solutions to math league contest 1 of 23 oct...

ACHS Math Team

Solutions to Math League Contest 1 of 23 Oct 2007

Peter S. Simon

Problem 1-1

If x − y = 2007 and y − z = 2008, what is the value of x − z?

2

Problem 1-1

If x − y = 2007 and y − z = 2008, what is the value of x − z?

x − z = (x − y) + (y − z) = 2007 + 2008 = 4015

2

Problem 1-2

One vertex of a square and the midpoints of

the two sides not containing this vertex are

vertices of the triangle shaded in the figure

to the right. If the area of the square is 16,

what is the area of the shaded triangle?

3

Problem 1-2

One vertex of a square and the midpoints of

the two sides not containing this vertex are

vertices of the triangle shaded in the figure

to the right. If the area of the square is 16,

what is the area of the shaded triangle?

2

4

2

2

4

2

Since the area of the square is 16, the length of each side is 4. The

unshaded right triangles have areas 4, 4, and 2, so the area of the

shaded triangle is

Shaded triangle area = 16 − (4 + 4 + 2) = 16 − 10 = 6

3

Problem 1-3

If 4x = 10

4, what is the value of 8

x?

4

Problem 1-3

If 4x = 10


x?

Note that 4x = (22)x = 2

2x = (2x)2.

4

Problem 1-3

If 4x = 10


x?


2x = (2x)2.

Method 1√

4x =√

104 =⇒ 2x = 102 =⇒ (2x)3 = (102)3 =⇒ 8x = 106

4

Problem 1-3

If 4x = 10


x?


2x = (2x)2.

Method 1√

4x =√

104 =⇒ 2x = 102 =⇒ (2x)3 = (102)3 =⇒ 8x = 106

Method 24

x = 104

and from above 2x = 10

2. Multiply equations to get 8

x = 106.

4

Problem 1-3

If 4x = 10


x?


2x = (2x)2.

Method 1√

4x =√

104 =⇒ 2x = 102 =⇒ (2x)3 = (102)3 =⇒ 8x = 106

Method 24

x = 104



x = 106.

Method 3(4x)3/2 = (43/2)x = 8

x,

4

Problem 1-3

If 4x = 10


x?


2x = (2x)2.

Method 1√

4x =√

104 =⇒ 2x = 102 =⇒ (2x)3 = (102)3 =⇒ 8x = 106

Method 24

x = 104



x = 106.

Method 3(4x)3/2 = (43/2)x = 8

x, so 8

x = (104)3/2 = 10

6.

4

Problem 1-3

If 4x = 10


x?


2x = (2x)2.

Method 1√

4x =√

104 =⇒ 2x = 102 =⇒ (2x)3 = (102)3 =⇒ 8x = 106

Method 24

x = 104



x = 106.

Method 3(4x)3/2 = (43/2)x = 8

x, so 8

x = (104)3/2 = 10

6.

Method 4Take the log (to base 10) of both sides to get x log 4 = 4 or

x = 4/ log 4 ≈ 6.6439. Use a calculator to compute this and you find that

8x = 8

4/ log 4 = 106.

4

Problem 1-4

Andy went bowling with Brandi and Candi. The sum of Andy’s score and

Candi’s score was twice Brandi’s score. The sum of Brandi’s score and

Candi’s score was three times Andy’s score. No one’s score was 0. Who

had the highest score?

5

Problem 1-4





Translate sentences into equations:

A + C = 2B, B + C = 3A

5

Problem 1-4






A + C = 2B, B + C = 3A

Subtracting equations, we get A − B = 2B − 3A or B = 43A > A.

5

Problem 1-4






A + C = 2B, B + C = 3A

Subtracting equations, we get A − B = 2B − 3A or B = 43A > A. Putting

this value for B into the second equation yields

C = 3A − 43A = 5

3A > B > A, so that

5

Problem 1-4






A + C = 2B, B + C = 3A

Subtracting equations, we get A − B = 2B − 3A or B = 43A > A. Putting

this value for B into the second equation yields

C = 3A − 43A = 5

3A > B > A, so that Candi had the highest score.

5

Problem 1-5

Write the digit 2 four times, positioning the 2s so that the resulting number

is as large as possible. (You may not write any symbol other than a 2.)

[For example, you could write 2222; but by repositioning the 2s, you can

write an even larger number.]

6

Problem 1-5





Note that ab

c

means a(bc)

.

6

Problem 1-5





Note that ab

c

means a(bc)

.

Since 2222> 2222 it’s clear that we should use exponentiation. Possible

bases for the expression are 2, 22, and 222. Let’s compare using 222

and 22 for the base:

2222 > 2222

= 224 > 2222.

But 2222 = (26)37 = 64

37 > 2222

so that we should choose 2 as the base.

Possible exponents are 222, 222,

6

Problem 1-5





Note that ab

c

means a(bc)

.




2222 > 2222

= 224 > 2222.

But 2222 = (26)37 = 64

37 > 2222


Possible exponents are 222, 222, and 2

22 = (211)2 = 20482 > 22

2, so

that the answer to the question is

6

Problem 1-5





Note that ab

c

means a(bc)

.




2222 > 2222

= 224 > 2222.

But 2222 = (26)37 = 64

37 > 2222


Possible exponents are 222, 222, and 2

22 = (211)2 = 20482 > 22

2, so

that the answer to the question is

2222

6

Problem 1-6

What are all ordered triples of positive integers (x, y , z) whose product is

four times their sum, if x < y < z?

7

Problem 1-6



xyz = 4(x + y + z) > 4z =⇒ xy > 4

Problem 1-6



xyz = 4(x + y + z) > 4z =⇒ xy > 4

xyz = 4(x + y + z) < 4(z + z + z) = 12z =⇒ xy < 12

so that 5 ≤ xy ≤ 11.

