Equations B

Adithya B., Brian L., William W., Daniel X.

5/6

Last week, the equations we focused on were many polynomial or rational in nature: they
only involved the four basic arithmetic operations of addition, subtraction, multiplication,
and division. In this handout, we discuss equations that involve other types of operations.

§1 Radicals
The general strategy in this section is to get rid of radicals.

Example 1.1 (2015 MPFG #17)

Let S be the sum of all distinct real solutions of the equation
√
x + 2015 = x2 − 2015.

Compute b1/Sc.

Solution. We begin by squaring both sides to remove radicals, giving

x + 2015 = (x2 − 2015)2

Now, if we suppose f (x) = x2 − 2015, we see that the above equation may be rearranged
into
f (f (x)) = x
Now, note that if f (x) = x, this equation is satisfied. Thus, we note that f (f (x)) − x
must be factorizable into (f (x) − x)g(x) for some quadratic g. Now, plugging back in
f (x) = x2 − 2015 gives

f (f (x)) − x = (x2 − x − 2015)(x2 + x − 2014)

Now, note that we only care about roots for which x2 − 2015 ≥ 0. Thus, we consider our
roots √ √
1 ± 8061 −1 ± 8057
x= ,
2 2
and we find that only √ √
1 + 8061 −1 − 8057
x= ,
2 2
work. Thus, √ √
8061 − 8057
S=
2

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Adithya B., Brian L., William W., Daniel X. (5/6) Equations B

so √ √
1 2 8061 + 8057
=√ √ =
S 8061 − 8057 2
1
Thus, S = 89.

§2 Floors

Example 2.1 (2019 HMMT November Team #5)

Compute the sum of all positive real numbers x ≤ 5 satisfying

dx2 e + dxe · bxc

x= .
dxe + bxc

Solution. We begin by noting that all x = 1, 2, 3, 4, 5 will work. For all other x, we have
that dxe = bxc + 1, so we do casework on what they are.
• bxc = 0, dxe = 1
Note that in this case, we must have 0 < x < 1, so 0 < x2 < 1 and dx2 e = 1. Thus,
we find that x = 1+1·0
0+1 = 1, which doesn’t satisfy bxc = 0, so we have no x in this
case.

• bxc = 1, dxe = 2
Note that in this case, 1 < x2 < 4, so dx2 e = 2, 3, 4. Plugging in each of these gives
x = 43 , 35 , 36 , respectively. We find that x = 43 , 53 both work, but x = 63 = 2 does not
in this case, since we need bxc = 1.

• bxc = 2, dxe = 3
Note that the possible values of dx2 e are 5, 6, 7, 8, 9. Plugging each of these in gives
x = 11 12 13 14 15 15
5 , 5 , 5 , 5 , 5 . Note that, as in the other cases, 5 doesn’t work because we
need bxc = 2, but all the other possible x can be verified to work.
We begin to notice a pattern, in particular, that all possibilities should work, except
for when x ends up being an integer.
Thus, we make the following conjecture:
k
Conjecture. x = n + 2n+1 works for all integer n > 0 and 0 ≤ k ≤ 2n.
2nk k2 2nk
Proof. We note that x2 = n2 + 2n+1 + (2n+1)2 . Note that 2n+1 > k − 1 (simple to show
2nk k k2
by cross multiplication) and k − 2n+1≥ = 2n+1 Thus, dx2 e = n2 + k. Now,
(2n+1)2
.
plugging in bxc = n and dxe = n + 1 satisfies the given equation.

Thus, we find the sum of our desired x in the interval [n, n + 1) is

2n   2n
X k 1 X
n+ = (2n + 1)(n) + k = 2n2 + 2n = 2n(n + 1),
2n + 1 2n + 1
k=0 k=0

so our answer is
4
X
2n(n + 1) + 5 = 85.
n=1

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Adithya B., Brian L., William W., Daniel X. (5/6) Equations B

Example 2.2 (2012 HMMT Algebra #5)

Find all ordered triplets (a, b, c) of positive reals that satisfy: bacbc = 3, abbcc = 4,
abbcc = 5.

