A Wheel of a deal
Some BASIC-ic addition
Three people painting a house
A clueless detective job
Where in the Square
A Diophantine Problem
Bus
Chameleons
Let A and B denote the two given points on the wheel, where A is above B
relative to the ground. Draw a line from A parallel to the ground to intersect
the wheel at C. Draw another line from B perpendicular to the ground to
intersect the cicle at D. Let P the point of intersection of the lines
constructed. From the center (denoted by O) of the wheel, draw a line
perpendicular to the ground to intersect AC at Q, and to intersect the wheel at
R and S with R above S. Let r denotes the radius of the wheel. Since A is 3
inches above the ground, so
QS=3,
OQ = OS-QS = r-3,
and
RQ = RO+OQ = 2r-3.
Since triangle APB is an isosceles right triangle, by symmetry angle OPQ is 45
degree. Thus
PQ = OQ = r-3
and
QC = AQ = AP+PQ = 2+(r-3) = r-1.
The chords AC and RS of the wheel intersect at Q, so
AQ * QC = RQ * QS,
(r-1) * (r-1) = (2r-3) * 3,
r2-8r+10 = 0,
r = (8+-sqrt(24))/2,
r = 4+-sqrt(6).
Since 0 < PQ = r-3, so r>3. This implies that r = 4+sqrt(6) is the only
solution. The diameter of the wheel is 2r = 8+2*sqrt(6) inches.
Let b denotes the base. Since there are at least 6 distinct letters, b>=6. Let
(1), (2), and (3) denote the given additions. In (2) there is a carry of at most
1 into the thousands column and A<>B, so
(4) B = A+1.
Also, there is a carry of at most 1 into the hundreds column, so
C+F+{0,1} = b+C,
F = b-{0,1}.
But F<b, so F=b-1. Now the units column in (2) gives
2F = B+{0,b},
B = 2(b-1)-{0,b} = 2b-2 or b-2.
Assume B=2b-2. Since B<b, so 2b-2<b or b<2 which is impossible. Therefore B=b-2
and A=b-3 by (4).
Next the units column of (3) gives
B+A = D+{0,b},
2b-5 = D+{0,b},
D = 2b-5 or b-5. Again D=2b-5 will lead to the contradiction that B<5, so D=b-5.
Furthermore the units column of (1) gives
D+F = E+{0,b},
2b-6 = E+{0,b},
E = 2b-6 or b-6. It is easy to see again that E=2b-6 will lead to b<6, so E=b-6.
Putting the values of A, B, D, E, and F into (1), we have:
(b-3) (b-2) C (b-5)
(b-6) (b-6) (b-1) (b-1)
-----------------------
(b-1) C C (b-6)
It is easy to see that there are carries of 1's into the tens and the hundreds
columns. For the hundreds column, we have
(b-2)+(b-6)+1 = C+{0,b},
C = 2b-7 or b-7.
In the first case, 2b-7<b or b<7. But b>=6, so b=6. In the second case,
b-7>=0 or b>=7. For the thousands column, we have, corresponding to these two
cases,
(b-3)+(b-6)+{0,1} = b-1,
b = 8 or 7.
It follows that the first case is not true, and C=b-7 and b=7.
Now we can substitute b=7 into the letters to get, A=4, B=5, C=0, D=2, E=1, F=6.
Finally, the desired addition becomes 6045+2114=11162.
Let a, b, and b be the number of hours taken by Adam, Beth, and Cory
individually to paint a room, respectively. If a room can be painted in x hours,
then 1/x of the room can be painted each hour. Also, if a portion x of a room
can be painted each hour, then the whole room can be painted in 1/x hours.
Using these two facts and the given information, we have the following
equations:
(1) 1/(1/a+1/b+1/c) = a-6 = b-1 = c/2.
The last three equal expressions in (1) give
(2) a = b+5,
(3) c = 2b-2.
Using (2) and (3) in the equality between the first and last expressions in (1),
we have
1/a+1/b+1/c = 2/c,
1/a+1/b = 1/c,
1/(b+5)+1/b = 1/(2b-2),
(2b+5)/(b^2+5b) = 1/(2b-2),
(2b+5)*(2b-2) = b^2+5b,
4b^2+6b-10 = b^2+5b,
3b^2+b-10 = 0,
(b+2)(3b-5) = 0.
