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• #1067936
Dr. Pepper
Participant

Mrs. “anon for this”,

Here’s how it works,

Instead of x and x + 2, let’s use (x – 1) and (x + 1). Now FOIL (Firsts, Outers, Inners and Lasts) gives us (x – 1)*(x + 1) = x^2 + x – x – 1^2 = x^2 -1.

So 6 x 8 = (7 – 1)*(7 + 1) = 7^2 – 1 = 48.

This will also work for (x – 2)*(x + 2) = x^2 – 4 and so on.

Here’s a trick I figured out for squaring an integer where the units digit is 5. Let’s square 75. Take all the digits besides for the last one (in this case it’s just the number 7) multiply it by by that number + 1 (7 * 8 = 56) and attach the numbers 25 at the end (5625). Please check that 75 * 75 = 5625.

Proof (10x + 5)^2 = 100x^2 + 50x + 50x + 25 = 100x^2 + 100x + 25 = 100(x^2 + x) + 25 = 100 * x * (x + 1) + 25.

Now try it for 45 and move the decimal two spots to the left for 4.5 ^ 2. You should get 20.25.

#1067937
anon for this
Participant

Dr. Pepper,

Also, the math for x+0.5 works like this:

x*(x+1)=x^2+x

(x+0.5)^2=(x+0.5)*(x+0.5)

#1067938
Joseph
Participant

Dr. Pepper,

Excellento! Correct mathematicaly on both questions. Though the only place in the world with equal Fahrenheit and Celsius is Antarctica.

#1067939
anon for this
Participant

Dr. Pepper,

The math for x+0.5 works like this:

x*(x+1)=x^2+x

and

(x+0.5)^2=(x+0.5)*(x+0.5)=x^2+0.5x+0.5x+0.5*0.5=x^2+x+0.25

So the square is 0.25 larger than the other product.

I did figure out the math when my daughter told me the trick because she was younger then and used smaller numbers, so I wanted to prove it would work for all numbers.

Thanks for the proof for squares of numbers ending in 5. I’d heard of that but didn’t know why it worked.

Note to mod: Would you please delete my previous reply to Dr. Pepper? I hit “send post” too soon. Thanks.

#1067940
Dr. Pepper
Participant

anon for this

(x+0.5)*(x+0.5) = x^2 +.5x + .5x +.25 = x^2 + x +.25.

=> 4.5 * 4.5 = 4^2 + 4 + .25 = 20.25

I agree that this will work but I found that my students had a hard time performing more than two mathematical operations mentally. (x+0.5)*(x+0.5) = x^2 +.5x + .5x +.25 reduces to three terms x^2 + x +.25 , by (x – 2)*(x + 2), the two middle terms cancel out so there are only two terms left, x^2 – 4.

If you (or your daughter) have no issue with this consider yourself gifted.

#1067941
GivPerf
Member

In a house with all southern exposure – that would have to be at the North Pole, so the bear would be a polar bear.

#1067942
anon for this
Participant

Dr. Pepper,

My daughter didn’t figure out the proof; she was only 5 years old at the time. She just figured out the rule using examples like 3*5 and 4^2. I did the proof on paper once to satisfy myself that it would always work. Now I just use the shortcut to quickly solve math problems (like estimating the area of a 9″ diameter pie pan).

#1067943
squeak
Participant

GivPerf – Right. The answer to the second one is the same (though the reasoning involves knowledge of physics). The answer “red” was given in my last discussion too. Insightful, but misses the point.

Joseph, have you never been to Yukon? Your statement is far from being correct. And I might add that the temperature at the South Pole can go South of -100 degrees.

#1067944
brooklyn19
Participant

just googled it. lowest recorded is -129F

#1067945
squeak
Participant

I think we’re seeing a whole new side to Joseph based on the comments of this day and the last. Is it possible that his screen name has been usurped by an imposter?

#1067946
SJSinNYC
Member

Dr. Pepper, are you also an engineer? I might have missed what you do.

I love all the math questions! It brings me back some good memories of math for fun ðŸ™‚

#1067948
Dr. Pepper
Participant

Here’s another one from an upper level math course (that can be answered in English with basic mathematics skills):

Does there exist a prime number “P” such that there is no prime number greater than “P”? (Is there a highest prime number?)

If “P” exists what is it, if not why does “P” not exist?

