American Math Competition Club

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Problems from AMC 12 (Solutions) 1) Find the number of ordered pairs of real numbers (a, b) such that (a + bi ) 2002 = a − bi . a) 1001

b)1002

c)2001

d)2002

e)2004

Solution: e)

a − ib, then this means that the 2002th of a + ib its conjugate. It is better to see If (a + bi ) 2002 = this geometrically. Instead of considering a complex number, z ∈ , in rectangular form, z= a + ib, consider it in polar form, z =rcis (θ ) =r[cos(θ ) + i sin(θ )] =reiθ .Then we can ask 2002 iθ 2002 z reiθ , has the property z 2002 = = e re − iθ . ourselves, what complex number (reiθ ) 2002 r= =

Since r 2002 = r and r ∈ [0, ∞), then this means r = 1. The fact that z is a nth root of unity, then we know 2π i

that e n = z , meaning that if we take z k , this would be equivalent to z k mod ( n ) . Therefore, you’re really asking yourself for which integers, n, will 2002 mod(n) = −1mod(n) .This is true if 2003 = 0 mod(n) . In other words, which are the factors of 2003? But 2003 is prime, so n = 2003. Therefore, the solutions to the question 2002 z= z is equivalent to = z 2003 1= or z 0. There are 2003 solutions to the first equation, and 1 solution to the second. So there are 2004 solutions total.

Written by L Marizza A Bailey with problems from the Art of Problem Solving Introduction to Number

American Math Competition Club

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2) How many three digit numbers are composed of three distinct digits such that one digit is the average of the other two? a) 96 b)104 c)112 d)120 e)256 Solution: b) A three digit number, n, is comprised of the ordered triple (a,b,c) such that a102 + b10 + c = n The added condition that the digits be distinct, just means that a, b, and c are distinct. The fact that these are coefficients of powers of 10 implies that 0 ≤ b, c ≤ 9 and 1 < a ≤ 9 . If one is the average of the other two, then, without loss of generality, we can assume that c is the average of a and b then permute that triple to get 3! = 6 other numbers with those digits, unless one of them is zero, in which case you can only get 2 numbers with those digits.

a+b =c 2 a+b = 2c This means that a + b < 18. The ordered triples that satisfy all of these conditions are No zero digits Amount of numbers One zero digit (1,3,2) 6 (2,0,1) (1,5,3) 6 (4,0,2) (1,7,4) 6 (6,0,3) (1,9,5) 6 (8,0,4) (2,4,3) 6 (2,6,4) 6 (2,8,5) 6 (3,5,4) 6 (3,7,5) 6 (3,9,6) 6 (4,6,5) 6 (4,8,6) 6 (5,7,6) 6 (5,9,7) 6 (6,8,7) 6 (7,9,8) 6 16(6) Total If you add them together, you will get a total of 96+8 = 104.

Amount of numbers 2 2 2 2

4(2)

This took a while, so how could we think of this combinatorically? If you see how I organized the triples, if we started with a = 1, then in order to ensure we were not going to double count, I only chose b so that it was greater than a. In order for for c to be an integer, a + b had to be even. Therefore, I counted how many integers, 1 < b ≤ 9 that were also odd; 4. For a = 2¸ we needed 2 < b ≤ 9 and even; 3 Then we do the same thing with a = 3, 4, 5, 6, 7, 8. Then answers decrease by 1 as a increases; 3, 2, 2, 1, 1,0. So you get (4+3+3+2+2+1+1)(6) = (16)(6). A similar counting argument can be made with b = 0.

Written by L Marizza A Bailey with problems from the Art of Problem Solving Introduction to Number

American Math Competition Club

3) For how many integers, n is a) 1

b) 2

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n the square of an integer? 20 − n c) 3

d) 4

e) 10

Solution: b) Of course, when I see n and 20 – n together, I always consider modular arithmetic. n = k2 20 − n = n k 2 (20 − n)

This means that if you mod out both sides of the equation by n you get 0 = 20k 2 mod(n) . Now we can further reduce this to

20 mod(n)·k 2 mod(n) = 0. This means that either k2 divides n, or 20 divides n. However, since 20 – n has to divide n, and the quotient can’t be negative, this means that 0 < n < 20. This means that k2 has to divide n. So this should be true for n = 1, 4, 16, 9, 12, 18, 8. However, the original statement is only true for 16 and 18. So for how many integers n, is there a solution to (20)(k2) = 0 mod (n)?