Since b>0, so b=5/3 and a=b+5=20/3.
We need to find 1/(1/a+1/b) which is 1/(3/20+3/5) = 1/(15/20) = 4/3.
It takes Adam and Beth 4/3 hour to paint the room.
Let n, d, and q denote the dividend, divisor, and quotient of the long division.
Then n=d*q, and after adjusting the decimal points, we have
(1) 10000(10n)=(100d)(1000q).
It is easy to see, from the long division, that 10n, 100d, and 1000q are 2-, 3-,
and 4-digit integers, respectively, having no ending zero. Therefore, by (1),
10000=(2^4)(5^4) divides (100d)(1000q). Since 100d and 1000q are integers with
no ending zero, each of these integers is not divisible by 2 and 5 at the same
time. So there are two cases to consider as follow.
(i) 5^4 divides 100d and 2^4 divides 1000q
Since 100d is a three digit integer, 100d=625 is the only possibility. Therefore
the 2-digit integer 10n has to be 63 by checking the locations of the digits
within the long division. So n=6.3 and d=6.25 and q=n/d=1.008. Redoing the long
division using (n,d,q)=(6.3,6.25,1.008), we find that this solution checks.
(ii) 2^4 divides 100d and 5^4 divides 1000q.
Each of the multiples of 54=625 ends with either 125, 375, 625, and 875, so
could the last three digits of the 4-digit integer 1000q. But the long division
tells that 1000q has 0 as both its hundreds' and tens' digit, and none of 125,
375, 625, and 875 having 0 as both its hundreds' and tens' digit. Therefore this
case could not happen.
In summary, (n,d,q)=(6.3,6.25,1.008) is the only solution.
Let ABCD be the given square such that PA = 7, PB = 35, PC = 49, and PD = x.
Assume A', B', C', D' be the feet of the perpendiculars from P to AB, BC, CD,
DA, respectively. Let PA' = a, PB' = b, PC' = c, and PD' = d. By the
Pythagorean theorem, we have
(1) d^2+a^2 = 7^2
(2) a^2+b^2 = 35^2
(3) b^2+c^2 = 49^2
(4) c^2+d^2 = x^2
Add (1) and (3) to get
(5) (a^2+b^2)+(c^2+d^2) = 7^2+49^2.
Using (1) and (4), (5) becomes
35^2+x^2 = 7^2+49^2,
x^2 = 7^2+49^2-35^2 = 49(1+49)-35^2 = 49*50-35*35 = 7*5*(7*10-5*7) = 5^2*7^2.
Therefore x = 5*7 = 35.
Let f(x) = x^4 + (x+1)^4. Then y^2 + (y+1)^2 = f(x), and so
2y^2+2y+(1-f(x)) = 0. The discriminant of this quadratic in y is
D = 4-8(1-f(x)) = 4(2f(x)-1) which must be a square, or simply 2f(x)-1 must be a
square. Now 2f(x)-1 = 4x^4+8x^3+12x^2+8x+1 = (2x^2+2x+2)^2-3 which is a square
iff (2x^2+2x+2)^2 = 4. Therefore 2x^2+2x+2 = 2, or x^2+x = x(x+1) = 0 whose only
nonnegative solution is x = 0. Thus f(x) = 1 and 2y^2+2y = 0. Therefore
y(y+1) = 0 and whose nonnegative solution is y = 0. Hence the only nonnegative
integer solution to the given equation is (x,y)=(0,0).
Let b, s, and t be the number of students, number of buses, and the number of
students per bus, respectively. The given conditions give:
(1) s = 33b+1
(2) s = t(b-1)
(3) t > 33
By (1) and (2), we have
(4) (t-33)b = t+1.
Clearly that t-33 divides t+1, or
t-33 divides (t-33)+34 by (3). It follows that t-33 also divides 34. Since the
factors of 34 are 1, 2, 17, and 34, so t-33 may be each of these factors. Hence
t = 34, 35, 50, or 67. By (4), b = 35, 18, 3, or 2. Finally, by (1), s = 1156,
595, 100, and 67.
