Download Victor Chen, ID 14868 USAMTS Round 2, Solution to Problem 1/2/22

Victor Chen, ID 14868
USAMTS Round 2, Solution to Problem 1/2/22
Label each row of the pyramid from top to bottom A-G. Then, we label each individual square 1-n, from
left to right in each nth row. So the top row (row A) can be expressed as A1, A2, A3, A4, A5, A6, and A7.
Next, we assign variable values to each space in the top row. Label A1 a, the blue squares (A2 & A3) b,
the yellow squares (A4 & A5) c, the next square (A6) d, and the final square on the right (A7) e. We then
sum to squares to get the squares in the next row, and so forth, as shown in the diagram below:
Since squares A2 and A3, A4 and A5, B3 and C1,
B6 and C3, and D3 and E1 are congruent, we just solve
a
b
b
a+b
2b
c
b+c
c
2c
d
c+d
e
d+e
the equations to find the variables. Note that we do not
need anything more after E3 yet, as there are no marked
a+3b 3b+c b+3c 3c+d
squares after E3. We then set the equations equal to each other,
a+6b+c
4b+4c
c+2d+e
b+6c+d 4c+3d+e
and solve them. So we then have:
b+c=a+3b
(a)
d+e=b+3c
(b)
b+6c+d=a+10b+5c
(c)
a+10b+5c
We simplify equation (a) to be:
a+2b=c
(d)
We then substitute and simplify this equation into equation (b):
3a+7b=d+e
(e)
Next, we substitute in our value of c in equation (d) into equation (c) and simplify to get
c+d=a+9b
=
a+2b+d=a+9b
=
7b=d.
Since b and d are nonzero digits, the only possible solution for this is b=1, d=7.
So now we have that
a+2=c.
We then substitute our values of b and d into equation (b) to get that 6+e=3c. So therefore e must be
divisible by 3. We also see that if e is 3, then c must be 3. However, this isn’t possible because e cannot
equal c. We then check the other two values of e, 6 and 9. When e is 6, c is 4, and a is 2. However, then
B3 and A1 both equal 2, and since they aren’t marked, they cannot equal each other. Therefore this isn’t
possible. Thus the only possible solution is e=9. When e is 9, c is 5, and a is 3. A complete diagram of the
solution is shown below:
Victor Chen, ID 14868
USAMTS Round 2, Solution to Problem 1/2/22
3
1
4
1
2
6
5
6
8
14
5
10
16
24
38
7
12
22
38
62
100
88
150
250
16
28
50
9
Victor Chen, ID 14868
USAMTS Round 2, Solution to Problem 2/2/22
(a) We see that it isn’t possible to get a tworrific sequence of length 1, or 2. Obviously we can’t have
a sequence of 1, as we need the terms 1 and 2011. If we have a tworrific sequence of length 2,
then we can only have the terms 1 and 2011. However, since 1+2011=2012, and 2012 isn’t a
power of 2, this isn’t possible. We then check for a tworrific sequence of length 3. Since 1 and
2011 cannot be next to each other, we must have a sequence of the form 1, a, 2011. However,
then a+1 and a+2011 must be powers of 2. Therefore a+1 = for some k, and a+2011= for
some l. We can then subtract the first equation from the second to get that 2010=
, or
, where k is the largest integer possible that factors into 2010 as a power of 2. This is
justifiable, because 2011 is greater than 1. However, since
, k must equal
1, which means must equal 2012. However, we know that this isn’t possible, so therefore there
are no tworrific sequences of length 3. We then check for tworrific sequences of length 4. Since
we know that a tworrific sequence of the form 1, a, 2011 isn’t possible, we know that the only
possible form of the tworrific sequence is 1, a, b, 2011. Therefore, we must have 1+a= for
some k, a+b= for some l, and b+2011=
for some m. We now subtract the second equation
from the first, to yield
. Therefore,
. Subtracting this from the
final equation and simplifying gives us
. We can factor this equation as
. So we now have
(
). This is because
(
). We note that a power of 2 is odd if and only if it is raised to the 0th power.
Therefore, for this equation to be true, we must have either m, l, or k equal to 2, such that we have
a one on the right hand side of the equation. Since
is negative in the equation, while
and
are positive, we have two different cases to consider (note that technically m and k are
interchangeable, at least in this equation.)
