We read your photo as:
Find the LCM of the given numbers.
Mathematics
ScanSolve AI Answer
Step-by-step solution
16 steps1. Find the LCM of the given numbers.
- i) 12 and 30
Find the prime factorization of each number.
$$ 12 = 2^2 \times 3 $$ $$ 30 = 2 \times 3 \times 5 $$Multiply the highest powers of all prime factors.
$$ \text{LCM}(12, 30) = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = \boxed{\text{60}} $$- ii) 16 and 50
Find the prime factorization of each number.
$$ 16 = 2^4 $$ $$ 50 = 2 \times 5^2 $$Multiply the highest powers of all prime factors.
$$ \text{LCM}(16, 50) = 2^4 \times 5^2 = 16 \times 25 = \boxed{\text{400}} $$- iii) 18, 20 and 36
Find the prime factorization of each number.
$$ 18 = 2 \times 3^2 $$ $$ 20 = 2^2 \times 5 $$ $$ 36 = 2^2 \times 3^2 $$Multiply the highest powers of all prime factors.
$$ \text{LCM}(18, 20, 36) = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = \boxed{\text{180}} $$- iv) 55, 70 and 77
Find the prime factorization of each number.
$$ 55 = 5 \times 11 $$ $$ 70 = 2 \times 5 \times 7 $$ $$ 77 = 7 \times 11 $$Multiply the highest powers of all prime factors.
$$ \text{LCM}(55, 70, 77) = 2 \times 5 \times 7 \times 11 = \boxed{\text{770}} $$- v) 45, 60, 90 and 120
Find the prime factorization of each number.
$$ 45 = 3^2 \times 5 $$ $$ 60 = 2^2 \times 3 \times 5 $$ $$ 90 = 2 \times 3^2 \times 5 $$ $$ 120 = 2^3 \times 3 \times 5 $$Multiply the highest powers of all prime factors.
$$ \text{LCM}(45, 60, 90, 120) = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = \boxed{\text{360}} $$2. Find the smallest number which on adding 11 becomes exactly divisible by 24, 36 and 60.
Let the required number be .
If is exactly divisible by 24, 36, and 60, then must be the LCM of these numbers.
Find the LCM of 24, 36, and 60.
$$ 24 = 2^3 \times 3 $$ $$ 36 = 2^2 \times 3^2 $$ $$ 60 = 2^2 \times 3 \times 5 $$ $$ \text{LCM}(24, 36, 60) = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360 $$Set up the equation and solve for .
$$ N + 11 = 360 $$ $$ N = 360 - 11 $$ $$ N = \boxed{\text{349}} $$3. Find the smallest number which leaves a remainder 7 when divided by 20, 50 and 75.
Let the required number be .
If leaves a remainder 7 when divided by 20, 50, and 75, then must be exactly divisible by 20, 50, and 75.
Find the LCM of 20, 50, and 75.
$$ 20 = 2^2 \times 5 $$ $$ 50 = 2 \times 5^2 $$ $$ 75 = 3 \times 5^2 $$ $$ \text{LCM}(20, 50, 75) = 2^2 \times 3 \times 5^2 = 4 \times 3 \times 25 = 300 $$Set up the equation and solve for .
$$ N - 7 = 300 $$ $$ N = 300 + 7 $$ $$ N = \boxed{\text{307}} $$
Was this helpful?
Still stuck on this one?
Ask a follow-up, and the answer starts from this question.
Got a different question?Ask your own question
More Mathematics Questions
QFind the LCM & HCF for the given numbers.QCalculate the LCM of two numbers and add/subtract fractions.QFind the Least Common Multiple (LCM) of the given algebraic expressions and numbers.QFind the LCM of 2,3,5 starting from 1 to 20QFind the LCM of 5,8 and 9
Browse by topic:Number Theory

Handwritten Step-by-Step Solution
Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.
Try on WhatsApp