If x = 1 < y , then 5 ≤ y ≤ 11.

7

Problem 1-6



xyz = 4(x + y + z) > 4z =⇒ xy > 4

xyz = 4(x + y + z) < 4(z + z + z) = 12z =⇒ xy < 12


If x = 1 < y , then 5 ≤ y ≤ 11.

If x = 2 < y , then 5/2 ≤ y ≤ 11/2 =⇒ 3 ≤ y ≤ 5.

7

Problem 1-6



xyz = 4(x + y + z) > 4z =⇒ xy > 4

xyz = 4(x + y + z) < 4(z + z + z) = 12z =⇒ xy < 12


If x = 1 < y , then 5 ≤ y ≤ 11.

If x = 2 < y , then 5/2 ≤ y ≤ 11/2 =⇒ 3 ≤ y ≤ 5.

If x = 3 < y , then 5/3 ≤ y ≤ 11/3 =⇒ 4 ≤ y ≤ 3 (impossible).

7

Problem 1-6



xyz = 4(x + y + z) > 4z =⇒ xy > 4

xyz = 4(x + y + z) < 4(z + z + z) = 12z =⇒ xy < 12


If x = 1 < y , then 5 ≤ y ≤ 11.

If x = 2 < y , then 5/2 ≤ y ≤ 11/2 =⇒ 3 ≤ y ≤ 5.


Method IWhen x = 1,

yz = 4 + 4y + 4z ⇐⇒ z(y − 4) = 4(y + 1) ⇐⇒ z = 4(y + 1)/(y − 4).Test (y , z) pairs satisfying this equation and y < z from among

y ∈ {5,6, . . . ,11}.

7

Problem 1-6



xyz = 4(x + y + z) > 4z =⇒ xy > 4

xyz = 4(x + y + z) < 4(z + z + z) = 12z =⇒ xy < 12


If x = 1 < y , then 5 ≤ y ≤ 11.

If x = 2 < y , then 5/2 ≤ y ≤ 11/2 =⇒ 3 ≤ y ≤ 5.


Method IWhen x = 1,


y ∈ {5,6, . . . ,11}. Solutions: (x, y , z) = (1,5,24), (1,6,14), (1,8,9).

7

Problem 1-6



xyz = 4(x + y + z) > 4z =⇒ xy > 4

xyz = 4(x + y + z) < 4(z + z + z) = 12z =⇒ xy < 12


If x = 1 < y , then 5 ≤ y ≤ 11.

If x = 2 < y , then 5/2 ≤ y ≤ 11/2 =⇒ 3 ≤ y ≤ 5.


Method IWhen x = 1,


y ∈ {5,6, . . . ,11}. Solutions: (x, y , z) = (1,5,24), (1,6,14), (1,8,9).When x = 2, then

2yz = 8 + 4y + 4z ⇐⇒ z(y − 2) = 2(y + 2) ⇐⇒ z =2(y+2)

y−2. Testing

using this equation and y < z from y ∈ {3,4,5} yields solutions

7

Problem 1-6



xyz = 4(x + y + z) > 4z =⇒ xy > 4

xyz = 4(x + y + z) < 4(z + z + z) = 12z =⇒ xy < 12


If x = 1 < y , then 5 ≤ y ≤ 11.

If x = 2 < y , then 5/2 ≤ y ≤ 11/2 =⇒ 3 ≤ y ≤ 5.


Method IWhen x = 1,


y ∈ {5,6, . . . ,11}. Solutions: (x, y , z) = (1,5,24), (1,6,14), (1,8,9).When x = 2, then

2yz = 8 + 4y + 4z ⇐⇒ z(y − 2) = 2(y + 2) ⇐⇒ z =2(y+2)

y−2. Testing

using this equation and y < z from y ∈ {3,4,5} yields solutions

(x, y , z) = (2,3,10) and (2,4,6).7

Problem 1-6 (Cont.)

xyz = 4(x + y + z), 5 ≤ xy ≤ 11

If x = 1, then 5 ≤ y ≤ 11.

If x = 2, then 3 ≤ y ≤ 5.

Method IISolve original equation for z:

z(xy − 4) = 4(x + y) ⇐⇒ z =4(x + y)

xy − 4(1)

8

Problem 1-6 (Cont.)

xyz = 4(x + y + z), 5 ≤ xy ≤ 11

If x = 1, then 5 ≤ y ≤ 11.

If x = 2, then 3 ≤ y ≤ 5.


z(xy − 4) = 4(x + y) ⇐⇒ z =4(x + y)

xy − 4(1)

Candidate (x, y) pairs are (1,5), (1,6), (1,7), . . . , (1,11), (2,3), (2,4),(2,5). Substitute each of these into Eq. (1) and check that the resulting z

value is an integer greater than y .

8

Problem 1-6 (Cont.)

xyz = 4(x + y + z), 5 ≤ xy ≤ 11

If x = 1, then 5 ≤ y ≤ 11.

If x = 2, then 3 ≤ y ≤ 5.


z(xy − 4) = 4(x + y) ⇐⇒ z =4(x + y)

xy − 4(1)

Candidate (x, y) pairs are (1,5), (1,6), (1,7), . . . , (1,11), (2,3), (2,4),(2,5). Substitute each of these into Eq. (1) and check that the resulting z

value is an integer greater than y . The five solution triples found in this

way are (x, y , z) = (1,5,24), (1,6,14), (1,8,9), (2,3,10), and (2,4,6).

8

solutions to math league contest 1 of 23 oct...

Documents