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?
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?
Note that 4x = (22)x = 2
2x = (2x)2.
Method 1√
4x =√
104 =⇒ 2x = 102 =⇒ (2x)3 = (102)3 =⇒ 8x = 106
4
Problem 1-3
If 4x = 10
4, what is the value of 8
x?
Note that 4x = (22)x = 2
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
4, what is the value of 8
x?
Note that 4x = (22)x = 2
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.
Method 3(4x)3/2 = (43/2)x = 8
x,
4
Problem 1-3
If 4x = 10
4, what is the value of 8
x?
Note that 4x = (22)x = 2
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.
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
4, what is the value of 8
x?
Note that 4x = (22)x = 2
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.
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
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?
Translate sentences into equations:
A + C = 2B, B + C = 3A
5
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?
Translate sentences into equations:
A + C = 2B, B + C = 3A
Subtracting equations, we get A − B = 2B − 3A or B = 43A > A.
5
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?
Translate sentences into equations:
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
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?
Translate sentences into equations:
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
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.]
Note that ab
c
means a(bc)
.
6
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.]
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
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.]
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, and 2
22 = (211)2 = 20482 > 22
2, so
that the answer to the question is
6
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.]
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, 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
What are all ordered triples of positive integers (x, y , z) whose product is
four times their sum, if x < y < z?
xyz = 4(x + y + z) > 4z =⇒ xy > 4
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?
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
What are all ordered triples of positive integers (x, y , z) whose product is
four times their sum, if x < y < z?
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.
If x = 2 < y , then 5/2 ≤ y ≤ 11/2 =⇒ 3 ≤ y ≤ 5.
7
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?
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.
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
What are all ordered triples of positive integers (x, y , z) whose product is
four times their sum, if x < y < z?
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.
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).
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
What are all ordered triples of positive integers (x, y , z) whose product is
four times their sum, if x < y < z?
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.
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).
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}. Solutions: (x, y , z) = (1,5,24), (1,6,14), (1,8,9).
7
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?
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.
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).
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}. 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
What are all ordered triples of positive integers (x, y , z) whose product is
four times their sum, if x < y < z?
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.
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).
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}. 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.
Method IISolve original equation for z:
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.
Method IISolve original equation for z:
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