SJSinNYC

My Ph.D. is not in engineering but I did take some engineering courses. If you have some questions I’ll be happy to try to help you.

#1067949

anon for this-

Dr. Pepper-

The puzzle was, “if you have a square wall made up of square bricks and remove one column from the side and add one row to the top, what will the difference in brick-count always be?”

SJSinNYC-

Math ? fun!

Math <> fun!

Math^2 = fun! (a negative squared)

Dr. Pepper-

One question:

#1067950

The numbers 3, 4 and 5 (used together) are very helpful for builders and contractors.

How?

#1067951
brooklyn19
Participant

squeak – i totally agree.

joseph – you could save yourself by saying that your wife or kid got on here today…

#1067952

Flights of Fancy

Airplanes, ships, cars, darts and many other objects are smooth to ensure less air resistance and faster speeds.

Why then are golf balls dimpled?

#1067953
Dr. Pepper
Participant

I can only try

(X + 2 * Y = 98) minus (X + Y = 74) => Y = 24.

We are subtracting one equation from another. (Cramer’s rule would also work.)

3^2 + 4^2 = 5^2 (is that what you were thinking of?) there are an infinite more sets of three integers that will fit that relationship.

The dimples in Golf balls allow the balls to travel further by trapping air inside of them and allowing the ball to ride or “float” on the air. A more detailed explanation is beyond the scope of this thread.

#1067954
Joseph
Participant

ICOT:

Perhaps this is easier to follow…

HM = Human, HR = Horse

HM + HR = 74

2HM + 4HR = 196

(2HM + 4HR) – (2 HM + 2HR) = 196 – 148

2HR = 48

HR = 24

HM + (24) = 74

HM = 74 – 24

HM = 50

24 horses and 50 humans.

#1067955

Dr. Pepper-

a) I got it – thanks a lot!

b) You’re headed in the right direction, but what does that have to do with builders, contractors and handymen? Also – I thought I remembered learning once that there are no other integers that when squared add up to the square of a third integer.

c) Flights of Fancy – Correct! (Don’t forget that as well as adding lift, it also decreases drag by carrying air in the dimples to the “rear” of the ball. Agreed – the full technical explanation is not for here).

Joseph-

Thank you for your explanation, too.

<<Math + me = Rebbe Akiva’s water + rock>>

#1067956
Joseph
Participant

ICOT: Put look what happened to that rock… and look where it got Rebbe Akiva!

#1067957
Joseph
Participant

Every month, a girl gets allowance. Assume last year she had no money, and kept it up to now. Then she spends 1/2 of her money on clothes, then 1/3 of the remaining money on games, and then 1/4 of the remaining money on toys. After she bought all of that, she had \$7777 left. Assuming she only gets money by allowance, how much money does she earn every month?

#1067958

Joseph-

Working our way backwards:

7777 x 1 1/3 = 10369.33

10369.33 x 1.5 = 15554

15554 x 2 = 31108

#1067959
oomis
Participant

“Assuming she only gets money by allowance, how much money does she earn every month?”

More than I do, apparently…

#1067960
Dr. Pepper
Participant

I can only try

There are actually an infinite amount of integers that will satisfy the equation a^2 + b^2 = c^2, another common example is 5, 12 & 13. There are no integers for a, b, c and n such that a^n + b^n = c^n for n > 2. (Please don’t ask me to prove Fermat’s Last Theorem here.)

Of the many uses of Pythagorean Triples, as they are called, the most common one is finding the length of the hypotenuse of a right triangle given the length of the legs. If the legs measure 3 and 4 respectively then the hypotenuse will measure 5.

#1067961
Dr. Pepper
Participant

Joseph

15,554,

7777 * 4 * 3 * 2 =186,648

186,648 / 12 = 15,554.

#1067962
anon for this
Participant

ICOT,

3,4, and 5 are a pythagorean triple. That is, they can form the sides of a right triangle, where 5 is the side opposite the right angle. Then, according to the Pythagorean Theorem, 3^2 + 4^2 = 5^2 (9+16=25). Another example of a unique Pythagorean Triple would be 5,12, and 13.

#1067963
Joseph
Participant

ICOT, correct. And divide the annual salary you quoted, 31,108, by 12 for the monthly allowance of 2592.33.

Dr. Pepper, you took another route to get there; I believe the final step is missing.

