The worlds first riddle!

Paula, how did you know my favourite color is green? I shall treasure it always. Perhaps one day I may be in a position to repay your kindness. In the meantime, can you spare a few bucks for a coffee?



MAP

Mark:
The duration of the trip is infinite.


I know what you mean, what with all the hold-ups, checks and double-booking. It takes longer to arrive.



Part 1:

Let y denote latitude in radians, so that
y=0 at the equator and y=pi/2 at the pole.
Let x denote longitude in radians, with x=0 at prime meridian.
Let t be total distance travelled, measured in radii.

Then dy/dt = cos(pi/4) = 1/sqrt(2).
So t=y*sqrt(2)+C, and initial conditions give C=0.
When we finish, y=pi/2, so t=pi/sqrt(2) radii.
Use (25000 miles = 2 pi radii) to conclude
t = 6250 sqrt(2) miles = 8839 miles.

Part 2:

dx/dt = (sin(pi/4))*(1/cos(y)).
Then dx/dy=1/cos(y).
x = log(|sec y + tan y|) + C.
Again initial conditions imply C=0.
We can solve to find y=(pi/2)-2*arctan(exp(-x)).
Substituting x=2*pi we find
y=(pi/2)-2*arctan(exp(-2*pi)),
so that the distance from the pole is:
(pi/2)-y=2*arctan(exp(-2*pi)) = 0.003735 radians = 14.86 miles.

Part 3:

If we change the heading to north by northwest, we find:
dy/dt = cos(pi/8).
dx/dt = (sin(pi/8))*(1/cos(y)).
dx/dy = (tan(pi/8))*(1/cos(y)).
x = (tan(pi/8))*(log(|sec y + tan y|)) + C.
y = (pi/2)-2*arctan(exp(-x/tan(pi/8))).
Recalling tan(pi/8)=sqrt(2)-1, and substituting x=2*pi, we find:
y = (pi/2)-2*arctan(exp(-2*pi/(sqrt(2)-1))),
so that the distance from the pole is:
(pi/2)-y=2*arctan(exp(-2*pi/(sqrt(2)-1)))=0.0000005167 radians
= 0.002056 miles = 10.855 feet.




"ABC = A^3 + B^3 + C^3 (A, B, and C are 1-digit numbers)"


If abc is perfect cube then abc = a^3 + b^3 + c^3
e.g., 153 = 1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153

ABC = A^3 + B^3 + C^3 (A, B, and C are 1-digit numbers)
a^3+b^3+c^3<=(a+b+c)(ab+bc+ac)

a^2*b+a^2*c+b^2*a+b^2*c+c^2*a+c^2*b+3*abc-a^3-b^3-c^3>=0

a^2*(b+c-a)+b^2*(a+c-b)+c^2*(a+b-c)+3*abc>=0


Evaluate if a integer number is a Perfect Cube or
number of Armstrong ( Number > 1 ).

//---------------------------------------------------------------------------
#pragma hdrstop
#include <iostream.h>
#include <conio.h>
//---------------------------------------------------------------------------
#pragma argsused
int main(int argc, char* argv[])
{
int number;
char ans;
do
{
do
{
cout<<"Enter a Integer Number greater than 1 : ";
cin>>number;
}while ( number <= 1);
int result;
int Naux;
result = 0;
Naux = number;
int dig;
while (Naux != 0)
{
dig = Naux % 10;
Naux = Naux / 10;
result = result + pow(dig,3);
}// end while (Naux != 0)
if (number == result)
cout<<"The number is PERFECT CUBE !!!"<<endl;
else
cout<<"The number is NOT PERFECT CUBE "<<endl;
cout<< "Continue (Y/N) : ";
cin>> ans;
}while(ans=='Y' || ans=='y');
getchar();
return 0;
}




Two players:
They start with an unlimited supply of coins, which have denominations 1, 5, 10, 25, 50, 100 cents. They take turns placing a coin into a pot, which is initially empty.

There is a target amount: $6.78 or 678 cents. The amount in the pot may not exceed this target.

The winner is the player who puts in the last coin, reaching the target.


Who wins



Kicky, an evil goblin is chasing us.
We are on a train, heading due east at 60 kilometres per hour.
The goblin starts out 30 kilometres due north of us.
He runs at a constant speed, but (fortunately for us) has lost the ability to plan ahead; he always heads directly towards our current position.

