s, 21p³qrs, 4pq²rs², 9p²s= L.C.M of numerical coefficients x L.C.M of literal coefficients

= 252 x p³q⁵r³s² = 252p³q⁵r³s².

(e) 36mn², 27m³no², 8mo

The L.C.M of numerical coefficients = L.C.M of 36, 27, 8

Since, 27=3x3x3, 36=2x2x3x3, 8=2x2x2

Thus, L.C.M of 36, 27, 8 = 3x3x3x2x2x2 = 192

The L.C.M of literal coefficients = L.C.M of mn², m³no², mo = m³n²o²

Since in, mn², m³no², mo

The highest power of m is m³.

The highest power of n is n²

The highest power of o is o².

Therefore, L.C.M of 36mn², 27m³no², 8mo= L.C.M of numerical coefficients x L.C.M of literal coefficients

= 192 x m³n²o² = 192m³n²o²

4. Find the L.C.M of each:

(a) 3a²bc²de, 5a³bc⁴de, 7a²bcde², 13ab³d²

(b) 12p²q²r, 14qs³t², 16u²tv

(c) 5x³y⁴, 10xz, 15xy⁴z², 20x²y², 25x³

Solution:

(a) 3a²bc²de, 5a³bc⁴de, 7a²bcde², 13ab³d²

The L.C.M of numerical coefficients = L.C.M of 3, 5, 7, 13

Since, 3=1×3, 5=1×5, 7=1×7, 1=1×13

Thus, Least common multiple of 3, 5, 7, 13 = 13x5x3x7 = 1365

The lowest common multiple of literal coefficients = L.C.M of a²bc²de, a³bc⁴de, a²bcde², ab³d² = a³b³c⁴d²e²

Since, in a²bc²de, a³bc⁴de, a²bcde², ab³d²

The highest power of a is a³

The highest power of b is b³

The highest power of c is c⁴

The highest power of d is d²

The highest power of e is e².

Therefore, least common multiple of 3a²bc²de, 5a³bc⁴de, 7a²bcde², 13ab³d² = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 1365 x a³b³c⁴d²e² = 1365a³b³c⁴d²e².

(b) 12p²q²r, 14qs³t², 16u²tv

The L.C.M of numerical coefficients = L.C.M of 12, 14, 16

Since, 12=2x3x2, 14=2×7, 16=2x2x2x2

Thus, least common multiple of 12, 14, 16 = 2x2x2x2x3x7x4 = 336

The L.C.M of literal coefficients = L.C.M of p²q²r, qs³t², u²tv = p²q²rs³t²v

Since, in p²q²r, qs³t², u²tv

The highest power of p is p²

The highest power of q is q²

The highest power of r is r

The highest power of s is s³

The highest power of t is t²

The highest power of v is v

Therefore, L.C.M of 12p²q²r, 14qs³t², 16u²tv = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 336 x p²q²rs³t²v = 336p²q²rs³t²v

(c) 5x³y⁴, 10xz, 15xy⁴z², 20x²y², 25x³

The L.C.M of numerical coefficients = L.C.M of 5, 10, 15, 20, 25

Since, 5=1×5, 10=2×5, 15=3×5, 20=4×5, 25=5×5

Thus, L.C.M of 5, 10, 15, 20, 25 = 5x2x3x4x5 = 600

The L.C.M of literal coefficients = L.C.M of x³y⁴, xz, xy⁴z², x²y², x³ = x³y⁴z²

Since in x³y⁴, xz, xy⁴z², x²y², x³

The highest power of x is x³

The highest power of y is y⁴

The highest power of z is z²

Therefore, L.C.M of 5x³y⁴, 10xz, 15xy⁴z², 20x²y², 25x³ = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 600 x x³y⁴z² = 600x³y⁴z².