Solution. Another sometimes helpful observation is that bxc is about x, and slightly
smaller. If we multiply all three equations, then we can approximate abc :

60 = a2 b2 c2 bacbbcbcc ≥ (bacbbcbcc)3 =⇒ bacbbcbcc ≤ 3.

Since all three floors are integers, we now know that at least two of bac, bbc, bcc are 1
(otherwise bacbbcbcc would be at least 4). So we can consider cases on which two floors
are 1. Before we jump into the cases, a quick review: if we know ab, bc, ca then we can
find a, b, c as follows. We multiply ab, bc, ca to find a2 b2 c2 , then take the square root to
find the value of abc. We can then divide this by ab, bc, ca to find a, b, c.

If bac = bbc = 1, then bc = 3 and ac = 4. Now bcc must equal 1, 2, or 3, so the third
equation gives that ab must equal 5, 52 , or 53 , respectively. Since bac = bbc = 1 we have
ab < 2 · 2 = 4 so ab = 5 is impossible. If ab = 52 then

√
r
p 5
abc = (ab)(bc)(ca) = · 3 · 4 = 30
2
so √ √ √
abc 30 abc 30 abc 2 30
a= = , b= = , c= = .
bc 3 ac 4 ab 5
We can verify that these indeed satisfy bac = 1, bbc = 1, and bcc = 2 (this is important!),
so this is a solution. On the other hand, if ab = 35 then we find that
√ √ √
2 5 5 6 5
a= , b= , c= ,
3 2 5
√ √ √ 
30 30 2 30
but this is not a solution because bcc = 2 instead of bcc = 3. Thus (a, b, c) = 3 , 4 , 5
is the only solution in this case.
Our other cases are very similar: we find that there are a couple of possible values for
ab, bc, ca, solve for a, b, c, and then check if they satify the floor conditions.

If bbc = bcc = 1 then ac = 4 and ab = 5. Since bac is 1, 2, or 3 we see that bc is 3, 32 , or

1, respectively. If bac = 1 and bc = 3 then we get
√ √ √
2 15 15 2 15
a= , b= , c=
3 2 5
3
but this is not a solution because bac =
6 1. If bac = 2 and bc = 2 then
√ √ √
2 30 30 30
a= , b= , c=
3 4 5
but this doesn’t work because bac =
6 2. Finally, if bac = 3 then bc = 1 and we get
√ √
√ 5 2 5
a = 2 5, b = , c=
2 5
which fails because bac =
6 3. Thus there are no solutions in this case.

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Adithya B., Brian L., William W., Daniel X. (5/6) Equations B

Our final case is if bcc = bac = 1, where ab = 5 and bc = 3. Here we have that bbc
must be 1, 2, or 3, corresponding to ac = 4, 2, 34 , respectively. The first case is impossible,
like in the first case: We have ac < 2 · 2 = 4. For the second case, we solve
√ √ √
30 30 30
a= , b= , c=
3 3 5
4
which works: we verify bac = bcc = 1 and bbc = 2. On the other hand, if ac = 3 then
√ √ √
2 5 3 5 2 5
a= , b= , c=
3 2 5
but this doesn’t work because bcc = 0.

To conclude, we have the two solutions

√ √ √ ! √ √ √ !
30 30 30 30 30 30
, , and , , .
3 4 5 3 2 5

Essentially, we multiplied the three equations at the beginning to place relatively tight
bounds on bac, bbc, and bcc and then used casework on these possibilities to find the
solutions.

Example 2.3 (2012 AIME II #10)

Find the number of positive integers n less than 1000 for which there exists a positive
real number x such that n = xbxc.

Solution. Again, the fact that bxc is always an integer is useful, specifically taking cases
on bxc. We can rewrite this as


 0 0≤x<1

x 1≤x<2



xbxc = 2x 2 ≤ x < 3

3x 3 ≤ x < 4





...