(A clever goblin would aim in front of us, since our train's future motion is so predictable; but our goblin is not clever.) He catches up to us after we have travelled 100 kilometres (that is, after 100 minutes).


How fast does the goblin move

Genius' pick green.

I just brewed a fresh pot, so here you go... (Warning: Rocket fuel!) And I thought you might like to read the paper as you sipped.

"e.g., 153 = 1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153 "
153 is one of the answers.

COINS
I think the first player wins. However, he can't open with 5 or 50.

GOBLIN
71.8 km/h

Paula, the coffee is just right. I like it strong enough to stick shingles on a barn roof. Now take it easy while I run you a bath.




ABC
3 main
153 370 371 407

4 main
1634 8208 9474

5 main
54748 92727 93084




Mark:
COINS
I think the first player wins. However, he can't open with 5 or 50.



The first player wins.
The second player will win if the target is equivalent to one of
0, 2, 4, 6 or 8 modulo 15; but 678 is equivalent to 3 modulo 15,
so the first player wins.

Put another way:
the nonnegative integers equivalent to 0, 2, 4, 6 or 8 modulo 15
will be called "winning numbers"; the other nonnegative integers
will be called "losing numbers".
You win if and only if after your turn the amount
still needed to reach the target is a "winning number".

If, at the start of your turn, the amount needed is a "losing number" (so that your opponent should lose), you can always contribute a coin which transforms it into a "winning number".
If it was 1, 3, 5, 7 or 9 modulo 15, you put in a 1-cent coin to reduce the amount needed to 0, 2, 4, 6 or 8 modulo 15
(without making the amount go negative).

If it was 10, 12 or 14 modulo 15, you put in a 10-cent coin, to reduce the amount needed to 0, 2 or 4 modulo 15.
If it was 11 or 13 modulo 15, you put in a 5-cent coin.
In all cases you will not make the amount go negative.

But if, at the start of your turn, the amount needed is a "winning number" (so that your opponent should win and you should lose), any coin you put in will either transform it into a "losing number"
or make the amount go negative; in either case you will lose.

If the amount started at 0, 2, 4, 6 or 8 modulo 15, contributing a 1-cent coin will transform it to
14, 1, 3, 5 or 7 modulo 15; contributing a 5-cent or 50-cent coin will transform it to
10, 12, 14, 1 or 3 modulo 15; contributing a 10-cent, 25-cent or 100-cent coin will transform it to
5, 7, 9, 11 or 13 modulo 15 (or make the amount go negative).



Mark:
GOBLIN
71.8 km/h (Close enough)

Let t denote time in hours.
Let (x,y) be the goblin's current position, with our initial position being (0,0).
Let u=60 be our speed, v the goblin's unknown speed, and r=u/v their ratio.
Let the goblin's initial position be (0,g) where g=30.
Let "^" denote exponentiation (as in LaTeX).
The goblin's motion satisfies the differential equations:

(dx/dt)^2 + (dy/dt)^2 = v^2 (the goblin moves at speed v);

x - y dx/dy = ut (he is always pointing at our current position).

The initial conditions: at y=g, x=t=0.

The solution: the goblin's position (x,y) at time t is given parametrically in terms of y:

t = (g/(v(1-r^2)) - (g/2v) ( (y/g)^(1+r)/(1+r) + (y/g)^(1-r)/(1-r) ).

x = (gr/(1-r^2)) + (g/2) ( (y/g)^(1+r)/(1+r) - (y/g)^(1-r)/(1-r) ).

One can calculate dt/dy and dx/dy, and verify that the differential equations are satisfied.

When y=g we get t=0 and x=0, which meets the initial conditions.
When y=0 we get t=g/(v(1-r^2)) and x=gr/(1-r^2).

The last equation gives 100=30r/(1-r^2), whence r=(sqrt(409)-3)/20 and v=60/r=3*(sqrt(409)+3)=69.67124525 km/hour.






Atlas is a country with 2001 citizens. Each carries a different national security number from 1 to 2001. In addition, there is the King of Atlas, Antonyms the First, who has been assigned the number 0, silent witness of his modest nature.

The king has 2001 boxes stuffed with money to celebrate the year 2001. The boxes are numbered from 1 to 2001. Box number n contains (inexhaustibly many) envelopes of n Atlas dollars each.

