from: category_eng

Integer Property

Absolute

Euqivalence

10.

11.

Manipulation

12.

13.

The positive integers $A, B, A-B,$ and $A+B$ are all prime numbers. The sum of these four primes is

$mathrm{(A)} mathrm{even}qquadmathrm{(B)} mathrm{divisible by }3qquadmathrm{(C)} mathrm{divisible by }5qquadm...$

14.

For how many integers $n$ is $dfrac n{20-n}$ the square of an integer?

$mathrm{(A)} 1qquadmathrm{(B)} 2qquadmathrm{(C)} 3qquadmathrm{(D)} 4qquadmathrm{(E)} 10$

15.

The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$ ?

$mathrm{(A) } 10qquad mathrm{(B) } 11qquad mathrm{(C) } 12qquad mathrm{(D) } 13qquad mathrm{(E) } 14$

16.

Suppose that $frac{2x}{3}-frac{x}{6}$ is an integer. Which of the following statements must be true about $x$ ?

$mathrm{(A)} ext{It is negative.}\qquadmathrm{(B)} ext{It is even, but not necessarily a multiple of 3.}\qquadmat...$

1.

2.

3.

4.

5.

6.

7. 0.45	Proof

8. 2	Complex

9. 328,245,326	Application

10. 6641	Complex

11. 2,000,001	Complex

12. 179

13.

Since $A-B$ and $A+B$ must have the same parity, and since there is only one even prime number, it follows that $A-B$ and $A+B$ are both odd. Thus one of $A, B$ is odd and the other even. Since $A+B > A > A-B > 2$ , it follows that $A$ (as a prime greater than $2$ ) is odd. Thus $B = 2$ , and $A-2, A, A+2$ are consecutive odd primes. At least one of $A-2, A, A+2$ is divisible by $3$ , from which it follows that $A-2 = 3$ and $A = 5$ . The sum of these numbers is thus $17$ , a prime, so the answer is $oxed{mathrm{(E)} ext{prime}}$ .

14.

Solution 1

Let $x^2 = frac{n}{20-n}$ , with $x ge 0$ (note that the solutions $x < 0$ do not give any additional solutions for $n$ ). Then rewriting, $n = frac{20x^2}{x^2 + 1}$ . Since $ext{gcd}(x^2, x^2 + 1) = 1$ , it follows that $x^2 + 1$ divides $20$ . Listing the factors of $20$ , we find that $x = 0, 1, 2 , 3$ are the only $4 Rightarrow mathrm{(D)}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ).

Solution 2

For $n<0$ and $n>20$ the fraction is negative, for $n=20$ it is not defined, and for $nin{1,dots,9}$ it is between 0 and 1.

Thus we only need to examine $n=0$ and $nin{10,dots,19}$ .

For $n=0$ and $n=10$ we obviously get the squares $0$ and $1$ respectively.

For prime $n$ the fraction will not be an integer, as the denominator will not contain the prime in the numerator.

This leaves $nin{12,14,15,16,18}$ , and a quick substitution shows that out of these only $n=16$ and $n=18$ yield a square.

15.

$AMC=frac{1234}{2}=617$

Since $A$ , $M$ , and $C$ are digits, $A=6$ , $M=1$ , $C=7$ .

Therefore, $A+M+C = 6+1+7 = oxed{mathrm{(E)} 14}$ .

16.

$frac{2x}{3}-frac{x}{6}quadLongrightarrowquadfrac{4x}{6}-frac{x}{6}quadLongrightarrowquadfrac{3x}{6}quadLongright...$ For $frac{x}{2}$ to be an integer, $x$ must be even, but not necessarily divisible by $3$ . Thus, the answer is $mathrm{(B)}$ .