Now taking cases on bxc, we can easily see what (real!) values xbxc can take:
• bxc = 0 : xbxc = 0.
• bxc = 1 : xbxc = x and 1 ≤ x < 2 so 1 ≤ xbxc < 2.
• bxc = 2 : xbxc = 2x and 2 ≤ x < 3 so 4 ≤ xbxc < 6.
• bxc = 3 : xbxc = 3x and 3 ≤ x < 4 so 9 ≤ xbxc < 12.
• ...
Thus, we see that the positive integers xbxc can take are n2 , n2 + 1, . . . , n2 + n − 1 for
each positive integer n. Thus there are n integers of the form xbxc in each interval
[n2 , (n + 1)2 ). Summing n = 1, 2, . . . , 30, we see that there are 1 + 2 + · · · + 30 = 465 such
integers between 1 and 960. For n = 31, we see that the 31 integers 961, 962, . . . , 991 are
also all expressible, and there are no other expressible integers from 992 to 1000, for a
total of 465 + 31 = 496.

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Adithya B., Brian L., William W., Daniel X. (5/6) Equations B

Example 2.4 (Adithya Balachandran)

Let a be a real number such that

bac4 − 4a2 + bac − 1 = 0.

Then, the sum of the squares of all such real numbers a can be written in the form
m
n , where m and n are relatively prime positive integers. Find the value of m + n.

Solution. We know that bac is an integer, so this motivates us to try to isolate all of the
terms containing bac on one side.

4a2 = bac4 + bac − 1

bac4 + bac − 1
a2 =
4
p
bac4 + bac − 1
a=±
2
The reason that this relation is useful is that when a is large, the RHS is approximately
bac2 , which is much larger than the LHS. Therefore, we expect that we can perform some
sort of bounding on the value of bac.
Let’s assume that we are first taking the positive root in the above equation. First,
note that 0 ≤ a − bac < 1, so we have that
p
bac4 + bac − 1
0≤ − bac < 1.
2
For positive values of a we see that bac = 2 is the smallest value that works, and for
bac ≥ 3, the fourth power terms grows faster, so the inequality will not be satisfied.
Therefore, no other value works. Now, lets take the negative value for the square root.
p
bac4 + bac − 1
0≤− − bac < 1
2
Now, for bac > 0, the equation is negative, so none of those values work. We now look into
negative integers. We see that bac = −2 works, and as bac decreases, the fourth power
term increases faster, so the inequality can never be achieved again. Therefore,
√ we have
bac4 +bac−1
that the solutions are bac = ±2. Substituting this into the equation a = ± 2 ,
we have that √ √
13 17
a=− ,
2 2
13 17 30 15
The sum of the squares of the solutions is 4 + 4 = 4 = 2 , so the answer to the
problem is 17.

§3 Exponents and Logarithms

We will first discuss some exponent and logarithm properties that you need to be familiar
with in order to solve problems on this topic.

1. Exponent addition when multiplying two numbers with the same base: ax ·ay = ax+y
ax
2. Exponent subtraction when dividing two numbers with the same base: ay = ax−y .

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Adithya B., Brian L., William W., Daniel X. (5/6) Equations B

3. Multiplication with the same exponent: ax · bx = (ab)x .

4. Logarithm addition: logb x + logb y = logb (xy).

 
5. Logarithm subtraction: logb x − logb y = logb xy .

logc a
6. Change of base: logc b = logb a.
1
7. Logarithm power rules: logb xa = a logb x and logba x = a logb x.
1
8. Reciprocal of a logarithm: logb a = loga b.

We’ll also mention another useful tip for switching back and forth between logarithms
and exponents. Logarithms and exponents are inverse functions. That is, if loga b = c,
then ac = b, which can also be written as aloga b = b. We will leave it as an exercise to
the reader to prove all of these properties.

Example 3.1 (2002 AIME I #6)

The solutions to the system of equations

log225 x + log64 y = 4
logx 225 − logy 64 = 1

are (x1 , y1 ) and (x2 , y2 ). Find log30 (x1 y1 x2 y2 ).

Solution. First, we note that we can use the logarithm power rules shown above to reduce
the equations to the following:
1 1
log15 x + log2 y = 4,
2 6
2 logx 15 − 6 logy 2 = 1.
In order to avoid rewriting the logarithms over and over again, we will let a = log15 x
and b = log2 y. Then, the first equation we are given is 12 a + 16 b = 4. By the reciprocal of
a logarithm property,
1 1
logx 15 = , logy 2 = .
a b
Therefore, the second equation we are given is a2 − 6b = 1. We can multiply this equation
by ab to obtain 2b − 6a = ab. We now have two equations for a and b. We can use the
equation 12 a + 16 b = 4 to solve for b and substitute this in to the second equation. We
getb = 24 − 3a. Therefore,

2(24 − 3a) − 6a = a(24 − 3a) =⇒ 3a2 − 36 + 48 = 0.

