Download AwesomeMath Admission Test A Solution 1. Page 1 of 2 Since min

AwesomeMath Admission Test A
Solution 1. Page 1 of 2
Since min(x2 , y 2 , z 2 ) is positive (The square of a positive number is also positive), we
see that,
x2 y 2 z 2 > 2016
(xyz)2 > 2016
Let n = xyz.
n2 > 2016
n ≥ 45
We can rewrite the original equation given as
x2 y 2 z 2 − 2016 = min(x2 , y 2 , z 2 )
(xyz)2 − 2016 = min(x2 , y 2 , z 2 )
n2 − 2016 = min(x2 , y 2 , z 2 ).
Proof by Contradiction
Let n > 45 and n2 − 2016 = c2 where c is a positive integer.
We see that n = 45 + x where x is also a positive integer.
Substituting results
(45 + x)2 − 2016 = c2
Expanding,
2025 − 2016 + 90n + n2 − c2
n2 + 90n + 9 = c2
p
n2 + 90n + 9 = c
√
n2 + 90n + 9 is not always a integer leading to a contradiction.
This means n 6> 45 so n = 45.
Going back to our
n2 − 2016 = min(x2 , y 2 , z 2 ).
Substituting,
452 − 2016 = min(x2 , y 2 , z 2 )
2025 − 2016 = min(x2 , y 2 , z 2 )
9 = min(x2 , y 2 , z 2 )
AwesomeMath Admission Test A
Solution 1. Page 2 of 2
Since x, y, z are all positive,
3 = min(x, y, z)
The possible combinations of (x, y, z) are,
(3, 3, 5)
(3, 5, 3)
(5, 3, 3)
AwesomeMath Admission Test A
Solution 2. Page 1 of 1
Working with the first equation,
ab − (a + b) = 2016
ab − a − b = 2016
ab − b = 2016 + a
b(a − 1) = 2016 + a
2016 + a
.
a−1
Guess and checking on some numbers result these integer pairs.
b=
(a, b) : (−2016, 0) (0, −2016) (2, 2018) (2018, 2)
But we can exclude the first two because a, b, c has to be positive.
Working with the second equation,
bc − (b + c) = 2026
bc − b − c = 2026
bc − c = 2026 + b
c(b − 1)2026 + b
2026 + b
.
b−1
Guess and checking on some number result these integer pairs.
c=
(b, c) : (−2026, 0) (0, 2026) (2, 2028) (2028, 2)
We can exclude the first two again because a, b, c has to be positive.
The only working set for (a, b, c) is (2018, 2, 2028)
Substituting the values into ca − (c + a) results,
(2018)(2028) − (2018 + 2028)
4092504 − 4046
4088458
AwesomeMath Admission Test A
Solution 3. Page 1 of 1
x
z
y
The grey area is the triangle with area of 2016.
This means,
xy = 2(2016) = 4032
and
If
p
x2 + y 2 is an integer, we have a right triangle with integer side lengths
4032 has 42 factors which means it has 21 pairs that might satisfy this problem.
(x,y): (1,4032), (2,2016), (3,1344), (4,1008), (6,672), (7,596), (8,504), (9,448), (12,336),
(14,288), (16,252), (18,224), (21,192), (24,168), (28,144), (32,126), (36,112), (42,96),
(48,84), (56,72), (63,64).
p
Calculating x2 + y 2 of these pairs, we find that only one pair, (32,196) work.
p
z = 322 + 1262
√
z = 1024 + 15876
√
z = 16900
z = 130
The only right triangle with integer side-lengths and area of 2016 has the sides,
32, 126, 130 .
AwesomeMath Admission Test A
Solution 4. Page 1 of 1
1
1
1
5bc + 2ac + ab
+
+
=
2a 5b 10c
10abc
Substituting,
5 log2 (2016)
2 log2 (2016)
log2 (2016)
+
+
log(3) log(7) log(2) log(7) log(2) log(3)
10 log3 (2016)
log(2) log(3) log(7)
Simplfying,
5 log(2) log2 (2016) + 2 log(3) log2 (2016) + log(7) log2 (2016)
log(2) log(3) log(7)
10 log3 2016
log(2) log(3) log(7)
(5 log(2) + 2 log(3) + log(7)) · log2 (2016)
log(2) log(3) log(7)
10 log3 2016
log(2) log(3) log(7)
(5 log(2) + 2 log(3) + log(7)) · log2 (2016)
10 log3 (2016)
We know that 5 log(2) + 2 log(3) + log(7) = log(2016).
So the equation can be rewritten as,
log(2016) · log2 (2016)
10 log3 (2016)
log3 (2016)
10 log3 (2016)
1
10
AwesomeMath Admission Test A
Solution 5. Page 1 of 1
2016 √
− x=2
x
Changing to the common denominator x gives,
√
2016 x x
2x
−
=
x
x
x
√
2016 − x x = 2x.
