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Numbers - the Property of Integers

For COMPETITION
Number of Total Problems: 16.
FOR PRINT ::: (Book)

Problem Num : 11

From : NCTM

Type: Complex

Section:Numbers

Theme:Manipulation
Adjustment# :

Difficulty: 1

Category the Property of Integers

Analysis

Solution/Answer

Problem Num : 12

From : NCTM

Type: None

Section:Numbers

Theme:None
Adjustment# :

Difficulty: 2

Category the Property of Integers

Analysis

Solution/Answer

Problem Num : 13

From : AMC10B

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
The positive integers $A, B, A-B,$ and $A+B$ are all prime numbers. The sum of these four primes is
$mathrm{(A)} mathrm{even}qquadmathrm{(B)} mathrm{divisible by }3qquadmathrm{(C)} mathrm{divisible by }5qquadm...$
'

Category the Property of Integers

Analysis

Solution/Answer
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 $oxed{mathrm{(E)} ext{prime}}$ .

Answer:

Problem Num : 14

From : AMC10B

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
For how many integers $n$ is $dfrac n{20-n}$ the square of an integer?
$mathrm{(A)} 1qquadmathrm{(B)} 2qquadmathrm{(C)} 3qquadmathrm{(D)} 4qquadmathrm{(E)} 10$
'

Category the Property of Integers

Analysis

Solution/Answer
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.

Answer:

Problem Num : 15

From : AMC10

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$ ?
$mathrm{(A) } 10qquad mathrm{(B) } 11qquad mathrm{(C) } 12qquad mathrm{(D) } 13qquad mathrm{(E) } 14$
'

Category the Property of Integers

Analysis

Solution/Answer
$AMC10+AMC12=123422$
$AMC00+AMC00=123400$
$AMC+AMC=1234$
$2cdot AMC=1234$
$AMC=frac{1234}{2}=617$
Since $A$ , $M$ , and $C$ are digits, $A=6$ , $M=1$ , $C=7$ .
Therefore, $A+M+C = 6+1+7 = oxed{mathrm{(E)} 14}$ .

Answer:

Problem Num : 16

From : AMC10

Type:

Section:Numbers

Theme:
Adjustment# : 0

Difficulty: 1

'
Suppose that $frac{2x}{3}-frac{x}{6}$ is an integer. Which of the following statements must be true about $x$ ?
$mathrm{(A)} ext{It is negative.}\qquadmathrm{(B)} ext{It is even, but not necessarily a multiple of 3.}\qquadmat...$
'

Category the Property of Integers

Analysis

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

Answer:

Array ( [0] => 3769 [1] => 3774 [2] => 7975 [3] => 7976 [4] => 7777 [5] => 7870 ) 161 2