# Problem of the Day

## Problem of the Day

Which of the following expressions yields an even integer for any integer ?

A.

B.

C.

D.

E.

[latexpage]

First, factor all of the expressions into a product of binomials:

A. $n^2-10n+21=(n-7)(n-3)$

B. $n^2-2n-24=(n-6)(n+4)$

C. $n^2+8n+7=(n+7)(n+1)$

D. $n^2+11n+18=(n+9)(n+2)$

E. $n^2-4n-60=(n-10)(n+6)$

In choice D, we have $n$ plus an odd number times $n$ plus an even number. Thus, if $n$ is even, $n+2$ will be even, and if $n$ is odd, $n+9$ will be odd, thus guaranteeing that the product will be even whether $n$ is odd or even. The other expressions have either both odd numbers or both even numbers and thus would not necessarily produce an even product.

## Problem of the Day

Let represent the number of prime numbers less than . What is the value of positive integer ?

(1)

(2)

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

See the Solution[latexpage]

First let’s explore how $T(n)$ works with a few examples. $T(4)=2$ because there are two prime numbers less than 4: 2 and 3. $T(9)=4$ because there are four prime numbers less than 9: 2, 3, 5, 7.

Statement 1:Â $\displaystyle \frac{T(a+1)}{T(a)}-1= \frac{1}{T(a)}$. We can rewrite this by multiplying both sides Â of the equation by $T(a)$ to get $T(a+1)-T(a)=1$. This implies that $a$ is prime, but it does not tell us the exact value of $a$. INSUFFICIENT.

Statement 2: $20 \leq a <29$.Â This is clearly insufficient, as it only tells us a range of possible values for $a$. INSUFFICIENT.

Statements 1&2: Combining the statements we know that $a$ is prime, and $20 \leq a <29$ The only possibility is $a=23$. SUFFICIENT.

## Problem of the Day

John and Deborah work together at their college library. If John works on a given day, he won’t work again until the day after that. If Deborah works on a given day, she won’t work again until the day after that. In days, the semester will end and John and Deborah will no longer work at the library. If they both work Â today, will they both work on the same day again before the semester ends?

(1) is even, and is odd

(2) Â None of the prime factors that divide evenly divide evenly.

[latexpage]

If John works the $j^{th}$ day after every day that he works, and he works today, then he will work on the day that is $j$, $2j$, $3j$… days from today. That is, the number of days until some future work day is always a Â multiple of $j$ if he works today. Similarly, Deborah’s work days will be some multiple of $d$ days from today. Thus, if they both work today, they won’t work together again until some number of days has passed that is a multiple of BOTH $j$ and $d$. The first such day will be the least common multiple of $j$ and $d$. If $j$ and $d$ have no factors in common, then their least common multiple will be $jd$. Otherwise, it will be smaller. For example, if $j=3$ and $d=8$, LCM$(3,8)=3 \cdot 8=24$, because 3 and 8 do not share any factors. However, if $j=2$ and $d=4$, then LCM$(2,4)=4$, which is not that same as $2 \cdot 4$.

Statement 1: $j$ is even, and $d$ is odd.Â If $j=3$ and $d=8$, then the semester will end in 17 days, but they won’t work together for another 24 days. However, if $j=3$ and $d=6$, then the semester will end in 11 days, but they will work together in 6 days. INSUFFICIENT.

Statement 2: Â Â Â None of the prime factors that divide $j$ evenly divide $d$ evenly. This tells us that $j$ and $d$ have no common factors, which means LCM$(j,d)=jd$. The semester ends in $jd-7$ days, but they won’t work together again for $jd$ more days. Thus, they definitely won’t work together again. SUFFICIENT.

## Problem of the Day

If and are positive integers, is divisible by 12?

(1) is a multiple of 3

(2)

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

See the Solution[latexpage]

If $a$ and $b$ are positive integers, is $4a+3b$ divisible by 12?

(1) $b!$ is a multiple of 3. Â This implies only that $b \geq 3$, and tells us nothing about the value of $a$. INSUFFICIENT.

(2) $2a+5b=26$. Â The only possible solutions of this equation where $a$ and $b$ are positive integers are $a=8, b=2$ or $a=3, b=4$. In the former case, $4a+3b=38$, which is not divisible by 12. In the latter case, $4a+3b=24$, which is divisible by 12. INSUFFICIENT.

Statements 1&2: Combining the two statements, we can eliminate $a=8, b=2$ as a possibility. Thus, it must be true that $a=3,b=4$, which means $4a+3b$ is divisible by 12. SUFFICIENT.

## Problem of the Day

A teacher administers a test to two classes. She adds 15 points to each student’s score in the first class and 10 points to each student’s score in the second class, which increases the overall average of the two classes by 13.5 points. If the ratio of the number of students in the first class to the number of students in the second class is where and are positive integers with no common factors, what is the value of ?