Case 1: l=2. When we have l=2, we now have
. Adding 1 to both sides
gives us
. We factor out 2, as
. So we divide by 2 on both
sides, giving us
. Noting the parities of the numbers, again we see that either
m or k must equal 3, giving us
. Since both m and k are positive, it doesn’t matter
which equals 1. So we let m equal 3, giving us
. So, therefore
.
However, no rational value of k will give us 250 as a power of 2, so this is impossible.
Case 2: m or k=2. When this happens, we have
. Subtracting 1 to both
sides gives us
. Once again, we factor out the greatest possible power of 2
out of this equation, which happens to be 1. So we divide both sides of the equation by , giving
us
. Once again, using parity we know that either m or k equal 3, such that
we get . Since all powers of 2 must be positive, we cannot have m equal 3, as then k must be
negative. Therefore k must equal 3, which gives us that
. However, then
must equal
, but this isn’t possible. Therefore, this case is impossible.
Since both cases don’t work, we see that there are no tworrific sequences of length 4.
Victor Chen, ID 14868
USAMTS Round 2, Solution to Problem 2/2/22
Also, the method used above is legitimate because if we have either m, l, or k less than or equal to
1 or 2 (we have negative exponents), there is no way to cancel out the fractions that result, as
there are no negative powers of 2 in a tworrific sequence. Therefore, we can use our division of
powers of 2 to factor and simplify all tworrific sequence.
We see now, that in general, for a sequence to be tworrific, we must take 2011, add or
subtract 1, divide by the largest power of 2 possible, again add or subtract 1, divide by the largest
power of 2 possible, and keep doing this until we get to 1. But what is the minimal number of
steps needed to do this? To minimize the number of steps needed, we need to add or subtract 1 to
get a number with as high a power of 2 in its prime factorization as possible. That way, we will
reduce the number of steps needed to get to 1. Here are the minimum number of steps needed to
get from 2011 to 1:
2011
Start
2012
Add 1
503
Divide by
504
Add 1
63
Divide by
64
Add 1
1
Divide by .
We see that there are three divisions, and since there is one division for each variable
between 1 and 2011, there are 3 variables, and therefore the shortest tworrific sequence consists
of 3+2=5 terms.
(b) Since the shortest possible tworrific sequence consists of 5 terms, it must be of the form 1, a, b, c,
2011. Why? 1 must be at the start of the sequence, and since a tworrific sequence of the form 1, a,
b, 2011 doesn’t work, nor a tworrific sequence of the form 1, a, 2011, nor a sequence of the form
1, 2011, a, b, c, 2011 therefore must be at the end of the sequence. Then, we have :
Subtracting the first equation from the second gives us that
, or
.
Then, subtracting the third equation from the fourth gives us that
, or
. Substituting our value for b from the first result gives us that
. Subtracting 1 from both sides gives us that
.
We then factor out and divide by 2, to get that
. Therefore,
one of the powers must equal 2! But which one? That is answered by casework:
Case 1: l or j equal 1. When this happens, we have
. Therefore,
we have
. We divide by 2 to get that
.
Yet again we see that either m or k or j must equal 2, so there is a 1 in the right hand side. We
Victor Chen, ID 14868
USAMTS Round 2, Solution to Problem 2/2/22
break up yet again into separate cases:
Case 1a: j=2. If this happens, we have that
. We factor out 8, getting that
. Then, we must have a 1 in the right hand side. However, if m or k is 6, then
the other variable must be
. This, however isn’t possible, since no positive integer
power of 2 results in 62. Therefore, there are no solutions for this case.
Case 1b: m or k=2. If this happens, we have
. We factor out 2, getting that
. However, then j must equal 3, as if m is 3 then
is negative. But when j
equals 3,
must equal 252, which isn’t possible, as no integer power of 2 results in 252.
Therefore, there are no solutions for this case.
Case 2: m or k equals 1. When this happens, we have
. We factor
out 4, yielding
. Then, either l or j, or m is 3. Again we branch off
into different cases:
Case 2a: m=3. When this happens, we have
. However, this isn’t possible,
as
and
must be positive. But from the equation above, we have that
! Therefore, there are no solutions for this case.
Case 2b: l or j =3. When this happens, we have
, or
Then, we factor out 4 out of the equation, yielding
. Since m cannot be 3
using the same reasoning from case 1b and 2a, by parity j must be 5. Then we have that
, and since
, m therefore is 11.