Subtracting 2016 from both sides,
√
−x x = 2x − 2016.
Squaring both sides,
x3 = (2x − 2016)2
x3 = 4x2 − 8064x + 406426
x3 − 4x2 + 8064x − 406426 = 0
Factoring results,
(x − 144)(x2 + 140x + 28224) = 0
One root of the equation is 144 .
(x2 + 140x + 28224) = 0
x2 + 140x = −28224
Completing the square results
x2 + 140x + 4900 = −23324
(x + 70)2 = −23324
√
x + 70 = ± −23324
√
x + 70 = ±14i 119
√
x + 70 = 14i 19
or
√
x + 70 = −14 19
Then the two solutions are
√
√
x = −70 + 14i 19 and x = −70 − 14i 19
√
This means the solutions are 144, −70 ± 14i 19
AwesomeMath Admission Test A
Solution 6. Page 1 of 2
The original equation can be rewritten as,
3a + 2b + c √
= 2016
6
√
3a + 2b + c = 6 2016
√
3a + 2b + c = 6 · 12 14
√
3a + 2b + c = 72 14
Since we are √
adding numbers that result an irrational number, it is highly unlikely
for us to get 72 14 by adding numbers (a, b, c) with different radicands.
√
√
√
Let a = a1 14, b = b1 14, c = c1 14.
We see that 3a1 + 2b1 + c1 = 72
Lets us test some numbers out to find the smallest combination of (a1 , b1 , c1 ) when
substituted into a2 + b2 + c2
Case 1. 3a1 = 72 where b1 , c1 = 0.
a1 = 24
√
a = 24 14, b = 0 c = 0
√ 2
a2 + b2 + c2 = 24 14 = 8064
Case 2. 2b1 = 72 where a1 , c1 = 0.
b1 = 36
√
a = 0, b = 36 14 c = 0
√ 2
a2 + b2 + c2 = 36 14 = 18144
Case 3. c1 = 72 where a1 , c1 = 0.
√
a = 0, b = 0 c = 72 14
√ 2
a2 + b2 + c2 = 72 14 = 72576
Case 4. 3a1 + 2b1 = 72. where c = 0
a1 = 12,
=0
√ b1 = 18, c1 √
a = 12 14, b = 18 14 c = 0
√ 2
√ 2
a2 + b2 + c2 = 12 14 + 18 14 = 6552
Case 5. 3a1 + c1 = 72. where b = 0
a1 = 12,
√ b1 = 0, c1 = 36 √
a = 12 14, b = 0 c = 36 14
√ 2
√ 2
a2 + b2 + c2 = 14 14 + 36 14 = 20160
It is clear to us that having a joint value of a1 and b1 , it will give us the smallest value.
AwesomeMath Admission Test A
Solution 6. Page 2 of 2
We must now test out combinations similar to Case 4.
Case 6. 3a1 + 2b1 = 72. where c = 0
a1 = 10,
=0
√ b1 = 21, c1 √
a = 10 14, b = 21 14 c = 0
√ 2
√ 2
a2 + b2 + c2 = 14 14 + 21 14 = 7574
Case 7. 3a1 + 2b1 = 72. where c = 0
a1 = 20,
=0
√ b1 = 6, c1 √
a = 20 14, b = 6 14 c = 0
√ 2
√ 2
a2 + b2 + c2 = 20 14 + 6 14 = 6104
Case 8. 3a1 + 2b1 = 72. where c = 0
a1 = 22,
=0
√ b1 = 3, c1 √
a = 22 14, b = 3 14 c = 0
√ 2
√ 2
a2 + b2 + c2 = 22 14 + 3 14 = 6902
We clearly see that Case 7 is the smallest possible value of a2 + b2 + c2 .
Answer: 6104
AwesomeMath Admission Test A
Solution 7. Page 1 of 2
2(x3 + y 3 ) −
xy
= 2016.
2
Multiplying by 2,
4(x3 + y 3 ) − xy = 4032.
Since x, y are both positive integers, xy is also positive. So we know that,
4(x3 + y 3 ) > 4032.
Dividing by 4,
x3 + y 3 > 1008.
This means the sum of different (or same) two perfect cubes is greater than 1008 but
by not that much because 103 = 1000.
Let us list some of the perfect cubes.
13 = 1
23 = 8
33 = 27
43 = 64
53 = 125
63 = 216
73 = 343
83 = 512
93 = 729
103 = 1000
113 = 1331
We see some pairs for (x,y) that might work.
*The order of x and y does not matter*.
(x, y) : (10, 2), (9, 7), (8, 8)
Let us substitute these pairs into our 2nd equation, 4(x3 + y 3 ) − xy = 4032.
—————————————————————————————————————–
Pair 1. (10,2)
?