A. 8

B. 9

C. 10

D. 11

E. 13

[latexpage]

There are at least a couple of ways we can approach this problem. One is to calculate the ratio directly from the definition of an average. To do this, let’s assume that there are exactly $p$ people in the first class and $q$ people in the second class. The increase in the total number of points would therefore be $15p+10q$. The average increase would then be $\displaystyle\frac{15p+10q}{p+q}$, which we are told is equal to 13.5:

$\displaystyle \frac{15p+10q}{p+q}=13.5$

$15p+10q=13.5p+13.5q$

$1.5p=3.5q$

$\displaystyle\frac{p}{q}=\frac{3.5}{1.5}$

$\displaystyle \frac{p}{q}=\frac{7}{3}$

So, $p:q=7:3$ and $p+q=10$

We can also find this ratio more quickly by using the concept of weighted averages. Note that the overall average increase is 3.5 away from the second class’s average increase, and 1.5 away from the first class’s average increase. This implies that the ratio of the number of people in each class is $3.5:1.5$, and because the overall average is closer to the average increase of the first class, the first class must have more people. Thus, $p:q=3.5:1.5=7:3$ and $p+q=7+3=10$

## Problem of the Day

Ted and Rali run one lap on a circular track. Ted runs in a circular path with a radius of , and Rali runs in a circular path with a radius of , where . Is a factor of ?

(1) Ted and Rali finish one lap in the same amount of time.

(2) Rali’s average speed is times greater than Ted’s, where is an integer.

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

See the Solution[latexpage]

Statement 1: Ted and Rali finish one lap in the same amount of time. Ted runs a total distance of $2\pi r$ and Rali runs a total distance of $2\pi(r+q)$. Let $S_T$ be Ted’s average speed and $S_R$ be Rali’s average speed. If they both complete one lap in the same amount of time, then $\displaystyle \frac{2\pi r}{S_T}=\frac{2\pi r +2\pi q}{S_R}$.

Rearranging:

$\displaystyle \frac{S_R}{S_T}=\frac{2\pi r +2\pi q}{2 \pi r}$

$\displaystyle \frac{S_R}{S_T}=1+\frac{q}{r}$ (Separating the right side out to two distinct fractions and simplifying).

This tells us that the ratio of their speeds is $\displaystyle 1+\frac{q}{r}$, but it doesn’t give us any information about $r$ being a factor of $q$. INSUFFICIENT.

Statement 2:Â Rali’s average speed is $x$ times greater than Ted’s, where $x$ is an integer, or $S_R=xS_T$. This tells us nothing about $q$ and $r$. Rali could run $x$ times faster than Ted regardless of the values of $q$ and $r$. INSUFFICIENT.

Statements 1&2: Statement 2 tells us that $\displaystyle \frac{S_R}{S_T}$ is an integer. Combining that with statement 1 tells us that $\displaystyle 1+\frac{q}{r}$ is an integer, which can only be true if $r$ is a factor of $q$. SUFFICIENT.

## Problem of the Day

David decides to start a business making gift baskets. He estimates that if he charges dollars per gift basket, he will sell baskets per day. If his estimate is accurate, which of the following sets describes all values of for which will he collect revenue in excess of $150 per day?

A.

B. or

C. or

D.

E. Â

[latexpage]

If David is selling $65-5x$ baskets at $x$ dollars per basket, then his total revenue is $x(65-5x)$. We want to know the values of $x$ for which this revenue will exceed $150$. Thus,

$x(65-5x)>150$

$-5x^2+65x>150$

$-5x^2+65x-150>0$

$-5(x^2-13x+30)>0$

$x^2-13x+30<0$

$(x-3)(x-10)<0$

By testing values, we can see that this final expression will be true for all values of $x$ such that $3<x<10$

## Problem of the Day

If and represent positive real numbers and and , what is the value of ?

A. 2.6

B. 3.0

C. 3.4

D. 4.0

E. 4.2

[latexpage]

First, square the expression $(a^4+b^2)^2$:

$(a^4+b^2)^2=$

$(a^4+b^2)(a^4+b^2)=$

$a^8+2b^2a^4+b^4$.

So now we have $a^8+2b^2a^4+b^4=198$

Rearranging the terms on the left:

$a^8+b^4+2b^2a^4=198$

Make a substitution with $a^8+b^4=36$ to get:

$36+2b^2a^4=198$

$2b^2a^4=162$

$b^2a^4=81$

Take the fourth root of both sides to get:

$a\sqrt{b}=3$

## Problem of the Day

Does Jim own at least one dog?

(1) Jim lives in an apartment complex that is home to 20 people (including Jim) who own a total of 39 dogs.

(2) No one in Jim’s apartment complex owns more than 2 dogs.

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

See the SolutionStatement 1: This tells us the total number of people and dogs at Jim’s apartment complex but tells us nothing about how the dogs are distributed. INSUFFICIENT.

Statement 2: This tells us the maximum number of dogs per person but doesn’t give us any information regarding Jim and his dog ownership status. INSUFFICIENT.

Statements 1&2: If no one at Jim’s apartment complex owns more than 2 dogs, then we could assign 2 dogs to the 19 people who are not Jim which would account for 38 of the dogs. The final dog could then only be assigned to Jim without violating statement 2. Therefore, Jim owns at least one dog. SUFFICIENT.

## Problem of the Day

If , is an integer?

(1)

(2) is an integer

[latexpage]

If $ab=c$, is $\displaystyle \frac {72}{b}$ an integer?

Statement 1: $c=36$. If $a=9$ and $b=4$, then $\frac{72}{b}=\frac{72}{4}=18$. If $a=\frac{36}{7}$ and $b=7$, thenÂ $\frac{72}{b}=\frac{72}{7}$, which is not an integer. INSUFFICIENT.

Statement 2: $\displaystyle \frac{72}{a}$ is an integer. This tells us nothing about $b$. INSUFFICIENT.

Statements 1&2: The examples used to prove Statement 1 insufficient also apply here. INSUFFICIENT.