Problem of the Day

A high school sells tickets to a basketball game. Adult tickets cost $9 and student tickets cost $5. If $n$ tickets are sold for a total revenue of $927, how many possible values are there for $n$ ?

A. 18
B. 19
C. 20
D. 21
E. 22

Reveal Answer

D. 21

See the Solution

From the given information we can set up the following equation to model the revenue, where $x$ represents the number of student tickets sold and $y$ represents the number of adult tickets sold: $5x+9y=927$ . First, note that $x$ and $y$ are both non-negative integers. Next, we can see that 927 is a multiple of 9 because its digits add to 18, and $9y$ is clearly a multiple of 9. This implies that $5x$ is also a multiple of 9, and since 5 does not contain any factors of 9, $x$ itself must be a multiple of 9. That is, all possible values of $x$ are of the form $x=9t$ for some non-negative integer, $t$ . Plugging this in to the original equation: $5(9t)+9y=927$ . Solve this for $y$ :

$45t+9y=927$

$5t+y=103$

$y=103-5t$

Values of $y$ must be non-negative integers, so it must be true that $103-5t >=0$ . Solving this inequality gives us $t<=20.6$ . We initially defined $t$ to be a non-negative integer, so $0<= t <= 20$ , which means there are 21 possible values for $t$ . To show that this means that there are 21 possible values of $n$ , note that $n=x+y$ , so $n=9t+103-5t$ or $n=103+4t$ . Clearly each of the possible 21 values for $t$ will produce a distinct value for $n$ .

Problem of the Day

The mean of a set of $n$ consecutive positive integers is $m_1$ . One of the integers is removed from the set to create a new set of positive integers with a mean of $m_2$ . Is $n$ even?

(1) $m_1=m_2$

(2) The least element of the original set of integers is odd.

Reveal Answer

A. Statement 1 alone is sufficient, but Statement 2 alone is not sufficient.

See the Solution

Problem of the Day

What is the sum of the digits of positive integer $q$ ?

(1) The sum of the digits of $q$ is an element of the set $\{226,313,447,617\}$

(2) $q=n^3-n$ for some positive integer $n$ .

Reveal Answer

C. Both statements together are sufficient, but neither statement alone is sufficient.

See the Solution

What is the sum of the digits of positive integer $q$ ?

Statement 1: The sum of the digits of $q$ is an element of the set ${226,313,447,617}$ . This is clearly insufficient, as it tells us the answer to the question could be any of these four elements. INSUFFICIENT.

Statement 2: $q=n^3-n$ for some positive integer $n$ . We can factor this expression as $q=n(n^2-1)=n(n+1)(n-1)=(n-1)n(n+1)$ . This expression is the product of three consecutive integers. Because every third integer is a multiple of 3, one of these integers must be a multiple of 3. Thus, $q$ is a multiple of 3. However, this tells us nothing about the sum of the digits of $q$ . INSUFFICIENT.

Statements 1&2: Remember that the sum of the digits of a multiple of 3 is also a multiple of 3. Thus, if $q$ is a multiple of 3 and the sum of its digits is either $226,313,447,$ or $617$ , it must be $447$ because that is the only multiple of 3 in the set. This is clear because $4+4+7=15$ , which is a multiple of 3.

Problem of the Day

A woman who has two sons enrolled in an elementary school brings a batch of $n$ cookies to the school. Her youngest son is in a class with 10 other students and her oldest son is in a class with 7 other students. If she divides the $n$ cookies evenly among the students in her youngest son’s class there will be 5 left over. If she divides the cookies evenly among the students in her oldest son’s class there will be 7 left over. If she has more than 200 cookies, which of the following is the sum of the digits of the smallest possible value of $n$ ?

A. 9
B. 11
C. 13
D. 17
E. 23

Reveal Answer

C. 13

See the Solution

We know that $n$ is 5 more than a multiple of 11 and 7 more than a multiple of 8. That is, $n=11a+5$ and $n=8b+7$ for some positive integers $a,b$ . Start by making a list of numbers that satisfies the first equation and stop when you get to one that also satisfies the 2nd. To generate this list, start with the remainder, 5, and add the divisor, 11: $5,16,27,38,49,60,71$ . 71 is the first number that is also 7 more than a multiple of 8. Thus, 71 is the smallest positive integer that satisfies both equations. We can obtain further solutions by adding the least common multiple of 11 and 8: LCM(11,8)=88. That gives us $71, 159, 247$ . 247 is thus the smallest possible value of $n$ that is greater than 200. The sum of its digits is $2+4+7=13$ .

Problem of the Day

Two farmers, $X$ and $Y$ , each surround a triangular plot of land with a fence on their respective properties. Farmer $X$ requires 240 meters of fencing and farmer $Y$ requires 60 meters of fencing. If the ratios of the lengths of the corresponding sides of the triangular plots of land are all equal to $k$ , then the area of the triangular plot of land on farmer $X$ ‘s property is how many times bigger than the area of the triangular plot of land on farmer $Y$ ‘s property?

A. $2$
B. $2k$
C. $8$
D. $8k$
E. $16$

Reveal Answer

E. 16

See the Solution

Problem of the Day

How many distinct factors does $14!$ have?