4(103 + 23 ) − 30 = 4032
?
4(1008) − 30 = 4032
?
4032 − 30 = 4032
4002 6= 4032
AwesomeMath Admission Test A
Solution 7. Page 2 of 2
Pair 2. (9,8)
?
4(93 + 83 ) − 30 = 4032
?
4(1241) − 72 = 4032
?
4964 − 72 = 4032
4892 6= 4032
Pair 3. (8,8)
?
4(83 + 83 ) − 64 = 4032
?
4(1024) − 64 = 4032
?
4096 − 64 = 4032
4032 = 4032
The only positive integer solution to the equation is x = 8, y = 8
AwesomeMath Admission Test A
Solution 8. Page 1 of 3
42x 48x
+
− 2016x
48
42
Changing the denominator to the common denominator, 2016,
42x+1 + 48x+1 − 2016x+1
2016
Let y = x + 1, then we can rewrite our equation as,
42y + 48y − 2016y
2016
—————————————————————————————————————–
Claim 1. If x is positive or 0, the value of
42x+1 + 48x+1 − 2016x+1
is negative.
2016
Proof
Let x = 1.
422 + 482 − 20162
−4060188
=
2016
2016
=
−112783
56
Let x = 2
423 + 483 − 20163
−8193355416
=
2016
2016
=
−113796603
28
We see that 482 + 482 < 20162 . Which means no matter how high we raise the power
to, since the three numbers share the same exponent x, the numerator will never become
positive.
—————————————————————————————————————–
AwesomeMath Admission Test A
Solution 8. Page 2 of 3
Claim 2. If x is negative, the value of
42x+1 + 48x+1 − 2016x+1
is positive.
2016
Proof
When x = −1, the expression will be positive because anything to the 0 power is 1.
When x < −1, the expression has negative exponents.
Then the equation can be rewritten as,
1
1
1
+ y −
y
42
48
2016y
2016
where y = −x − 1 (y is always positive).
Changing the denominators of the numerator to 2016y ,
48y + 42y − 1
2016y
2016
Simplifying,
48y + 42y − 1
2016y+1
Since y is always positive, the fraction above will also be always positive.
Claim 3. When x gets smaller in Claim 2, the value of the expression also gets
smaller.
Proof
The fraction to express the ratio between x − 1 and x
(Numerator is x − 1 and the denominator is x)
48y+1 + 42y+1 − 1
2016y+2
y
48 + 42y − 1
2016y+1
Simplifying,
2016y+1 · (48y+1 + 42y+1 − 1)
2016y+2 · (48y + 42y − 1)
48y+1 + 42y+1 − 1
2016(48y + 42y − 1)
The denominator will be always greater than the numerator because,
48y+1 + 42y+1 − 1
≈ 48 + 42 − 1 ≈ 87
48y + 42y − 1
AwesomeMath Admission Test A
Solution 8. Page 3 of 3
2016 > 87
From all of this, we can conclude that the biggest value can be achieved when x is
biggest negative integer*, which is -1.
Going back to the simplified equation in the beginning,
42y + 48y − 2016y
.
2016
When we substitute,
421 + 481 − 20161
1+1−1
=
2016
2016
=
1
2016
*In this problem fractions were excluded as a choice for exponents because it would
lead to roots in the fractions making it impossible to simplify.*
AwesomeMath Admission Test A
Solution 9. Page 1 of 1
I first approached the idea of 2016 being written as a sum of consecutive integers.
m(m + 1)
The formula for the sum of consecutive numbers from 1 to m is
2
m(m + 1)
= 2016
2
Multiply both sides by 2
m(m + 1) = 4032
m = 63
The numbers did not exceed 77 but m, 63 was not a square number!
The next smallest square number was 49 (72 ). This meant we have to make 2016 out
of 49 numbers, so we had to tweak the numbers, like to not start at 1.
Since we lost 15 terms (64 − 49), we had to start somewhere around 15.
We see that we can sum up consecutive numbers that sum up to about 2016 and we
can take out one or two from the consecutive numbers.
15 + 16 + 17 + · · · + 65 results 2040.
2040 − 2016 = 24. We can just take out 24 and the numbers will sum up to 2016.
(15 + 16 + 17 + · · · + 23) + (25 + 26 + 27 + · · · + 65) = 2016
But there are 50 terms (15 + 23 − 1 + 27 − 25 + 1 = 50) in this set so this doesn’t work.
17 + 18 + 19 + · · · + 66 results 2075.
2075 − 2016 = 59. We can take 59 from the set.
(17 + 18 + 19 + · · · + 58) + (60 + 61 + 62 + · · · + 66) = 2016
Now, there are 49 terms (58 − 17 + 1 + 66 − 60 + 1 = 49).
We have been able to write 2016
√ as a sum of 49 distinct positive integers which means
the answer to this problem is 49 = 7