We only find solutions for case 2b. Since m and k are interchangeable, we can interchange the
values that we found earlier between them. Therefore, m and k are 1 and 11, in some order, and
similarly l and j are 5 and 3, in some order. We now substitute these values into the original
equations as shown:
We now solve each case of equations, and find that there are 4 tworrific sequences of length 5.
The sequences are as shown:
.
Victor Chen, ID 14868
USAMTS Round 2, Solution to Problem 3/2/22
Label each vertex of the cube, except for Richard and the
F
E
Gortha beast A, B, C, D, E, and F. Label the
vertices with Richard and the Gortha beast R
R
G
(behind) A
and G, respectively. We now calculate the
respective probabilities of Richard throwing
D
B
C
A
Let the probability of Richard throwing the potato to the Gortha be a when Richard starts out with the
potato. Note that the probability of person C starting out with the potato and feeding the Gortha is also a,
as his position is the exact same as Richards, except that the cube is flipped. We can apply this same logic
to person E, so his probability of feeding the Gortha is also a. Now we consider persons F, B, and D. Note
that if person B starts out with the potato, Richard has a 1/3 chance of getting the potato and then feeding
the Gortha, and so does person C, but person E has a 1/27 chance of getting the potato at the start and then
feeding the Gortha as it will take three passes at minimum to get the potato to person E and each pass
occurs with probability 1/3. Since the infinite sequence is equal for everyone as only persons R, C, and E
can feed the potato, the probability of Richard feeding the Gortha if person B starts with the potato is a/3.
This same logic can be applied to persons F and D because of the symmetry of flipping around the
vertices. Our next case is if person A starts with the potato. Since Richard, B, and C are equally distanced
from person A, and it takes a minimum of 2 passes to get the potato to Richard, the probability of Richard
feeding the Gortha is a/9. The final case is if the Gortha starts out with the potato. In this case, since no
one fed the Gortha, the probability of this happening is 0.
the potato to the Gortha beast.
We now add up our respective probabilities to get that
, as we note that the
probability that either Richard, person B, or person C feeds the Gortha is already probability 1. The
covers the cases where there are infinite sequences, and covers any possibilities of both cases.
We solve this equation to get that a=9/19, so the probability that Richard feeds the Gortha is ⁄
.
Victor Chen, ID 14868
USAMTS Round 2, Solution to Problem 4/2/22
D
A
F p
I
s
o
k
B
m
r
h
G
q
n
l
E j
C
Consider DBC and ABC. Since they both share base BC, and AD||BC, which means that their heights
are equal, we see that both triangles have equal areas. We can also express the area of triangle ABC as rs.
Therefore, we have by substitution
, where point E is the altitude from point D of
triangle DBC. We now see that if we can prove DE=3r or DI=2IE, then we will have that AB+AC=2BC.
To prove this, we will let the altitude from C intersect DB at point F, and we will also let the altitude from
B intersect DC at G. Also, since angles DFI, CEI, and DGI are right angles, and by vertical angles DIF
and CIE, DIG and BIE, and FIB and CGI are congruent. Therefore, DIF and CIE, FIB and GIC,
and BIE and AIG are similar triangles by AA Similarity. Since IE has length r, if we let BE have
length h, let CE have length j, BI have length k, CI have length l, DI have length m, GI have length n, IF
have length o, BF have length s, DF have length p, and DG have length q, we have that:
We can cross multiply these equations to get that
Therefore, we have that
(a)
Also by cross multiplying we have that
Victor Chen, ID 14868
USAMTS Round 2, Solution to Problem 4/2/22
Therefore,
and
Substituting this back into equation (a) gives us that
Dividing both sides by r and cross multiplying yields
Also, by the Pythagorean theorem we have that
We can substitute and subtract the equations to yield 3 new equations:
We can now substitute and simplify to get that
so DE=3r. Therefore, we have
as desired.
