from: category_eng

Three positive integers are each greater than $1$ , have a product of $27000$ , and are pairwise relatively prime. What is their sum?

$extbf{(A)} 100qquad extbf{(B)} 137qquad extbf{(C)} 156qquad extbf{(D)}} 160qquad extbf{(E)} 165$

Manipulation

For how many positive integers $n$ is $n^2 - 3n + 2$ a prime number?

$mathrm{(A)} ext{none}qquadmathrm{(B)} ext{one}qquadmathrm{(C)} ext{two}qquadmathrm{(D)} ext{more than tw...$

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 $n$ ?

$extbf{(A)} 6 qquad extbf{(B)} 14 qquad extbf{(C)} 21 qquad extbf{(D)} 28 qquad extbf{(E)} 42$

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 $n$ ?

$extbf{(A)} 6 qquad extbf{(B)} 14 qquad extbf{(C)} 21 qquad extbf{(D)} 28 qquad extbf{(E)} 42$

Which of the following numbers is a perfect square?

$mathrm{(A) } 98! cdot 99! qquad mathrm{(B) } 98! cdot 100! qquad mathrm{(C) } 99! cdot 100! qquad mathrm{(D) ...$

10.

What is the probability that a randomly drawn positive factor of $60$ is less than $7$ ?

$mathrm{(A) } frac{1}{10}qquad mathrm{(B) } frac{1}{6}qquad mathrm{(C) } frac{1}{4}qquad mathrm{(D) } frac{...$

11.

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?

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

12.

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 $n$ ?

$mathrm{(A) } 12qquad mathrm{(B) } 15qquad mathrm{(C) } 18qquad mathrm{(D) } 21qquad mathrm{(E) } 24$

13.

What is the probability that an integer in the set ${1,2,3,...,100}$ is divisible by $2$ and not divisible by $3$ ?

$mathrm{(A) } frac{1}{6}qquad mathrm{(B) } frac{33}{100}qquad mathrm{(C) } frac{17}{50}qquad mathrm{(D) } ...$

14.

15.

Integer Property

16.

17.

How many positive cubes divide $3! cdot 5! cdot 7!$ ?

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

18.

Suppose that $m$ and $n$ are positive integers such that $75m = n^{3}$ . What is the minimum possible value of $m + n$ ?

$ext{(A)} 15 qquad ext{(B)} 30 qquad ext{(C)} 50 qquad ext{(D)} 60 qquad ext{(E)} 5700$

19.

20.

Integer Property

21.

For $k > 0$ , let $I_k = 10ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$ ?

$extbf{(A)} 6qquad extbf{(B)} 7qquad extbf{(C)} 8qquad extbf{(D)} 9qquad extbf{(E)} 10$

1.

2.

3. 7

4.

The prime factorization of $27000$ is $2^33^35^3$ . These three factors are pairwise relatively prime, so the sum is $2^3+3^3+5^3=8+27+125=$ $oxed{ extbf{(D) }160}$

5.	Complex

6.

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 $oxed{mathrm{(B)} ext{one}}$ .

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 $oxed{ extbf{(D)} 28}$ .

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 $oxed{ extbf{(D)} 28}$ .

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

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.

10.

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 $frac{6}{12} = oxed{mathrm{(E)} frac{1}{2}}$

11.

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 $oxed{mathrm{(E)} 6}$

12.

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:

The largest possible value of $n$ is $1533$ .

So, the sum of the digits of $n$ is $1+5+3+3=12 Rightarrow A$

13.

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.

14. 20	Understanding

15.	Understanding

16.

17.

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 : $oxed{E}$

18.

$3 cdot 5^2m$ 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 $60 mathrm{(D)}$ .

19. 38	Complex

20.	Complex

21.

The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}cdot 2^{k+2} + 2^6$ .

For $kin{1,2,3}$ we have $I_k = 2^{k+2} left( 5^{k+2} + 2^{4-k} ight)$ . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2leq 5$ .

For $kgeq 5$ we have $I_k=2^6 left( 5^{k+2}cdot 2^{k-4} + 1 ight)$ . For $k>4$ the value in the parentheses is odd, hence $N(k)=6$ .

This leaves the case $k=4$ . We have $I_4 = 2^6 left( 5^6 + 1 ight)$ . The value $5^6 + 1$ is obviously even. And as $5equiv 1 pmod 4$ , we have $5^6 equiv 1 pmod 4$ , and therefore $5^6 + 1 equiv 2 pmod 4$ . Hence the largest power of $2$ that divides $5^6+1$ is $2^1$ , and this gives us the desired maximum of the function $N$ : $N(4) = oxed{7}$ .

Alternate Solution

Notice that 2 is a prime factor of an integer $n$ if and only if $n$ is even. Therefore, given any sufficiently high positive integral value of $k$ , dividing $I_k$ by $2^6$ yields a terminal digit of zero, and dividing by 2 again leaves us with $2^7 * a = I_k$ where $a$ is an odd integer.
Observe then that $oxed{7}$ must be the maximum value for $N(k)$ because whatever value we choose for $k$ , $N(k)$ must be less than or equal to 7.