#1067964

Dr. Pepper-

anon for this-

What I should have said is that there are no sequential numbers where a^2 + b^2 = c^2 except 3, 4 and 5.

I read in the news several years ago that someone claimed he had discovered a proof for Fermat’s Last Theorem, but a mistake was uncovered. Then, a couple of years ago, someone else claimed he had discovered it, and mathematicians were very excited and preliminary indications were that he was correct, but it would take months to review and verify. I don’t remember if it was actually verified or disproven. What was the end of that story?

Builders and contractors often use the 3, 4, 5 measurement to ensure something is being built correctly at right angles (measure the sides at 3′ and 4′ from a corner – a straight line between those two points should be 5′). For all newlyweds building bookcases, this is an easy way to confirm that your parallelogram is also a rectangle ðŸ™‚

Joseph-

I didn’t divide the final result by 12, because I was misupek if we should count from Tishrei or January (just kidding – I simply forgot).

#1067965

For (and from) Kids

Kid: Daddy, can you jump higher than a house?

Father (chuckling): No, of course not.

Kid: Daddy, do you have a green key?

Father: No.

Kid: A red key?

Father: No.

Kid: A blue key?

Father: No.

Kid: A pink key?

Father: No.

Kid: Yes you do. You have one on each hand!

Kid: Hot. You can’t catch a hot!

#1067966
anon for this
Participant

ICOT & Dr. Pepper,

I can prove a^n+b^n=c^n is false for all n>2. Oops, just ran out of space.

#1067967
Dr. Pepper
Participant

Joseph,

I read the question too fast too late at night. I thought it said that half was remaining, then one third was remaining, then one quarter was remaining…

I can only try

Andrew Wiles announced a proof in 1993 which was almost complete but a colleague of his, I think his last name is Katz, found a “hole” in the proof. Andrew Wiles spent about another year with one of his students “plugging the hole” and I far as I know the proof has been accepted.

This might be a legend but I heard that in the 1800s and early 1900s there were large rewards set aside for the first one to prove (or disprove) the theorem. Unfortunately for Wiles the rewards were in German Marks held in German banks and after the record inflation Germany suffered after World War I, the rewards were practically worthless.

#1067968
noitallmr
Participant

Re: I can only try

Shouldn’t that be in the humor thread (for kids)?

#1067969
intellegent
Member

Joseph,

Alright, here it is:

( [3x – 3x^2 +1]^744 ) x ( [- 3x + 3x^2 +1]^745 ) ?

Actually I’m not sure what you mean by the sum of the coefficient (btw, wouldn’t that be product?)

((3 – 9)^744)((-3 + 9)^745)

(-6)^744

x 6^745

_______

1

or do you mean

( [3x – 3x^2 +1]^744 ) x ( [- 3x + 3x^2 +1]^745 ) ?

((3x – 9x +1)^744)((-3x + 9x +1)^745)

((-6x +1)^744)((6x+1)^745)

Well actually I guess I get one in any case. or is that 2? I really don’t remember.

or is it 6x+1?

(copied and pasted the post from the other thread.)

#1067970
SJSinNYC
Member

Dr. Pepper – thanks for the engineering help/offer, but I’m already an engineer! Anon is too (if I’m not mistaken).

#1067971
anon for this
Participant

SJS, yes, but I don’t work outside the home now.

#1067973
Joseph
Participant

Can someone be kind enough to respond to intellegent? She wants the sum of the coefficient of the equation she put in the beginning of her post (3rd line. I’m away from a computer now.)

#1067974
SJSinNYC
Member

Anon, it doesnt mean you arent still an engineer ðŸ™‚

#1067975
anon for this
Participant

SJS, you’re right, I definitely still think like one.

#1067976
squeak
Participant

Joseph:

Assuming she only gets money by allowance, how much money does she earn every month?

I disagree and say you are all wrong, including Joseph. The correct solution to this riddle is \$0. It doesn’t say stipend.

#1067977
SJSinNYC
Member

Anon, given your aptitude for math you can surely tell ðŸ™‚

Plus, as a friend of my mother’s used to say you are now a “domestic engineer.” ðŸ™‚

#1067978

noitallmr-

Since the father who played the straight-man was yours truly, they were riddles from my perspective.