A. 2592
B. 2718
C. 3142
D. 3654
E. 3660

Reveal Answer

A. 2592

See the Solution

$14!=14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1$

Break down all of these numbers on the right into their prime factors:

$=(2*7)(13)(2^2*3)(11)(2*5)(3^2)(2^3)(7)(2*3)(5)(2^2)(3)(2)$

Simplify the expression using exponent laws:

$=2^{11}*3^5*5^2*7^2*11*13$

Any factor of this number must be of the form $2^a3^b5^c7^d11^e13^f$ where $a,b,c,d,e,f$ are integers and $0<=a<=11,0<=b<=5,0<=c<=2,0<=d<=2,0<=e<=1,0<=f<=1$ .

Thus, when constructing a factor, we have 12 choices for $a$ , 6 choices for $b$ , 3 choices for $c$ , 3 choices for $d$ , 2 choices for $e$ and 2 choices for $f$ .

Therefore there are $12*6*3*3*2*2=2592$ factors of $14!$

Problem of the Day

A group of burglars is trying to decide whether to target the houses in a particular neighborhood in which there are 100 households. $x$ of the households own a high-definition television, $y$ of the households own a valuable piece of artwork, $z$ of the households own a luxury car, and all of the households own at least one of these items. The burglars will target the houses in the neighborhood unless the probability of any given household owning all three items exceeds 50%, as this is correlated with tighter security. If $z<y<x$ , do the burglars target the houses in the neighborhood?

(1) $x+y=120$

(2) $z=55$

Reveal Answer

A. Statement 1 is sufficient alone, but statement 2 is not sufficient alone.

See the Solution

Statement 1: If $x+y=120$ , and $z<y<x$ , then the biggest value $z$ could have is if $x=61, y=59, z=58$ . In this case $x+y+z=178$ . But there are only $100$ households, so there is an excess of 78 households that must have been counted multiple times. Every household that is in all three categories gets counted two extra times when you add $x,y$ and $z$ . So there could be at most $39$ households that have all three items. If at most $39$ out of $100$ households have all three items, then the probability of any given household having all three items is no greater then $39$ percent. Thus, they will target the houses in this neighborhood. SUFFICIENT.

Statement 2: If $z$ is 55, then if $x=65,y=60$ , then the maximum overlap would be $40$ , in which case they would target the houses. If $x=90,y=85$ , then the maximum overlap is $55$ in which case they would not target the houses. INSUFFICIENT.

Problem of the Day

If $x,y\neq0$ and $x\sqrt{12}+y\sqrt{51}=\sqrt{z}(2x+y\sqrt{17})$ , what is the value of $z$ ?

A. $\sqrt{3}$
B. $2$
C. $3$
D. $\sqrt{5}$
E. 7

Reveal Answer

C. 3

See the Solution

$x\sqrt{12}+y\sqrt{51}=\sqrt{z}(2x+y\sqrt{17})$

On the left side, simplify $\sqrt{12}$ to $2\sqrt{3}$ . Also, note that $\sqrt{51}$ can be written as $\sqrt{3}\sqrt{17}$ . So the left side becomes $2x\sqrt{3}+y\sqrt{3}\sqrt{17}$ . Now factor a $\sqrt{3}$ from the left side to get $\sqrt{3}(2x+y\sqrt{17})$ . To make this match the right side, it must be true that $z=3$ .

Problem of the Day

Four families of three are lining up for a photo. How many ways can they line up if all of the members in each family must stand together?

A. $12$
B. $4!3!$
C. $7!$
D. $4!6^4$
E. $12!$

Reveal Answer

D. $4!6^4$

See the Solution

First calculate the number of ways to arrange the four families: $4!=24$ . Then, note that each family can be rearranged 3! ways. For each of the 24 ways we can arrange the families, the members within the families can be arranged $3!3!3!3!=6^4$ ways. Hence, there are $4!6^4$ ways to arrange the people.

Problem of the Day

Andy and Bob play a game in which a computer randomly selects two real numbers between 0 and 10. Andy’s score is the sum of the numbers and Bob’s score is one more than the product of the numbers. If the person with the higher score wins the game, what is the probability that Bob wins?

A. 18%
B. 19%
C. 68%
D. 82%
E. 90%

Reveal Answer

D. 82%

See the Solution

Two numbers are selected randomly from the interval $(0,10)$ . Let the first number selected by $x$ and the second number be $y$ . Andy’s score will be $x+y$ and Bob’s score will be $xy+1$ . Bob will be the winner when $xy+1>x+y$ . Solving this inequality:

$xy+1>x+y$

$xy-x-y+1>0$

$x(y-1)-(y-1)>0$

$(x-1)(y-1)>0$

This will be true if either $(x-1)$ and $(y-1)$ are both positive OR if $(x-1)$ and $(y-1)$ are both negative:

$x-1>0$ AND $y-1>0$

$x>1$ AND $y>1$

Because the numbers chosen are between 0 and 10, each number has a $\frac{9}{10}$ probability of being greater than one. The probability that they both are greater than one is $\frac{9}{10}\cdot\frac{9}{10}=\frac{81}{100}$ .

The other possibility is that both numbers are less than 1, which, through similar analysis we can show happens with probability $\frac{1}{10}\cdot\frac{1}{10}=\frac{1}{100}$

Adding the two probabilities gives us $\frac{82}{100}$ or 82%