, or in other words,
, so
. Therefore, DI=2IE,
, so we have
Victor Chen, ID 14868
USAMTS Round 2, Solution to Problem 5/2/22
We note that when Zara guesses a number between 1 and 2011, if she doesn’t guess the number on the
spot, then she divides the possible of choices for Ada into two sections. For example, if Zara guesses
1347, if Ada’s number was not 1347, then Ada’s number was in 1 of 2 possible ranges: 1-1346, and 13482011. However, due to Ada’s rule, if Zara doesn’t guess Ada’s number on the spot, then Ada will add or
subtract 1. Therefore, if Ada’s number wasn’t 1347, then her new number ranges can be 1-1347 or 13472011. However, this isn’t the best way to narrow down Ada’s number, as one of the number ranges above
is much larger than the other number range. Since we are looking for the minimum number of guesses
required to find Ada’s number, we want to split the 2 ranges equally. Therefore, Zara’s first guess should
be directly in between 1 and 2011, so Zara should guess 1006, the middle number, as her first guess.
Since we want the minimum number of guesses guaranteed to find Ada’s number, we try to find Ada’s
number in the case that takes the most steps. Therefore, unless Ada’s number is guaranteed (e.g. the
number range of her possible numbers is 1), then we will ignore all cases in which Zara guesses Ada’s
number directly on the spot.
We know that Ada’s number is in 1 of 2 number ranges:1-1006, or 1006-2011. However, we do not have
to use casework here, as due to symmetry these cases are the same. Therefore, we only have to examine
one of the two cases, as the other case is exactly the same due to symmetry. As I deal better with lower
numbers, I will just analyze the 1-1006 case.
As usual, we want to split the number range of 1-1006 into two equal sections. Therefore, we should pick
the middle number, 503. We now split the ranges of Ada’s second number into 2 sections again: 1-502
and 504-1006. But which one to pick this time? Note that in this range we can’t just pick any case, as 1 is
on the original boundary of 1-2011 (e.g. Ada cannot pick an integer less than 1 as her number), whereas
we can pick a number greater than 1006, as long as it is less than or equal to 2011. Therefore, it will take
less guesses to pinpoint Ada’s number in the range of 1-503 than in the range of 504-1006. Since we want
to take the case in which it takes more cases as it guarantees that we will find Ada’s number in a
minimum of n guesses, we will now go with the range of 504-1006. Since we didn’t correctly guess
Ada’s number, Ada’s new number will be 1 more or less than her second number, and therefore her new
number can be in the range of 503-1007. Here again, as usual we guess the middle number of this number
range, which is 760. By guessing 760 we split the number range of Ada’s third number into 2 groups,
503-759, and 761-1007.
We now need to decide which number range to pick next. We know that we want to pick cases closer to
the center, as they are not close to any of the boundaries on the ranges. Therefore it will take more steps
to guaranteedly narrow down Ada’s number. Nothing that 503 is closer to one of the bounds of Ada’s
entire number range (1) than 1007 is to 2011, this means that the range for Ada’s 3rd number as 503-759
is closer to the side boundary of 1 than 761-1007 is. This means that it will take more guesses to guess
Ada’s number if Ada’s number is in the range of 761-1007, than if Ada’s number is in the range of 503759. We now let Ada’s 3rd number be in the range of 761-1007, and following our strategy, we guess the
middle number, 884. Since 761 is closer to 1 than 1007 is to 2011, we now let Ada’s 4th number be in the
range of 761-1008. Again we try to follow our strategy, picking the middle number, but here there are two!
Victor Chen, ID 14868
USAMTS Round 2, Solution to Problem 5/2/22
Which one do we pick? We note that the two middle numbers here are 884 and 885. We now note, that if
we pick 884, the largest range available to pick from is 885-1008(before adding or subtracting), whereas
if we pick 885, the largest range available to pick from is 761-884. However, the number range of 761884 is closer to the lower boundary of 1 than the number range of 885-1008 is to either 1 or 2011.