Now, we can divide by 3 to obtain a2 − 12a + 24 = 0. Note that the roots of this
polynomial are not integers, so we will not solve for the roots directly. The roots of this
polynomial are log15 x1 and log15 x2 . We want to determine the value of the product
x1 x2 . By the logarithm addition property and Vieta’s,

log15 (x1 x2 ) = log15 x1 + log15 x2 = 12 =⇒ x1 x2 = 1512 .

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Adithya B., Brian L., William W., Daniel X. (5/6) Equations B

Now, we can also determine the product y1 y2 . As we had the substitution b = 24 − 3a

above,

log2 (y1 y2 ) = log2 y1 + log2 y2 = 24 − 3 log15 x1 + 24 − 3 log15 x2

= 48 − 3(log15 x1 + log15 x2 )
= 12.

Therefore, y1 y2 = 212 . Now, this means

log30 (x1 y1 x2 y2 ) = log30 (212 512 ) = log30 (3012 ) = 12.

Example 3.2 (2016 PUMaC Algebra #3)

For positive real numbers x and y, let f (x, y) = xlog2 y . Compute the sum of the
solutions to the equation
4096f (f (x, x), x) = x13

Solution. The function xlog2 y may seem unusual, but this might remind you of the
property aloga b = b. However, the problem here is that the base of the exponent is x
instead of 2. Therefore, this motivates us to try to make the substitution x = 2a , so that
the base becomes 2. Therefore, we have the following nice simplification for f (x, y):
a
f (x, y) = f (2a , y) = 2a log2 y = 2log2 y = y a
2
Therefore, f (x, x) = f (2a , x) = xa = 2a by substituting y = x in the relation above for
f (x, y). From the same relation, we also ahve
2 2 3
f (f (x, x), x) = f (2a , x) = xa = 2a .

Noting that 4096 = 212 ,

3 3 +12
4096f (f (x, x), x) = 212 2a = 2a = x13 = 213a

The function y = 2x is strictly increasing, so the exponents must be equal. This means
that a3 + 12 = 13a. We can factor this cube as (a − 1)(a − 3)(a + 4). Therefore,
a = −4, 1, 3. The values for x are 2−4 = 16 1
, 21 = 2, and 23 = 8. The sum of these
1 161
solutions is 16 + 2 + 8 = 16 .

Example 3.3 (2017 AIME I #14)

Let a > 1 and x > 1 satisfy loga (loga (loga 2)+loga 24−128) = 128 and loga (loga x) =
256. Find the remainder when x is divided by 1000.

Solution. In this solution, we will demonstrate the technique of repeated substitution

and use logarithm properties. First, note that we can actually clear the outer logarithm
by using the fact that logarithms and exponents are inverse functions.

loga (loga 2) + loga (24) − 128 = a128

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Adithya B., Brian L., William W., Daniel X. (5/6) Equations B

Motivated by the substitution in the previous problem, we will let a = 2b . This is because
this substitution will allow us to use many logarithm properties to simplify the equation.
The equation simplifies to

2128b = log2b (log2b 2) + log2b (24) − 128

 
1
= log2b + log2b (24) − 128
b
 
24
= log2b − 128
b
 
24
= log2b − log2b (2128b )
b
 
24
= log2b
b2128b

Now, if we again use the inverse property of logarithms and exponents,

128b 128b 24
(2b )2 = 2b2 =
b2128b
Now, we see the term b2128b on both sides, so we will make the substitution c = b2128b .
Therefore,
24
2c = =⇒ c2c = 24.
c
Now, if we look at this equation, we see by inspection that the only solution is c = 3
because both c and 2c are monotonically increasing functions. Therefore, c = 3 and
b2128b = 3. In order for this equation to work, we must have 3b be a power of 2. Therefore,
let b = 3 · 2k . Therefore,
k k
3 · 2k 2128·3·2 = 3 =⇒ 2k+384·2 = 1