Let 0, 1, 2 denote the colors red, green, and blue. The sum of the colors of any
two chameleons before the meet and after the meet are the same modulo 3 (i.e
0+1 = 2+2 (mod 3)). Now the sum of the colors mod 3 is 13*0+15*1+17*2 = 1 (mod
3). If it is ever possbile for all chameleons to become the same color, then
the sum modulo 3 should be (13+15+17) * 0, 1, or 2, which is always 0 and never
1. Hence it is not possible.
Let s be the desired speed, so 2/60 = 1/30+1/s. Since there is no solution, so it is impossible.
It takes (1.5/4+15/4)/(15+1.5) hour = 1/4 hour = 15 minutes.
Take k coins from the kth box, k = 1,...,10, and put onto balance to get the total weight t. If the weight of one copper is c, that of one gold coin is g, and the nth box contains the copper coins. Then t = nc+(1+2+...+10-n)g, t = nc+(55-n)g, n = (t-55g)/(c-g).
Let b be the number of blue balls, so C(b,5)/C(n,5) = 1/2. So (1) b(b-1)(b-2)(b-3)(b-4)/n(n-1)(n-2)(n-3)(n-4) = 1/2, 2b(b-1)(b-2)(b-3)(b-4) = n(n-1)(n-2)(n-3)(n-4) Since the right side is a product of 5 consecutive integers, one possiblility is that 2(b-4), b, b-1, b-2, and b-3 are consecutive integers. Therefore we can solve b and n using the equations 2(b-4) = n and b-3 = n-4. now n = 2b-8 = b+1, b = 9, n = b=1 = 10. By checking, there are no solutions for (1) if n = 6, 7, 8, and 9. Hence n = 10 is the desired answer.
There is a 2/6 = 1/3 chance to die if spinning the barrel again, and 1/4 chance to die if pulling the trigger. So I would pull the trigger.
The expected return (1+...+6)/6*100 = 2100/6 = 350 dollars.
Clearly if the dice is rolled below 4, you want to continue rolling it at any
time. So the desired charge is >= $4.
Let S_k be the event that you want to roll the third time if you have rolled k
or above on the second roll. So S_k exists only when you rolled once and want
to continue, and also we are only interested in S_k when k >= 4.
Now E{S_4} = (3/6)((4+5+6)/3) + (3/6)((1+2+3+4+5+6)/6) = (1/2)(5+21/6) =
(1/2)(51/6) = 51/12 = 17/4,
E{S_5} = (2/6)((5+6)/2) + (4/6)((1+...+6)/6) = (1/3)(11/2+42/6) = (1/3)(75/6) =
75/18,
and E{S_6} = (1/6)6 + (5/6)((1+...+6)/6) = (1/6)6 + (5/6)(21/6) =
(1/6)(6+105/6) = (1/6)(141/6) = 141/36.
Therefore E(S_4) > E(S_5) > E(S_6).
Let T_k be the event that you want to roll the second time if you have rolled k
or above on the first roll. Clearly we are only interested in T_k where k >= 5
since E{S_4} = 17/4 < 5.
Now E{T_5} = (2/6)((5+6)/2) + (4/6)E(S_4) = (1/3)(11/2) + (2/3)(17/4) =
(1/3)(11/2+34/4) = (1/3)(56/4) = 14/3 = 4.67.
and E{T_6} = (1/6)(6) + (5/6)E(S_4) = (1/6)(6 + 5*17/4) = (1/6)((24+85)/4) =
(1/6)(109/4) = 109/24 = 4 13/24 < 4.67.
Hence the desired charge is $4.67 and the strategy is:
If the first roll is 5 or above we stop. Otherwise we continue and if the
second roll is 4 or above we stop. if the second roll is below 4, we continue
the third roll.
Let x and y be the positions of the first and second breaks on the stick from
its left. Then the feasible breaks are within the regions of the two triangles
{(0,1/2), (1/2,1), (1/2, 1/2)} and {(1/2, 1/2), (1, 1/2), (1/2, 0)} in the
xy-plane. Hence the desired probability is the area of these triangles which is
1/4.
end.
Let f_n(x) = (...(x^x)^x)...) with n x's. Assume lim f_n exists, so it is = 2. Then lim_n f_(n+1)(x) = lim_n (f_n(x)^x) = 2, 2^x = 2, x = 1. But f_n(1) = 1 and lim f_n = 1 <> 2 which is impossible. Hence there is no solution for x.