Therefore, we will pick the case of 885-1008 over the case of 761-884. Since Ada hasn’t added or
subtracted 1 yet to her 4th number, we will do so now, creating her 5th number. Therefore the new range of
possibilities is 884-1009. Yet again here there are two middle numbers, so we analyze which range is
closer to the middle. The middle numbers here in this range are 946 and 947. If we take the largest cases,
we get 947-1009 and 884-946. We note that the range of 884-946 is at minimum 883 numbers away from
1, or a minimum 1066 numbers away from 2011, whereas the range of 947-1009 is 946 numbers away
from 1, or a minimum 1002 numbers away from 2011. Since we want the minimum, we see that 884-946
is closer to 1 than 947-1009 is. Therefore, the range of 947-1009 is closer to the center. We now see that
Ada’s 6th number is in the range of 946-1010. Again we cut this in half to get a middle number of 978. As
usual, since 977 is closer to one of the limits that 1010 is, we pick the range of 979-1010 as the possible
range of Ada’s 6th number. Ada’s 7th number is now in the range of 978-1011. The middle numbers here
are 994 and 995. Since the range of 978-995 is closer to 1 than the range of 994-1010 is to 2011, we take
the range of 994-1010 as the possible range of Ada’s 7th number. Following her rule, Ada’s 8th number
can be in the new range of 993-1011. We take the middle number again, which is 1002. Here we have two
different ranges, 993-1001, and 1003-1011. Since 1011 is 1000 numbers away from 2011 at minimum,
while 993 is 992 numbers away from 1 at minimum, we take the range of 1003-1011 as the possible range
of Ada’s 8th number. Ada’s 9th number can now be in the range of 1002-1012. The middle number here is
1007. Since 1012 is 997 numbers away from 2011, the upper bound, while 1002 is 1001 numbers away
from 1, the lower bound, we take the lower range of 1007-1012 as the possible range of Ada’s 9th number.
Ada’s 10th number now can be in the range of 1006-1013, and here again there are two middle numbers:
1009 and 1010. The range of 1006-1009 is closest to 2011, by 1002 numbers, whereas the number range
of 1010-1013 is closest to 2011, by a minimum of 998 numbers. Therefore we pick the range of 10061009 as the range of Ada’s 10th number. Her 11th number can now be in the range of 1005-1010.
Here there are two cases. The two middle numbers are 1007 and 1008. If we guess 1007, the maximum
range is 1007-1011 (for Ada’s 12th number), with 6 numbers still, but it is still closer to one of the
boundaries (namely 2011). Here we can repeat this method until we hit the boundary, when then we will
be able to correctly determine Ada’s number. However, if Ada’s number is less than 1007, we will be
able to narrow down her new number range to 4 numbers instead of 6. Note that we can do the same thing
with 1008, but 1011 is closer to 2011 than 1005 is closer to 1. Therefore, we will consider these two cases
with 1007 only.
Case 1:Ada’s 11th number was in the range of 1008-1010. Therefore, her 12th number is in the range of
1007-1011. We guess 1009, and as usual we go for the case closer to the center. Therefore, Ada’s 12th
number was in the range of 1007-1008, and her 13th number is now in the range of 1006-1009. There are
two middle numbers here again, 1007 and 1008. We branch off into cases again, but first use 1007, as
1009 is closer to 2011 than 1006 is to 1:
Victor Chen, ID 14868
USAMTS Round 2, Solution to Problem 5/2/22
Case 1a: Ada’s 13th number is 1006. Her 14th number is now either 1005 or 1007. We guess 1007, and if
we guess wrong, this means that her 14th number is 1005. Her 15th number is now either 1004 or 1006.
We guess the higher number 1006, narrowing down Ada’s 15th number to 1004. Ada’s 16th number is
now either 1003 or 1005. At this point we see that every time we make a guess, we narrow down the two
possibilities of Ada’s new number by 1. We notice that we cannot narrow down the guesses until one of
the numbers is 1. It will take less guesses if the lower number is 1, so we need to count how many guesses
will it take to reach 1 from 1003. Since 1003-1=1002, we will need 1002 additional guesses until Ada’s
number is either 1 or 3. At this point we notice that this is Ada’s 1018th number. We guess 3 here, and if
Ada’s 1018th number was 1, then her 1019th number is 2 since 1 is the lower boundary of all of Ada’s
possible numbers. We now just guess 2, and we have used 1019 guesses to find Ada’s number in this case.
Case 1b: Ada’s 13th number is either 1008 or 1009. Her 14th number is now in the range of 1007-1010.
We guess 1008 here, as 1010 is closer to 2011 than 1007 is to 1, and if Ada’s 14th number was 1007, then
her 15th number is either 1006 or 1008. Here we proceed with our strategy in case 1a, except going
upwards, and find that it still takes 1019 guesses to pinpoint Ada’s number. If Ada’s 14th number is in the
range of 1009-1010, her 15th number is now in the range of 1008-1011. Note that this range is the same as
the range for Ada’s 14th number, except all the numbers have 1 added to them. We can keep repeating this,
as if Ada’s number is in the section by itself, then it will still always take 1019 guesses to pinpoint it
using the strategy from case 1a. Otherwise, we keep doing this until 1011 becomes the upper bound of
2011, which takes 1000 more guesses. Here, Ada’s 1014th number (after adding or subtracting 1) is in the
range of 2008-2011. We now guess 2009, and if Ada’s 1014th number was 2008, her new number is either
2007 or 2009. We use our strategy from case 1a, and use 4 more guesses to guarantee Ada’s number, at
2010. This takes 1019 guesses as well. However, if Ada’s 1014th number is either 2010 or 2011, we just
guess 2010 twice to guarantee Ada’s number, in 1016 guesses.
