| Username |
Post |
| Robert Frost |
Posted
on 04-Mar-04 11:39 AM
A young couple is planning for the education of their two children. They plan to inverst the same amount of money at the end of each of the next 16 years. The first contribution will be made at the end of the year and the final contribution will be made at the end of the year the older child enters college.
The money will be invested in securities that are certain to earn a return of 8 percent each year. The older child will begin college in 16 years and the second child will begin college in 18 yeras. The parents anticipate college costs of $25,000 a year (per child). These costs must be paid at the end of each year. If each child takes four years to complete their college degree, then how much money must the couple save each year??
|
| DWI |
Posted
on 04-Mar-04 12:01 PM
Excellent (in Mr. Burns style) question Bob.
Are we allowed to use Recursive Loops? In any case it will take a while to come up with the final answer or the formula to get the final answer. Will do on leasure time.
Great question. Are you in investment banking?
|
| Professor Moriarity |
Posted
on 04-Mar-04 01:13 PM
Hah.. Time Value of Money question. Yeah, Will do that in leisure.
|
| bhunte |
Posted
on 04-Mar-04 01:52 PM
Hey Bob, i m bit rusty but will give you some clues. Your life becomes easier if you compute the annuities separate for two childs.
Equate Future value of a 16 year annuity at t=16 with the Present value of a 4 year annuity at t=4. With little algebraic manipulation, you will come up with an annuity to save for the first child. do same for second child....
|
| thaag |
Posted
on 04-Mar-04 02:35 PM
i thnk
2472.41/yr for small child(18 years left)
3053.41/yr for big child (16 years left)
using formula
Balance(n) = P(1 + r)n + c[((1 + r)n + 1 - (1 + r))/r]
|
| thaag |
Posted
on 04-Mar-04 02:38 PM
oops I didn't read question properly .........
|
| nepali_angel |
Posted
on 04-Mar-04 04:20 PM
Okay, this question is kind of confusing, but from my calculation, they need to invest $5925.886 for 16 years. However, I haven't taken into account the interest received after the first kid goes to college.
How close am I?? Missed by a mile probably.
|
| Gokul |
Posted
on 04-Mar-04 04:39 PM
Using Excel, one can easily solve this problem.
It is interesting to notice that the simple sounding problems are the most difficult to solve. Consider the four color problem which states that one does not need more than four colors to paint any map: The rules being - adjacent areas should have different color. If two areas meet only at apoint, they can have similar color.
This problem so simple sounding remained elusive for a quite time until a few mathematicians solved it using computer algorithms. The analytic proof is still not found.
|
| DWI |
Posted
on 04-Mar-04 09:12 PM
Gokul, Excel Guru.
I think I can break it down to a polynomial function, Gokul can then solve it in the Excel (which DOES solve the polynomials). Since the function is of order 18, I decided not to solve it myself.
The equation should be something like:
200,000 = (1.08y)(1+n(0.08) to the power n)
Where n represents the polynomial function of the order 18 where x=0.08 and a=n.
(Note n represents a function not power).
Where y is the amount invested.
For example, in the 5th year, the equation would look like this:
200,000 = (1.08y)(1+3(0.08)+3(0.08)sqrd+ (0.08)cube)
(Ofcourse solving for y here won't give right value as around 5th year, the LHS won't be 200,000).
I am pretty sure there has to be an easier formula for such problems.
Speak out Bob.
|
| Robert Frost |
Posted
on 04-Mar-04 09:40 PM
DWI,
You can easily solve it using excel, however, my problem is that I am not quite familiar with Excel, also you might be able to do it with a financial caculator. As for now, I can give you the choice. But, actually I want somebody to come up with the right answer before giving the choices out.
I am hoping for the answer. Lets go sajha.
|
| Rusty |
Posted
on 05-Mar-04 06:47 AM
Bob, I think you should consider Bhunte's suggestion. He's given some hints. You should be able to solve this problem by using financial calculator. But I don't know how!! I did similar maths four years ago, and now I completely forgot. However, I don't think you need to use EXCEL or theory of polynomial function for this problem...
|
| Professor Moriarity |
Posted
on 05-Mar-04 09:04 AM
It is a Sinking Fund Deposit of Time Value of Money problem. Since we know that by the end of the 16th year the couple need to come up with total of $200,000. Now lets draw the good ole' Time Line:
-----------I----------I------------/\/\/\/\/\/\-----------I
t=0 t=1 t=2 t=16
Present Value(PV)=$0.00
Future Value @ end of 16th year(FV)=$200,000
Interest Rate(I)=5%
Number of periods(N)=16
if you are using a finacial calculator (HP 10B) or (Texas Instrument BA II Plus), you might want to have 1 for payment(contribution) per year(P/Y) and 1 for the compounding year(c/y).
If you plug in all the datas in financial calculator or Excel spreadsheet(using Function),
the couple should deposit $6,595.97 per year.
