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Numbers - Factor and Multiple

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

Problem Num : 11

From : AMC10

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?

$mathrm{(A) } 2qquad mathrm{(B) } 3qquad mathrm{(C) } 4qquad mathrm{(D) } 5qquad mathrm{(E) } 6$
'

Category Factor and Multiple

Analysis

Solution/Answer
Solution 1

Let the squares be labeled $A$ , $B$ , $C$ , and $D$ .
When the polygon is folded, the "right" edge of square $A$ becomes adjacent to the "bottom edge" of square $C$ , and the "bottom" edge of square $A$ becomes adjacent to the "bottom" edge of square $D$ .
So, any "new" square that is attatched to those edges will prevent the polygon from becoming a cube with one face missing.
Therefore, squares $1$ , $2$ , and $3$ will prevent the polygon from becoming a cube with one face missing.
Squares $4$ , $5$ , $6$ , $7$ , $8$ , and $9$ will allow the polygon to become a cube with one face missing when folded.
Thus the answer is $oxed{mathrm{(E)} 6}$ .

Solution 2

Another way to think of it is that a cube missing one face has $5$ of its $6$ faces. Since the shape has $4$ faces already, we need another face. The only way to add another face is if the added square does not overlap any of the others. $1$ , $2$ , and $3$ overlap, while squares $4$ to $9$ do not. The answer is $oxed{mathrm{(E)} 6}$
Answer:

Problem Num : 12

From : AMC10

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
Let $n$ be the largest integer that is the product of exactly 3 distinct prime numbers $d$ , $e$ , and $10d+e$ , where $d$ and $e$ are single digits. What is the sum of the digits of $n$ ?
$mathrm{(A) } 12qquad mathrm{(B) } 15qquad mathrm{(C) } 18qquad mathrm{(D) } 21qquad mathrm{(E) } 24$
'

Category Factor and Multiple

Analysis

Solution/Answer
Since $d$ is a single digit prime number, the set of possible values of $d$ is ${2,3,5,7}$ .
Since $e$ is a single digit prime number and is the units digit of the prime number $10d+e$ , the set of possible values of $e$ is ${3,7}$ .
Using these values for $d$ and $e$ , the set of possible values of $10d+e$ is ${23,27,33,37,53,57,73,77}$
Out of this set, the prime values are ${23,37,53,73}$
Therefore the possible values of $n$ are:
$2cdot3cdot23=138$
$3cdot7cdot37=777$
$5cdot3cdot53=795$
$7cdot3cdot73=1533$
The largest possible value of $n$ is $1533$ .
So, the sum of the digits of $n$ is $1+5+3+3=12 Rightarrow A$

Answer:

Problem Num : 13

From : AMC10

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
What is the probability that an integer in the set ${1,2,3,...,100}$ is divisible by $2$ and not divisible by $3$ ?
$mathrm{(A) } frac{1}{6}qquad mathrm{(B) } frac{33}{100}qquad mathrm{(C) } frac{17}{50}qquad mathrm{(D) } ...$
'

Category Factor and Multiple

Analysis

Solution/Answer
There are $100$ integers in the set.
Since every 2nd integer is divisible by $2$ , there are $lfloorfrac{100}{2} floor=50$ integers divisible by $2$ in the set.
To be divisible by both $2$ and $3$ , a number must be divisible by $lcm(2,3)=6$ .
Since every 6th integer is divisible by $6$ , there are $lfloorfrac{100}{6} floor=16$ integers divisible by both $2$ and $3$ in the set.
So there are $50-16=34$ integers in this set that are divisible by $2$ and not divisible by $3$ .
Therefore, the desired probability is $frac{34}{100}=frac{17}{50} Rightarrow C$

Controversy

Due to the wording of the question, it may be taken as "Find the probability that an integer in said set is divisible by 2 and not 3 EXISTS". One example would be 2, which is not a multiple of 3, thus the probability is 1.

Answer:

Problem Num : 14

From : NCTM

Type: Understanding

Section:Numbers

Theme:None
Adjustment# : 0

Difficulty: 1

Category Factor and Multiple

Analysis

Solution/Answer

Problem Num : 15

From : NCTM

Type: Understanding

Section:Numbers

Theme:Integer Property
Adjustment# : 0

Difficulty: 1

Category Factor and Multiple

Analysis

Solution/Answer

Problem Num : 16

From : NCTM

Type: None

Section:Numbers

Theme:None
Adjustment# : 0

Difficulty: 2

Category Factor and Multiple

Analysis

Solution/Answer

Problem Num : 17

From : AMC10

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
How many positive cubes divide $3! cdot 5! cdot 7!$ ?
$mathrm{(A) } 2qquad mathrm{(B) } 3qquad mathrm{(C) } 4qquad mathrm{(D) } 5qquad mathrm{(E) } 6$
'

Category Factor and Multiple

Analysis

Solution/Answer
Solution 1

$3! cdot 5! cdot 7! = (3cdot2cdot1) cdot (5cdot4cdot3cdot2cdot1) cdot (7cdot6cdot5cdot4cdot3cdot2cdot1) = 2^{8...$
Therefore, a perfect cube that divides $3! cdot 5! cdot 7!$ must be in the form $2^{a}cdot3^{b}cdot5^{c}cdot7^{d}$ where $a$ , $b$ , $c$ , and $d$ are nonnegative multiples of $3$ that are less than or equal to $8$ , $4$ , $2$ and $1$ , respectively.
So:
$ain{0,3,6}$ ( $3$ posibilities)
$bin{0,3}$ ( $2$ posibilities)
$cin{0}$ ( $1$ posibility)
$din{0}$ ( $1$ posibility)

So the number of perfect cubes that divide $3! cdot 5! cdot 7!$ is $3cdot2cdot1cdot1 = 6 Rightarrow mathrm{(E)}$

Solution 2

If you factor $3! cdot5 ! cdot 7!$ You get
$2^7 cdot 3^4 cdot 5^2$
There are 3 ways for the first factor of a cube: $2^0$ , $2^3$ , and $2^6$ . And the second ways are: $3^0$ , and $3^3$ .
$3 cdot 2 = 6$ Answer : $oxed{E}$

Answer:

Problem Num : 18

From : AMC10

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
Suppose that $m$ and $n$ are positive integers such that $75m = n^{3}$ . What is the minimum possible value of $m + n$ ?
$ext{(A)} 15 qquad ext{(B)} 30 qquad ext{(C)} 50 qquad ext{(D)} 60 qquad ext{(E)} 5700$
'

Category Factor and Multiple

Analysis

Solution/Answer
$3 cdot 5^2m$ must be a perfect cube, so each power of a prime in the factorization for $3 cdot 5^2m$ must be divisible by $3$ . Thus the minimum value of $m$ is $3^2 cdot 5 = 45$ , which makes $n = sqrt[3]{3^3 cdot 5^3} = 15$ . These sum to $60 mathrm{(D)}$ .

Answer:

Problem Num : 19

From : NCTM

Type: Complex

Section:Numbers

Theme:None
Adjustment# : 0

Difficulty: 1

Category Factor and Multiple

Analysis

Solution/Answer

Problem Num : 20

From : NCTM

Type: Complex

Section:Numbers

Theme:Integer Property
Adjustment# : 0

Difficulty: 1

Category Factor and Multiple

Analysis

Solution/Answer

Array ( [0] => 7776 [1] => 7780 [2] => 7781 [3] => 3687 [4] => 3698 [5] => 3700 [6] => 7817 [7] => 7858 [8] => 3013 [9] => 3018 ) 211 2 3