Case 2: Ada’s 11th number was in the range of 1006-1007. Ada’s 12th number is now in the range of
1005-1008. Noting that 1007 is 1003 numbers away from 2011 whereas 1006 is 1004 numbers away from
1, we guess 1006 to narrow the number of guesses required. If Ada’s 12th number was 1005, then her 13th
number is either 1006 or 1004, and we follow our strategy from case 1a, guessing Ada’s number in 1017
guesses. However, if Ada’s 12th number was either 1007 or 1008, then her 13th number is in the range of
1006-1009. We keep guessing the second lowest number here, following our strategy from case 1b, until
1009 eventually becomes 2011, in 1002 guesses. Ada’s range for her 1015th number is now 2008-2011,
and we proceed from case 1b, which takes an additional 5 guesses at maximum, for a maximum of 1020
total guesses.
We see that the maximum case takes 1020 guesses to guarantee Ada’s number, so therefore, if Zara
follows the optimal strategy of picking the middle number and pushing the range as close as possible to
one of the boundaries, it will take her at least 1020 guesses.
Victor Chen, ID 14868
USAMTS Round 2, Solution to Problem 6/2/2
Lemma: The rule that the robot follows gets the same result for a fraction , whether or not is in
simplest form.
Proof: Let
and
. The robot’s rule for yields
for results in
can factor
. Therefore,
. We now test the rule for and . The robot’s rule
. Since x and y both have a common factor of a, we
. Since the two fractions are equal, the robot’s rule gets the same
, giving that
result for fractions, whether in simplest form or not.
We now start by finding numbers that can get to 2 in one move. We try to find a fraction , already
simplified and in lowest terms and a positive integer , p and q being integers, such that
is possible, then
. Getting all the terms on the left hand side, we see that
. If this equals 0, then there are 2 different cases:
and
.
Case 1:
and
true then
. If this
, or
. Since q cannot equal 0, as is a common fraction, if this is
. However, then
and therefore
. Therefore,
However, as p is 0, q
can be any non-zero integer! We then explore this further, so we test this with p being 0, q being left as it
is , and n being left as it is. We have
. Therefore, it is possible to jump from 0 to any
number! Therefore, this means that if we can get from 2011 to 0, then you can get from 0 to 2011! But
first, we consider case 2.
Case 2:
. This equation can be simplified to
. We will
prove that this has no cases by a proof by contradiction. We will assume
p and q are relatively prime. Therefore,
this equation as
divides q, and
is in lowest terms, so therefore
divides p. However, we can express
, as cross multiplying gets the above equation. Therefore, this equation isn’t in
simplest form! Then what is? If we let
and
, then we have that
.
Therefore,
and
. However, then and aren’t relatively prime to each other!
This contradicts our original statement of assuming that p and q are relatively prime, so this case is
impossible.
Seeing as Case 1 has a possibility of getting to 2 from 2011, we try this case first. We know that it is
possible to get from 0 to 2, but is it possible to go from 2011 to 0? We find out, by expressing 2011 as
. Now, we need to find an integer n such that
. However, it is only possible for a
fraction to equal 0 if the numerator equals 0, so
. Therefore,
. However, since n
must be positive, this is impossible. Therefore, it is impossible to get from 2011 to 0 directly! Now, the
question is whether there exists a fraction that can possibly get to 0.
Victor Chen, ID 14868
USAMTS Round 2, Solution to Problem 6/2/2
Again here we will use a proof by contradiction. We will assume there is a fraction in lowest terms and
an integer n such that
by
, getting that
. Since
cannot equal 0, we can multiply both sides of the equation
. However, then
, and p is divisible by q! This contradicts
what was assumed of (that it’s relatively prime), so therefore it is impossible for any fraction get to 0!
Since 0 is the only number that can get to 2, we can conclude that it is impossible for the roving
rational robot to get to 2 from 2011.