Categories

## Problem based on LCM | AMC 8, 2016 | Problem 20

Try this beautiful problem from Algebra based on LCM from AMC-8, 2016.

## Problem based on LCM – AMC 8, 2016

The least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$. What is the least possible value of the least common multiple of $a$ and $c$?

• $26$
• $20$
• $28$

### Key Concepts

Algebra

Divisor

multiplication

AMC-8, 2016 problem 20

Challenges and Thrills of Pre College Mathematics

## Try with Hints

We have to find out the least common multiple of $a$ and $c$.if you know the value of $a$ and $c$ then you can easily find out the required LCM. Can you find out the value of $a$ and $c$?

Can you now finish the problem ……….

Given that the least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$ .then b must divide 12 and 15. There is only one possibility that b=3 which divide 12 and 15. therefore $a$=$\frac{12}{3}=4$

can you finish the problem……..

so$b$=3. Given that LCM of $b$ and $c$ is 15. Therefore c=5

Now lcm of $a$ and $c$ that is lcm of 4 and 5=20

Categories

## Linear Equations | AMC 8, 2007 | Problem 20

Try this beautiful problem from Algebra based on Linear equations from AMC-8, 2007.

## Linear equations – AMC 8, 2007

Before the district play, the Unicorns had won $45$% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?

• $40$
• $48$
• $58$

### Key Concepts

Algebra

linear equation

multiplication

AMC-8, 2007 problem 20

Challenges and Thrills of Pre College Mathematics

## Try with Hints

At first, we have to Calculate the number of won games and lost games. Unicorns had won $45$% of their basketball game.so we may assume that out of 20 unicorns woned 9.

Can you now finish the problem ……….

Next unicorns won six more games and lost two.so find out the total numbers of won game and total numbers of games i.e won=$9x+6$ and the total number of games become $20x+8$

can you finish the problem……..

Given that Unicorns had won $45$% of their basketball games i.e $\frac{45}{100}=\frac{9}{20}$

During district play, they won six more games and lost two,

Therefore they won$9x+6$ and the total number of games becomes $20x+8$

According to the question, Unicorns finish the season having won half their games. …

Therefore,$\frac{9x+6}{20x+8}=\frac{1}{2}$

$\Rightarrow 18x+12=20x+8$

$\Rightarrow 2x=4$

$\Rightarrow x=2$

Total number of games becomes $20x+8$ =$(20 \times 2) +8=48$

Categories

## Divisibility | AMC 8, 2014 |Problem 21

Try this beautiful problem from Algebra based on multiplication and divisibility of two given numbers.

## Multiplication and Divisibility- AMC 8, 2014

The  7-digit numbers 74A52B1 ana 326AB4C are each multiples of 3.which of the following could be the value of c ?

• 1
• 2
• 3

### Key Concepts

Algebra

Division algorithm

Integer

AMC-8, 2014 problem 21

Challenges and Thrills of Pre College Mathematics

## Try with Hints

Use the rules of Divisibility ……..

Can you now finish the problem ……….

If both numbers are divisible by 3 then the sum of their digits has to be divisible by 3……

can you finish the problem……..

Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 7 + 4 + 5 + 2 + 1 = 19. In order to be a multiple of 3, A + B has to be either 2 or 5 or 8… and so on. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add two of the selected values, 5 to 15, to get 20. We then see that C = 1, 4 or 7, 10… and so on, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7… and so on. In order to be a multiple of three, we select a few of the common numbers we got from both these equations, which could be 1, 4, and 7. However, in the answer choices, there is no 7 or 4 or anything greater than 7, but there is a 1. so the answer is 1

Categories

## LCM – AMC 8, 2016 – Problem 20

The least common multiple of a and b is 12 .and the lest common multiple of b and c is 15.what is the least possible value of the least common multiple of a and c?

• 30
• 60
• 20

### Key Concepts

Algebra

Division algorithm

Integer

AMC-8, 2016 problem 20

Challenges and Thrills of Pre College Mathematics

## Try with Hints

Find greatest common factors

Can you now finish the problem ……….

Find Least common multiple….

can you finish the problem……..

we wish to find possible values of a,b and c .By finding the greatest common factor 12 and 15, algebrically ,it’s some multiple of b and from looking at the numbers ,we are sure that it is 3.Moving on to a and c ,in order to minimize them,we wish to find the least such that the LCM of a and 3 is 12,$\to 4$.similarly with 3 and c,we obtain 5.the LCM of 4 and 5 is 20 .