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Numbers - Factor and Multiple

For COMPETITION
Number of Total Problems: 21.
FOR PRINT ::: (Book)

Problem Num : 1

From : NCTM

Type: None

Section:Numbers

Theme:None
Adjustment# :

Difficulty: 1

Category Factor and Multiple

Analysis

Solution/Answer

Problem Num : 2

From : NCTM

Type: None

Section:Numbers

Theme:None
Adjustment# :

Difficulty: 1

Category Factor and Multiple

Analysis

Solution/Answer

Problem Num : 3

From : NCTM

Type: None

Section:Numbers

Theme:None
Adjustment# :

Difficulty: 1

Category Factor and Multiple

Analysis

Solution/Answer

Problem Num : 4

From : AMC10B

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
Three positive integers are each greater than $1$ , have a product of $27000$ , and are pairwise relatively prime. What is their sum?
$extbf{(A)} 100qquad extbf{(B)} 137qquad extbf{(C)} 156qquad extbf{(D)}} 160qquad extbf{(E)} 165$
'

Category Factor and Multiple

Analysis

Solution/Answer
The prime factorization of $27000$ is $2^3*3^3*5^3$ . These three factors are pairwise relatively prime, so the sum is $2^3+3^3+5^3=8+27+125=$ $oxed{ extbf{(D) }160}$

Answer:

Problem Num : 5

From : NCTM

Type: Complex

Section:Numbers

Theme:Manipulation
Adjustment# : 0

Difficulty: 1

Category Factor and Multiple

Analysis

Solution/Answer

Problem Num : 6

From : AMC10B

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
For how many positive integers $n$ is $n^2 - 3n + 2$ a prime number?
$mathrm{(A)} ext{none}qquadmathrm{(B)} ext{one}qquadmathrm{(C)} ext{two}qquadmathrm{(D)} ext{more than tw...$
'

Category Factor and Multiple

Analysis

Solution/Answer
Factoring, we get $n^2 - 3n + 2 = (n-2)(n-1)$ . Exactly $1$ of $n-2$ and $n-1$ must be $1$ and the other a prime number. If $n-1=1$ , then $n-2=0$ , and $1 imes0=0$ , which is not prime. On the other hand, if $n-2=1$ , then $n-1=2$ , and $1 imes2=2$ , which is a prime number. The answer is $oxed{mathrm{(B)} ext{one}}$ .

Answer:

Problem Num : 7

From : AMC10B

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
Suppose that $n$ is the product of three consecutive integers and that $n$ is divisible by $7$ . Which of the following is not necessarily a divisor of $n$ ?
$extbf{(A)} 6 qquad extbf{(B)} 14 qquad extbf{(C)} 21 qquad extbf{(D)} 28 qquad extbf{(E)} 42$
'

Category Factor and Multiple

Analysis

Solution/Answer
Whenever $n$ is the product of three consecutive integers, $n$ is divisible by $3!$ , meaning it is divisible by $6$ .
It also mentions that it is divisible by $7$ , so the number is definitely divisible by all the factors of $42$ .
In our answer choices, the one that is not a factor of $42$ is $oxed{ extbf{(D)} 28}$ .

Answer:

Problem Num : 8

From : AMC10B

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
Suppose that $n$ is the product of three consecutive integers and that $n$ is divisible by $7$ . Which of the following is not necessarily a divisor of $n$ ?
$extbf{(A)} 6 qquad extbf{(B)} 14 qquad extbf{(C)} 21 qquad extbf{(D)} 28 qquad extbf{(E)} 42$
'

Category Factor and Multiple

Analysis

Solution/Answer
Whenever $n$ is the product of three consecutive integers, $n$ is divisible by $3!$ , meaning it is divisible by $6$ .
It also mentions that it is divisible by $7$ , so the number is definitely divisible by all the factors of $42$ .
In our answer choices, the one that is not a factor of $42$ is $oxed{ extbf{(D)} 28}$ .

Answer:

Problem Num : 9

From : AMC10B

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
Which of the following numbers is a perfect square?
$mathrm{(A) } 98! cdot 99! qquad mathrm{(B) } 98! cdot 100! qquad mathrm{(C) } 99! cdot 100! qquad mathrm{(D) ...$
'

Category Factor and Multiple

Analysis

Solution/Answer
Using the fact that $n! = ncdot (n-1)!$ , we can write:

$A=98! cdot (99cdot 98!) = 99 cdot (98!)^2 = 9cdot 11cdot(98!)^2$
$B=100 cdot 99 cdot (98!)^2 = 9cdot 11cdot (10cdot 98!)^2$
$C=100cdot (99!)^2 = (10cdot 99!)^2$
$D=101cdot 100cdot (99!)^2 = 101 cdot(10cdot 99!)^2$
$E=101cdot (100!)^2$

Clearly $oxed{mathrm{(C) } 99! cdot 100!}$ is a square, and as $9$ , $11$ , and $101$ are primes, none of the other four are squares.

Answer:

Problem Num : 10

From : AMC10

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
What is the probability that a randomly drawn positive factor of $60$ is less than $7$ ?
$mathrm{(A) } frac{1}{10}qquad mathrm{(B) } frac{1}{6}qquad mathrm{(C) } frac{1}{4}qquad mathrm{(D) } frac{...$
'

Category Factor and Multiple

Analysis

Solution/Answer
Solution 1

For a positive number $n$ which is not a perfect square, exactly half of the positive factors will be less than $sqrt{n}$ .
Since $60$ is not a perfect square, half of the positive factors of $60$ will be less than $sqrt{60}approx 7.746$ .
Clearly, there are no positive factors of $60$ between $7$ and $sqrt{60}$ .
Therefore half of the positive factors will be less than $7$ .
So the answer is $oxed{mathrm{(E)} frac{1}{2}}$ .

Solution 2

Testing all numbers less than $7$ , numbers $1, 2, 3, 4, 5$ , and $6$ divide $60$ . The prime factorization of $60$ is $2^2cdot 3 cdot 5$ . Using the formula for the number of divisors, the total number of divisors of $60$ is $(3)(2)(2) = 12$ . Therefore, our desired probability is $frac{6}{12} = oxed{mathrm{(E)} frac{1}{2}}$

Answer:

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