The King, who does not dislike a bit of math, has also divided his citizens, including himself, into two groups.

The first group passes a box. If two people in the group have national security numbers whose sum equals the number of the box, they jointly take an envelope out of the box, and each takes an amount of Atlas dollars out of the envelope, equal to their national security number. If one citizen has a number which is half the number of the box, one can take an envelope with all n dollars for oneself. The group goes on to the next box and does accordingly. The second group does likewise.

For instance, if Mr. 1200 and Mrs. 300 are in the same group, they step forward the moment their group reaches box 1500, and take an envelope of 1500 dollars. Mr. 1200 takes 1200 dollars and Mrs 300 takes 300 dollars. Mrs. 750 passes box 1500, and takes the envelope with all 1500 dollars.

The groups are conceived so that at each box, both groups will have taken the same number of envelopes out of the box.


Question 1:

Does citizen number 2001 belong to the King's group


Question 2:

Which citizen(s) will have received the highest amount of Atlas dollars, once both groups have passed all boxes




The sentence below feature three numbers. All read either backwards or forwards.

"The hen opened the door but a bucket ended the escape when the hen knocked it over. Now that has ended your escape, said the farmer when he caught the hen."


What are the numbers



Complete the following sequences:

a. Cat, cat, dog, cat, cat, cat, dog, cat, cat, cat, cat, dog, cat, cat, cat, cat, cat...

b. 1, 2, 6, 24, 120, 720...

c. red, orange, yellow, blue, purple...

d. a, b, d, g, k, p...

e. 1, 2, 2, 3, 3, 3, 4, 4, 4...

SENTENCE
one, two, ten

SEQUENCES
a. dog
b. 7!
c.
d. v
e. 4

Tryagain wrote:
" Now take it easy while I run you a bath."

You will? Do you have a candle you can light?

< the thought of a softly illumed room with warm running water makes paula's eye's grow, heavy...and heavier...heavier..... zzzzzzz....... "wake up self!" >

Would you let me know when it's ready?

153 and 371 are what I was looking for.

However, the second part calls for three 2-digit numbers such that (AB)^3 + (CD)^3 + (EF)^3 = ABCDEF.

c. red

ATLAS
1) no
2) #997 ($502,488)

Paula, "Bath - "Would you let me know when it's ready?"

Sure thing. First, I've gotta run to git the tin bath from the barn. Then, I gotta git the water from the kreek…in the meantime, perhaps you may like to start plowing the north field?


"However, the second part calls for three 2-digit numbers such that (AB)^3 + (CD)^3 + (EF)^3 = ABCDEF."

Damn, back to the abacus.


Mark:
SENTENCE
one, two, ten

SEQUENCES
a. dog
b. 7!
c. DrewDad - Red
d. v
e. 4



1. The hen opened the door but a bucket ended the escape when the hen knocked it over. "N[/B]ow that has ended your escape," said the farmer when he caught the hen.
T
he words are highlighted in the sentence - one, then ten, then two. One and two are backwards.

2.a. Dog.

b. 5040 - The first number is multiplied by two to give the second, then the second multiplied by three to give the third etc.

c. Red - The first and third colour mixed together make the second. Therefore, the fourth (blue) and the sixth (which was the answer to the question) make the fifth (purple).

d. V.

e. 4



Mark:
ATLAS
1) no
2) #997 ($502,488)


Question 1:
Does citizen number 2001 belong to the King's group?
Answer: No

Group 1 = 0,2,5,6,8,11,13,14,17,18,20, ...
Group 2 = 1,3,4,7,9,10,12,15,16,19,21, ...

Procedure:

Write the number in binary.
Odd number of zero's in binary expansion: with the King.
Even number of zero's in binary expansion: not with the King.
2001 = 11111010001 even number of zero's.
Let n be a nonzero number and B one of the two groups.
n is element of B (if and only if) 2n is not an element of B (if and only if) 2n+1 is an element of B.

Proof:
Assign people to groups inductively, starting with 0 (automatically in Group 1) and working up. When assigning Person n, we consider the fate of envelopes in Box n. The group chosen for Person n will forced (if indeed it is possible); that is, it cannot be the case that, given the assignments so far of people 0,1,...,n-1, that Person n could either be in Group 1 or Group 2 and that either way the condition at Box n would be satisfied, since assignment to Group 1 gives one more envelope to Group 1 (0+n=n) and no change to Group 2. So we only have show that the proposed global assignment is consistent.