4) The sum of 18 consecutive positive integers is a perfect square. The smallest possible value of this sum is a) 169 b) 225 c) 289 d) 361 e) 441

Solution: b) The sum of 18 consecutive positive integers can be written as

(k + 1) + (k + 2) + (k + 3) +  + (k + 18) (18)(18 + 1) = k18 + 2 = 18k + 9(19 = ) 18k + 171 If the number is a perfect square that means that 18k + 171 is a perfect square. Trying the numbers in the options above, 169 169 = 18k+171 k is not positive 225 225=18k + 171 k =3 18k = 225 – 171 = 54 Since this is the smallest number of the remaining possibilities, it is the answer. You could also plug in numbers for k, starting at 1 until you get a perfect square. Written by L Marizza A Bailey with problems from the Art of Problem Solving Introduction to Number

American Math Competition Club

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5) We call a number “prime-looking” if it is composite, but not divisible by 2, 3, or 5. The smallest prime-looking numbers are 49, 77, and 91. There are 168 prime numbers less than 1000. How many prime looking numbers are there less than 1000? a) 100 b) 102 c) 104 d) 106 e) 108 Solution: Now we have to omit all multiples of 2, 3, and 5 without double counting the numbers divisible by 6, 15, 10. So we use the floor function to find the number of multiples of each number.

 999 

 999 

 999 

Multiplies of 2 =  = 499 , Multiples of 3 =  = 333 , Multiples of 5 =  = 199 .  2   3   5  But then we counted the multiples of 6, 15, and 10 twice.

 999 

 999 

Multiples of 10 = 99, Multiples of 6 =  = 66 = 166 , Multiples of 15,   15   6  499+333+199-99-166-66 = 700 However, we counted multiples of 30 three times, but then took them out three times, so we have to take those back out. There are 33 of them, that makes the total count 733. So there are 999 – 733 = 266 numbers that are not divisible by 2,3 or 5. However, 165 of those are prime. So that leaves 101 numbers that are not prime, nor divisible by 2, 3, or 5. However, this includes the number 1, which isn’t composite nor prime. So it’s not prime looking, and we haven’t taken it out yet. So that leaves 100 “prime-looking” numbers. Modular Arithmetic II Solving Linear Conguences Multiplication, Addition, Division and Subtraction are all a little different mod(n). We call mod a homomorphism, and it’s important to understand the structure of the integers modulo n,  n . First of all

 n = {0,1, 2,3,.....n − 1} and -3 mod(n) = n-3. Also, the multiplicative inverse of k mod(n) doesn’t necessarily have to exist. Example: Consider the group 10 2 + 8 = 0 mod(10) means that -2 mod (10 ) = 8 Also, although 3(7) = 1 mod(10), two does not have a multiplicative inverse, because 2(5) = 0 mod(10). We call 2 a zero divisor mod(10). It is also important to note (a + b) mod(10) =a mod(10) + b mod(10) and (ab) mod(10) = a mod(10)b mod(10) . Why do you think this is true?

Written by L Marizza A Bailey with problems from the Art of Problem Solving Introduction to Number

American Math Competition Club

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Exercise: Write all the additive and multiplicative inverses in 10 , if any exist. Answer:

Now let us consider a more complicated problem. What are the solutions to the modular equation

1233 x + 45 = 9090 mod(24) We can still subtract 45 from both sides

1233 x = 9045 mod(24) But this means that

1233mod(24) x mod(24) = 9045 mod(24) 9 x = 21mod(24) By definition of mod(24), this means that there is an integer k such that

= 9 x 24k + 21 We can divide both sides of this equations by 3 to achieve

3= x 8k + 7 This means 3x ∈ {..., −33 − 25, −17, −9, −1, 7,15, 23,31,39, 47,55, 63, 71, 79,87,} . The numbers for which there is a solution are 3x = 15, 39, 63, 87, ..... In which case, the values of x would be {..,-11,-3, 5, 13, 21, 29, ... } = 5 + 8 .N Exercises:

1. 2. 3. 4.

1235 x + 45 = 9090 mod(24) 1235 x + 45 = 9090 mod(11) 1235 x + 45 = 9087 mod(11) 1235 x + 45 = 9090 mod(24)

5. Solve the simultaneuos equations

= 3 x 4= mod(7), 4 x 5 mod(8), = and 5 x 6 mod(9)

Written by L Marizza A Bailey with problems from the Art of Problem Solving Introduction to Number