We must have k + 384 · 2k = 0. By inspection, we see that k = −6 is the solution to this

equation. We notice this because 384 only has 7 powers of 2 so it suffices to try values of
3
k from 0 to 7. Therefore, b = 3 · 2−6 = 64 3
, which implies a = 2 64 .
Now, we can look at the equation loga (loga x) = 256 to determine x. We can use the
inverse property to obtain a256 = loga x. If we use the same property once again we
obtain
256 256· 3 12 3 12
x = aa = a2 64 = a·2 = 2 64 ·2 = 2192 .
To find this mod 1000, it suffices to find this mod 8 and mod 125. 2192 is trivially 0
(mod 8). For mod 125, we can utilize Euler’s theorem ( 2100 ≡ 1 (mod 125)), so
1 1 1
2192 ≡ = = ≡ 21 (mod 125).
28 256 6
By the Chinese Remainder Theorem, we obtain x ≡ 876 (mod 1000).

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Adithya B., Brian L., William W., Daniel X. (5/6) Equations B

§4 Problems
Problem 4.1 (2020 AMC 12A #13). There are integers a, b, and c, each greater than
1, such that r q
a b √c
√
36
N N N= N 25
for all N > 1. What is b?
Problem 4.2 (2005 AIME I #8). The equation 2333x−2 + 2111x+2 = 2222x+1 + 1 has
three real roots. Given that their sum is m
n where m and n are relatively prime positive
integers, find m + n.
Problem 4.3 (2014 HMMT Algebra #3). Let
1
A = ((log2 (3))3 − (log2 (6))3 − (log2 (12))3 + (log2 (24))3 ).
6
Compute 2A .
Problem 4.4 (2002 AIME II #8). Find the least positive integer k for which the
equation 2002n = k has no integer solutions for n. (The notation bxc means the greatest
integer less than or equal to x.)
Problem 4.5 (2020 AMC 12A #25). The number a = pq , where p and q are relatively
prime positive integers, has the property that the sum of all real numbers x satisfying

bxc · {x} = a · x2

is 420, where bxc denotes the greatest integer less than or equal to x and {x} = x − bxc
denotes the fractional part of x. What is p + q?
Problem 4.6 (2017 CMIMC Algebra #5). The set S of positive real numbers x such
that    
2x 3x
+ + 1 = bxc
5 5
can be written as S = ∞
S
j=1 Ij , where the Ii are disjoint intervals of the form [ai , bi ) =
{x | ai ≤ x < bi } and bi ≤ ai+1 for all i ≥ 1. Find 2017
P
i=1 (bi − ai ).

Problem 4.7 (2012 Baltic Way/3). Show that the equation

bxc(x2 + 1) = x3 ,

where bxc denotes the largest integer not larger than x, has exactly one real solution in
each interval between consecutive positive integers.
Problem 4.8 (2013 USAMTS Round 1 #4). Show that for all integers k and m ≥ 3,
there exists exactly one pair of positive integers (n, i) with i ≤ m such that
 m 
2 −1 m−i
n−2 +1 =k
2i−1

Problem 4.9 (Canada 1999/1). Find all real solutions to the equation 4x2 −40bxc+51 =
0.
Problem 4.10 (Saint Petersburg 2013/11/1). Find the minimum positive noninteger
root of sin x = sinbxc.

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Adithya B., Brian L., William W., Daniel X. (5/6) Equations B

Problem 4.11 (2006 MOP). Find all pairs (a, b) of positive real numbers such that
babbncc = n − 1 for all positive integers n.

Problem 4.12 (Pan Africa 2012/3). Find all real solutions x to the equation bx2 −
2xc + 2bxc = bxc2 .

Problem 4.13 (Vietnam 2007/1). Solve the system of equations:

 1 + 12 √x = 2
 
3x+y
 1 − 12 √y = 6
 
3x+y

Problem 4.14 (Romania TST 2012/1). Prove that for any positive integer n ≥ 2 we
have that
n n
X √ X
b k nc = blogk nc.
k=2 k=2

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