Consider odd n=2k+1.
Since (2j in group 1) iff (2j+1 in group 2), we know size(group 1 intersect {0,1,...,2k+1}) = size(group 2 intersect {0,1,...,2k+1}) = k+1. Now of the k+1 pairs of people, each of which could collect an envelope, say that B are entirely in group 1, C are entirely in group 2, and D pairs are split between the two groups. Then k+1=2B+D=2C+D, so that B=C, and the condition is satisfied at n.

Now consider even n=2k.
Let Person k temporarily clone himself into k' and k", and think of (k',k") as being one of the pairs of people who collects an envelope. Now there are k+1 pairs of people with numbers adding to n=2k, namely (0,2k), (1,2k-1), ..., (k-1,k+1) and (k',k"). Again the two groups are equally sized: the people in {0,1,...,2k-1} are split evenly between the two groups; there is an extra person represented by k" in one group; and the unmatched person at n=2k is in the opposite group. So each group has an intersection of size k+1 with the set {0,1,...,k-1,k',k",k+1,...,2k}.
As before k+1=2B+D=2C+D so that B=C, and the condition is again satisfied.

Question 2:
Which citizen(s) will have received the highest amount of Atlas dollars, once both groups have passed all boxes?

Answer: Citizen number 997.
Let F(n) be the amount that citizen n receives.
If n is even, then exactly half the citizens with numbers between 0 and 2001-n are in the same group with citizen n, so that citizen n shares in (2002-n)/2 envelopes. Each such envelope gives him n dollars, except that if 2n<2001, then envelope 2n gives him an extra n dollars.
So:
F(n) = n*(2002-n)/2 if n is even and 2n>2001;
F(n) = n*(2002-n)/2+n if n is even and 2n<2001.

If n is odd, then exactly half the citizens with numbers between 0 and 2000-n are in the sane group with citizen n, so that citizen n shares in (2001-n)/2 envelopes, each giving him n dollars. As before, box 2n will give him an extra n dollars if 2n<2001. Also, box 2001 will give him an extra n dollars if citizen 2001-n is in the same group with n. Therefore:

F(n) = n*(2001-n)/2 if n is odd, 2n>2001, and
2001-n is in the opposite group;
F(n) = n*(2001-n)/2+n if n is odd, 2n>2001, and
2001-n is in the same group;
F(n) = n*(2001-n)/2+n if n is odd, 2n<2001, and
2001-n is in the opposite group;
F(n) = n*(2001-n)/2+2n if n is odd, 2n<2001, and
2001-n is in the same group.


We find that 997 is the largest odd number n in this latter group, namely 2n<2001 and (997,1004) are in the same group.
F(997)=502488.
Numbers n in the other groups give smaller values of F(n).

Facts:
The King's celebration will cost his treasury a paltry KU$669132464.
Box #1999 will have yielded more money to the citizens than any other box.
Box #1919 will have yielded more envelopes to the citizens than any other box. (598) Person 2001 will go broke.





If you improve your capabilities by 1 percent per day, in how many days will you be twice as capable as you are today


Spot the odd one out:

a. 1, 5, 11, 15, 18, 27, 33, 61

b. Soccer, American Football, Tennis, Rugby, Netball, Basketball, Badminton, Baseball.

c. Cat, Dog, Human, Snake, Bear, Mole, Fox, Wolf.

d. J, F, M, A, F, J, J, A, S, O, N, D.

e. Michigan, Florida, Detroit, Wyoming, Texas, Washington.




A house has been robbed and five suspects are being questioned. At least one of them is guilty. All those involved will tell lies, but the innocent ones will tell the truth. Below are their statements. How many of them are telling the truth

Suspect A: One of us did it.
Suspect B: Two of us did it.
Suspect C: Three of us did it.
Suspect D: Four of us did it.
Suspect E: Five of us did it.

b. 7! = 5040

Tryagain wrote:

"…in the meantime, perhaps you may like to start plowing the north field?"

I'll let you know when I'm finished. Could be a while, I will have to use my bare hands from lack of tools. Perhaps you should pick yourself up a 12 pack,...make that a case (for the long wait.)

