Categories
AMC 8 Math Olympiad USA Math Olympiad

Ratio of LCM & GCF | Algebra | AMC 8, 2013 | Problem 10

Try this beautiful problem from Algebra based on the ratio of LCM & GCF from AMC-8, 2013.

Ratio of LCM & GCF | AMC-8, 2013 | Problem 10


What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?

  • 310
  • 330
  • 360

Key Concepts


Algebra

Ratio

LCM & GCF

Check the Answer


Answer:$330$

AMC-8, 2013 problem 10

Challenges and Thrills in Pre College Mathematics

Try with Hints


We have to find out the ratio of least common multiple and greatest common factor of 180 and 594. So at first, we have to find out prime factors of 180 & 594. Now…….

\(180=3^2\times 5 \times 2^2\)

\(594=3^3 \times 11 \times 2\)

Can you now finish the problem ……….

Now lcm of two numbers i.e multiplications of the greatest power of all the numbersĀ 

Therefore LCM of 180 & 594=\(3^3\times2^2 \times 11 \times 5\)=\(5940\)

For the GCF of 180 and 594, multiplications of the least power of all of the numbers i.e \(3^2\times 2\)=\(18\)

can you finish the problem……..

Therefore the ratio of Lcm & gcf of 180 and 594 =\(\frac{5940}{18}\)=\(330\)

Subscribe to Cheenta at Youtube


Categories
Algebra AMC 8 Math Olympiad

Unit digit | Algebra | AMC 8, 2014 | Problem 22

Try this beautiful problem from Algebra about unit digit from AMC-8, 2014.

Unit digit | AMC-8, 2014|Problem 22


A $2$-digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the unit digit of the number?

  • 7
  • 9
  • 5

Key Concepts


Algebra

Multiplication

integer

Check the Answer


Answer:$9$

AMC-8, 2014 problem 22

Challenges and Thrills in Pre College Mathematics

Try with Hints


Let the ones digit place be y and ten’s place be x

Therefore the number be \(10x+y\)

Can you now finish the problem ……….

Given that the product of the digits plus the sum of the digits is equal to the number

can you finish the problem……..

Let the ones digit place be y and ten’s place be x

Therefore the number be \(10x+y\)

Now the product of the digits=\(xy\)

Given that the product of the digits plus the sum of the digits is equal to the number

Therefore \(10x+y=(x\times y)+(x+y)\)

\(\Rightarrow 9x=xy\)

\(\Rightarrow y=9\)

Therefore the unit digit =\(y\)=9

Subscribe to Cheenta at Youtube