2010 AMC 12B Problem 15

Question

For how many ordered triples

(x, y, z)

of nonnegative integers less than

20

are there exactly two distinct elements in the set

\{i^x, (1+i)^y, z\},

where

i=\sqrt{-1}?

149

205

215

225

235

Answer: D

Concepts:complex number casework basic counting

Difficulty rating: 2070

Small hint:

|i^x|=1

always, while

(1+i)^y

has magnitude

2^{\frac{y}{2}}

Big hint:

Split into three cases by which two of the three entries coincide, keeping the third distinct

Solution:

We need exactly two of

i^x,

(1+i)^y,

z

equal, with the third different. The three cases are the three possible equal pairs.

Case $i^x=(1+i)^y:$ since

|i^x|=1

but

|(1+i)^y|=2^{\frac{y}{2}}\gt1

for

y\ge1,

we need

y=0,

so

(1+i)^0=1

and

i^x=1,

i.e.

x\in\{0,4,8,12,16\}.

Then

z

is any of the

19

values other than

1.

This gives

5\cdot19=95

triples.

Case $i^x=z:$ the only nonnegative-integer value of

i^x

is

1

(with

x

a multiple of

4

), so

z=1

and

(1+i)^y\ne1,

meaning

y\ge1.

This gives

5\cdot19=95

triples.

Case $(1+i)^y=z:$ since

(1+i)^2=2i,

the power

(1+i)^y

is a nonnegative integer below

20

only for

y=0

(value

1

) or

y=8

(value

16

). If

y=0,\ z=1,

we need

i^x\ne1,

so

x

is not a multiple of

4

(

15

values). If

y=8,\ z=16,

then

i^x

is never

16,

so

x

is free (

20

values). This gives

15+20=35

triples.

Altogether

95+95+35=225.

Thus, the correct answer is D.

2010 AMC 12B Problem 15

Answer: D

Difficulty rating: 2070

Small hint:

Big hint:

Solution:

Problem 15 in other years