SEQUENCES
b. Since when is 7! <> 5040? (thanks sjig)

IMPROVEMENT
log(2)/log(1.01) = 69.66 days

ODD ONE OUT
a) 18 is even
b) I'll say netball because I've never heard of it
c) snakes aren't mammals
d) second F should be an M
e) Detroit is a city

ROBBERY
Only D is telling the truth

Here's one I got at another site:

Adam is 8kg heavier than Eve, and Cain is 4kg heavier than Abel. The sum of the weights of the heaviest and lightest is 2kg less than the sum of the weights of the other two. The sum of all four weights is 402kg. How much does each person weigh?

Paula complains, "I will have to use my bare hands from lack of tools."

Where I ask, is the pioneer sprit? Circle the wagons, trim the topsail! I put forkandles in the barn and you could take your pick.

That is what passes for humour in these here parts - Four candles = Fork handles = Pick!

Oh well. Back to chopping the wood for the fire. :wink:





Sjig:
b.7! = 5040


Welcome Sjig. I guess you must hail from the Nederlands. A big Hi to those from the old country. Very few can claim to help Markr out. Now, I have to sort out your Eggs, eggs etc problem.



SEQUENCES
b. Since when is 7! <> 5040? (thanks sjig)





Mark:
IMPROVEMENT
log(2)/log(1.01) = 69.66 days .


The surprising answer is a little more than 69 days.
If x is your present capability and n is the number of days, you simply need to solve:
x(1.01)^n = 2x
So
1.01^nx = 2x
0 = 2x-1.01^nx
0 = (2-1.01^n)x
So either x is zero, which means you have NO capabilities today, which is a possibility you probably shouldn't explore too strenuously in the job interview. Luckily, there is another possibility:
2-1.01^n=0
2=1.01^n
log2 = log(1.01^n)
log 2 = n*log(1.01)
log2/log1.01 = n
n = 69.66 days




Mark:
ODD ONE OUT
a) 18 is even
b) I'll say netball because I've never heard of it
c) snakes aren't mammals
d) second F should be an M
e) Detroit is a city

ROBBERY
Only D is telling the truth




3.a. 18 - It is the only even number.
b. Badminton - All the other sports have balls involved.
c. Snake - All of the others are mammals.
d. The second F - It is a sequence of all the months first letters (January, February etc.) except for the second F which should be an M for May.
e. Detroit - all the others are American states.


4. 1 - The question asks how many of them are telling the truth. There must have been at least one person committing the crime, and the maximum is five, so all options are covered. This means that at least one of them must be telling the truth. There cannot be more than one telling the truth as they have all said different things.


Mark:
Here's one I got at another site:

Nice one.




John and Fred have a dilemma. They need to get back to town as quickly as possible, but they only have one bicycle between them. And, of course, you can travel faster on the bike than you can by walking.

John says, "We'll leave at the same time. You start walking, and I'll ride the bike for a mile. Then I'll leave it by the side of the road, while you keep walking. When you get to the bike, you get on it, and you ride for a mile, and then you leave it by the side of the road. When I get to it, I'll ride another mile. We'll keep leapfrogging like this until we get to town.
"Since we'll be riding part of the way and walking part of the way, each of us will have a higher average speed than if we walked, so we'll get there faster," John explains.

Fred says, "That isn't going to do any good, you dope! Think about it. Between the two of us, one or the other is going to walk every inch of the way from here to town. So, we can't possibly get there any faster than if we just walked all the way and didn't use the bike at all."

Who is right, John, who think they will get there faster or Fred, who thinks it doesn't make a difference

Or are they both wrong



Susie, the best student in arithmetic in her grade school, started a gourmet lemonade stand with a unique payment policy.
Instead of taking payment for the lemonade.

Susie had this policy: a customer would open the cash drawer, match whatever amount of money was in there, and then take out 20 cents.

The first customer of the day does just that: matches whatever is in the cash drawer and then takes out 20 cents.

The second customer of the day does the same, as does the third.

The fourth customer, however, looks at the cash drawer and calls out, "Hey, the cash drawer is empty. There's not a cent in here."