#1067979
squeak
Participant

Here’s a tough one from a magazine back in the ’50’s. If anyone here can solve it without using brute force I will be very, very impressed. I might even bestow upon you all of my designations:

5 men are shipwrecked on a desert island. On the first day, they begin collecting food. The only food source on the island is coconut. The men spend the whole day collecting coconuts and at the end of the day have quite a nice pile. Exhausted, they all lie down to sleep.

After the others have fallen asleep, one man realizes that any of the others could steal from his “share” while the group is sleeping. So he gets up and divides the coconuts into 5 even piles. But there is one extra coconut – so he gives it to a monkey who is jumping around in the nearby tree. The man takes his pile and hides it, moves the other 4 piles back together to hide what he did, and then goes to sleep.

A short while later, one of the other men wakes up, and reasons the same as the first man, that he had better take his share now so that he does not get cheated. He divides the big pile into 5 even ones, but there is one left over. He gives it to the aforementioned monkey. He then takes his pile, pushes the other 4 piles back into 1, hides his and goes back to sleep.

A short while later, another man wakes up and does the same as the man before him. Again there is an extra coconut when he divides by 5, which he gives to the monkey. This process continues until every man has done the same deed. That said, each man now has a hidden pile of coconuts, there is a diminished pile left for the community, and the monkey has 5 coconuts.

In the morning, they all wake up and divide the pile into 5. This time, it divides evenly and the monkey walks away dejected. Each man must have noticed that the pile was much smaller than last night, but his guilt prevented him from bringing up the subject.

How many coconuts were there originally, and how many were there when it was finally divided?

There are of course mulitple solutions to this (I found over 100), but they are all multiples of the minimum solution. So there is really only one solution of importance – the minimum. Using brute force requires plenty of skill in this problem so it is legitimate to answer it that way, but an elegant solution would be amazing!

#1067980
Joseph
Participant

squeak – she does the laundry to earn her allowance. Sorry for skipping that crucial detail. ðŸ™‚

#1067981
squeak
Participant

Joseph – then it’s not a riddle – it’s just a math problem (9th grade, I would say)

#1067982
Joseph
Participant

squeak – then how’d 2 (very) kluger yidden not get a 9th grade Q.?

#1067983
squeak
Participant

Are you saying that it is a riddle because a couple of people got the wrong answer? Can’t argue with logic like that….

#1067984
Dr. Pepper
Participant

squeak

This is not as innocent as it looks.

Let the original pile = y and the final pile (before it’s divided into 5) = x.

We know that x must be divisible by 4 and 5 (the last person split it into 4 even groups and the 5 people split it into 5 even groups).

y = ((((((x)*5/4+1)*5/4+1)*5/4+1)*5/4+1)*5/4+1) which simplifies to (3125*x)/1024 + (8404/1024).

(8404/1024) = 8 + 53/256 therefore the fractional part of (3125*x)/1024 must equal 203/256 (so we are not left with a fraction at the end).

=> (3125*x)/1024 = an integer (henceforth denoted as “I”) + 203/256

=> (3125*x)/1024 = I + 203/256

=> 3125*x = 1024*I + 812

So we need MOD(3125*x,1024) = 812

MOD(3125,1024) = 53

MOD(812,53) = 17

MOD(53,17) = 2

=> x = 4 * 5 * MOD(812,53) * (MOD(53,17) + 1) = 4 * 5 * 17 * 3 = 1020

=> x = 1020 and y = 3121

If we weren’t looking for the minimum value of x the correct answer would be 1020 + 1024c where c is a non-negative constant.

#1067985
squeak
Participant

………………………………………..

<closes mouth>

Dr. P, please tell me that you’ve already seen this problem. I guess that would make you an old timer like me (with quite a memory), but that’s better than believing that you came up with that in 1.5 hours (and that’s if I assume you were working on it the whole time).

Now that we’ve been introduced, I’d like to ask a favor. Can you stop using the clam stuff please?

#1067986

squeak-

Good puzzle.

The original number can end with either a 1 or a six (subtract 1, get a number divisible by 5)

the next four numbers must all end with 16, 36, 56, 76 or 96 (divisible by 4, subtract 1 divisibe by 5)

the last number must be 0 ot 5 (divisible by 5)

From that point on, brute force produces the following:

3121

2496

1996

1596

1276

1020

#1067987
noitallmr
Participant

I may be wrong but I think this is the highest number of posts a thread has ever had…

…and I started it!

Autographs- \$5

Signed Photos- \$7

ðŸ™‚

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