Categories

## Combinations | AIME I, 2009 |Problem 9

Try this problem from American Invitational Mathematics Examination, AIME, 2019 based on Combinations

## Combinations- AIME, 2009

A game show offers a contestant three prizes A B and C each of which is worth a whole number of dollars from $1 to$9999 inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A B and C. As a hint the digits of three prizes are given. On a particular day the digits given were 1,1,1,1,3,3,3. Find the total number of possible guesses for all three prizes consistent with the hint.

• 110
• 420
• 430
• 111

### Key Concepts

Combinations

Theory of equations

Polynomials

## Check the Answer

Answer: 420.

AIME I, 2009, Problem 9

Combinatorics by Brualdi .

## Try with Hints

Number of possible ordering of seven digits is$\frac{7!}{4!3!}$=35

these 35 orderings correspond to 35 seven-digit numbers, and the digits of each number can be subdivided to represent a unique combination of guesses for A B and C. Thus, for a given ordering, the number of guesses it represents is the number of ways to subdivide the seven-digit number into three nonempty sequences, each with no more than four digits. These subdivisions have possible lengths 1/2/4,2/2/3,1/3/3, and their permutations. The first subdivision can be ordered in 6 ways and the second and third in three ways each, for a total of 12 possible subdivisions.

then total number of guesses is 35.12=420

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

## Check the Answer

Answer:20

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 .