Since dx/(2-x) = 2dt, -d(x-2)/(x-2) = 2dt, -ln(x-2) = 2t+c' for some constant c, x-2 = ce^(-2t) where c = e^(-c'), x = 2+ce^(-2t). Also 1 = x(0) = 2+c, so c = -1. Therefore x = 2-e^(-2t).
Let u = x and dv = f''dx, so du = dx and v = f'.
By integration by parts,
integral{x=0,2}{xf''}dx = uv{x=0,2} - integral{x=0,2}{v}du =
xf'|{x=0,2}-integral{x=0,2}{f'}dx = 2f'(2)-f(2)+f(0).
Let u = F(x) and dv = f(x)dx, so du = f(x) and v = integral{x=-inf,inf}{f(x)}*dx = 1.
By integration by parts,
E{F(x)} = integral{x=-inf,inf}{F(x)*f(x)}dx
= integral{x=-inf,inf}{u}dv = uv - integral{x=-inf,inf}{v}du
= F*1|{x=-inf,inf} - integral{x=-inf,inf}{1*f(x)}dx
= 1 - 1 = 0.
Let H be the event that a coin picked is being flipped 10 times to be heads; let D be the event that a coin picked has double head. By Bayes' theorem, p(D|H) = p(H|D)*p(D)/p(H), where p(H|D) = 1, p(D) = 3/6 = 1/2, and p(H) = (3/6)*1 + (2/6)*(1/2)^10 + (1/6)*0 = 3/6+1/(3*2^10) = 1/(3*2^10)*(3*2^9+1) = 1537/3072. So p(D|H) = 1*(3/6)/(1537/3072) = 1536/1537.
Here are the steps to compute y: (1) A(n-1)*x (2) (A(n-2)+A(n-1)*x)*x ... (n-1) [(A(1)+A(2)*x+...+A(n)*x^(n-1)]*x This needs only n = O(n) multiplications and n-1 = O(n) additions while the usual way needs 1+2+...+(n-1) = O(n^2) multiplications and n-1 = O(n) additions.
If there is one tiger, then it will eat the sheep. If there are two tigers, neither of them will eat the sheep to avoid being eaten later on by the other tiger. If there are 3 tigers, then all of them will try to eat the sheep. Once one tiger eat the sheep and turned into another sheep, there remains two tigers and one sheep. As explained before, none of these two tigers will eat the one only sheep. If there are 4 tigers and one sheep, then none of the tigers will eat the sheep. For if not, the tiger who ate the sheep will turn into another sheep and there would be 3 tigers and one sheep. As seen before, this new sheep will be eaten by another tiger afterwards. But the tigers are smart, so this is not possible. Repeat the same reasoning, it is easy to see only one tiger will eat the sheep and turn into another sheep if the number of tigers is odd; and no tiger will eat the sheep if the number of tigers is even.
If there was one person wearing a red hat, then he sees no other wearing one and so he left. If there were two people wearing red hats, then each of them only sees each other wearing red hat. Since they are no sure if each of themself wearing red hat the first time door is opened, they stay in the room. But on the second time the door is opened, each of them know that he is wearing a red hat and both of them leave. Similarly, if there are n people wearing red hats, then each of them sees n-1 other people wearing red hats. So they are no sure if they wear red hats the first n-1 time the door is opened. But on the nth time the door is opened, each of them know he is wearing a red hat and so these n people leave. Hence n = 6 people went out the room.
I will pick one of the guard and ask the compound question: Is it true or false that (((you are a liar) or (this door is to heaven)) and ((you are honest) or (this door is to hell)))? (i) The guy is honest and this door is to heaven. The guard's answer is ((F or T) and (T or F)) = T. (2) If the guy is lying and this door is to heaven: The guard's answer is not ((T or T) and (F or F)) = not F = T. (3) If the guy is honest and this door is to hell: The guard's answer is ((F or F) and (T or T)) = F. (4) If the guy is lying and this door is to hell: The guard's answer is not ((T or F) and (F or T)) = not T = F. Hence, if the answer is T then the guard being asked is for heaven, otherwise the guard is for hell.