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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.