The question is how much money did Susie have in the cash drawer to start




In a custom microchip processing plant, workers shape 1 gram packages of Silicon semiconductor material into custom microprocessor wafers. During the manufacturing process, not all the Silicon is used. For every five wafers the plant fabricates, there is enough extra Silicon to make one additional wafer. Suppose a worker is presented with 25 grams of Silicon.

What is the maximum number of wafers she can make



A gold bar balances with nine-tenths of 1 pound and nine-tenths of a similar gold bar.

How much does each gold bar weigh

BICYCLE
John, of course.

Fred is correct that every inch of the way will be walked by one or the other. However, there are times when they are walking different sections at the same time.

WAFERS
31

LEMONADE
17.5 cents (I don't know how she managed that!)

GOLD BAR
9 pounds

Tryagain wrote:
"Paula complains, "I will have to use my bare hands from lack of tools."
Oh well. Back to chopping the wood for the fire. :wink:"

I wasn't complaing silly. Need help chopping the wood?

Paula writes, "I wasn't complaining silly. Need help chopping the wood?"

Yes, that wood be very nice but, I could not axe you to spoil your hands, leave it to me. I have a packet of firelighters. - got a light?




ABCDEF;


16^3+50^3+33^3=165033

34^3+10^3+67^3=341067

44^3+46^3+64^3=444664

48^3+72^3+15^3=487215

98^3+28^3+27^3=982827

98^3+32^3+21^3=983221

40^3+70^3+01^3=407001

22^3+18^3+59^3=221859







Mark:
BICYCLE
John, of course.
Fred is correct that every inch of the way will be walked by one or the other. However, there are times when they are walking different sections at the same time.




John is right; his method will get them there faster.
Suppose they are traveling 2 miles. They each walk 5 mph, or 1 mile every 12 minutes. Suppose they can ride 10 mph, or 1 mile every six minutes.

If they walk the two miles together, the trip will take them 24 minutes.
Watch what happens if they use John's method with John riding first:
After 6 minutes, John will have gone a mile with the bike and Fred a half mile. John leaves the bike and walks. After another 6 minutes, Fred will have walked a mile and will be at the bike. John will have traveled 1.5 miles (1 mile riding and a half walking). After another 6 minutes, each will be at the 2 mile marker, having covered the distance in 18 minutes as opposed to 24 walking together.


Mark:
LEMONADE
17.5 cents (I don't know how she managed that!)


Working backward, since the fourth customer found the drawer empty, the third customer had to have seen $10 (which he matched, making $20, and then withdrew per the instructions, leaving the drawer empty.)
Working backward one more step, in order for the third customer to have left $10, the second customer must have seen $15 (matching it, the cash drawer had $30, from which the second customer removed $30, leaving $10 for the third customer to find.)

From that logic, the first customer must have left $15. Since he took $20 from the drawer, there must have been $35, which was made up of equal parts of the original cash plus the match from the first customer. Half of $35 is $17.50.


Mark:
WAFERS
31



This is a class remainder problem because the by-products of the process can be reused. First, the 25 grams of Silicon make twenty-five microchip wafers. This process generates 25/5 or 5 grams of excess Silicon. This extra Silicon is used to make five additional wafers. While making these five additional wafers, one additional gram of Silicon is generated which can be used to create one more wafer. The maximum number of wafers the plant can fabricate is 25 plus 5 plus 1 or 31 wafers.
Solution: Thirty-one wafers.



A gold bar balances with nine-tenths of 1 pound and nine-tenths of a similar gold bar. How much does each gold bar weigh?

Mark:
GOLD BAR
9 pounds


Nine Pounds. If we looked at the balance pan containing the two bars, we'd see that one-tenth of the gold bar is absent. In its place we have nine-tenths of a pound. From this we can infer that one-tenth of a gold bar weighs nine-tenths of a pound. Therefore, a complete gold bar would weigh ten times as much. 9/10 pound x 10= 90/10 or 9 pounds.





A one-armed surgeon needs to operate on three patients, one after another. But the surgeon has only two individual surgical gloves.

How can the surgeon operate on the three patients in turn without risking infection for the patients or for himself



You are locked into an empty room with only a working refrigerator plugged into a standard electric outlet. The room is uncomfortably warm, and your goal is to cool the room to the maximum extent possible.

What can you do




What is the biggest number which when divided into 364 and 414 and 539 leaves the same remainder