The last part of the question is irrelevant as the final contribution is to be made at the end of 16th year and it is also irrelevant when the second child starts his college as the parents has to come up with $200,000 by the end of 16th year.
So the right answer should be $6,595.97
Did I make any sense. If not let me know when I went wrong.
|
| Professor Moriarity |
Posted
on 05-Mar-04 09:07 AM
N=8% not 5%
|
| AdjunctProf |
Posted
on 05-Mar-04 09:44 AM
Assume, money deposited each year = P
Money available at the end of 1st year = P
Money available at the end of 2nd year
= P*1.08 +P
=P(1.08+1)
Money available at the end of 3rd year
= 1.08[P(1.08+1)]+P
=P(1.08^2 + 1.08 +1)
Money available at the end of 4th year
= 1.08[P(1.08^2 + 1.08 +1)] +P
=P(1.08^3 +1.08^2+1.08+1)
By induction,
Money available at the end of 16th year
= P(1.08^15 + 1.08^14 + 1.08^13 + ..... 1.08^2 + 1.08 + 1)
Putting the terms inside the paranthesis in reverse order
= P(1+1.08+1.08^2+ .... +1.08^14+1.08^15)
The terms inside the paranthesis make a Geometric Progression with
First term 'a' = 1
Ratio 'r' = 1.08
Number of terms 'n' = 16
The sum of a Geometric Progression, which can also be derived by simple induction, is,
Sum = [ar^(n-1)]/(r-1)
= [1*1.08^15]/.08 = 39.65211
Total money needed at the end of 16th year is $200,000.
Now,
39.65211*P = $200000
Therefore P = $5043.87
|
| Robert Frost |
Posted
on 05-Mar-04 10:05 AM
Professor, did you plug in 5% instead of 8% on the financial calculator, because its not the right answer, lets make sure if you actually made that mistake.
Adjuntprof's calculations makes a lot of sense to me, but thats just me, the answer is quite different.
Even with Bhunte's hint, I could not solve it, I did made them calculations, but it was not the right answer.
Anyways, I will lay out the choices,
a) $9612.10
b) $5477.36
c) $12507.29
d) $5329.45
e) $4944.84
|
| thaag |
Posted
on 05-Mar-04 10:33 AM
I think the question is bit complicated;
Investment should be made for 16 years
then 25000 is taken out each year for 2 years
50000 for next two and again 25000 for another 2 years.
the question is,
will the investment grow during the education years (6 years)??
Since the money is taken out at the end of year, there is time for growth
So money required at end of 16 years is ----153793.54
which requires contribution of 4695.95 per year
So the college year begins..........
at the end of 1st college year the money will grow to ----166097.03
take out 25000 for payment then left over is ------------141097.03
Next year it will grow (8 percent) to ----------------------------152384.78
take out 25000 and left over--------------------------------------127384.78
next year it will grow to ---------------------------------------------137575.57
take out 50000 and left over is----------------------------------- 87575.57
at the end of next year it will grow to ----------------------------94581.62
take out 50000 and remainder------------------------------------44581.62
will grow to---------------------------------------------------------------48148.19
take out 25000 and remainder------------------------------------23148.148
which with 8 percent growth at the end will be ..................25000
So the answer is 4695.95 per year (I guess)
|
| Professor Moriarity |
Posted
on 05-Mar-04 11:00 AM
No, I plugged in 8% but and 5% is a typo.
We all agree that it is $200,000 th couple need to have at the end of 16th year, right?
So with the five answers for the payment you provided would not jive with Future Value $200,000, That is for sure.
If the FV is not $200,000 then that is a whole different ball game. From the question, I do not see why $200,000 is not the total amount they need by the 16th year.
|
| john doe |
Posted
on 05-Mar-04 11:01 AM
Step 1:
At t=16, find the PV of $25,000 annual payments at t=16,17,18,19 (first child) and also at t=18,19,20,21 (second child) assuming the 8% interest rate.
Step 2:
Find the 16-yr annuity amount with its FV equal to the answer from Step 1.
Happy mathematicking!
|
| john doe |
Posted
on 05-Mar-04 11:05 AM
Prof Moriarty, since they are not paying the entire tuition in a lump sum, they do not need to have $200 K by t=16. Time value for the next 6 years after t=16 needs to be taken into account.
|
| Newtien |
Posted
on 05-Mar-04 02:09 PM
Roberto,
I am pretty confident the answer is A. ie $9612.10
|
| gham-pani |
Posted
on 05-Mar-04 03:37 PM
The answer is
C)12507.29
|
| Professor Moriarity |
Posted
on 06-Mar-04 09:31 AM
Okay let me try with your method, John Doe bro
Cashflow at the end of 16 year = $25,000
17 year = $25,000
18 year = $50,000 (payments for both kids)
19 year = $50,000 (1st kid finishes college)
20 year = $25,000
21 year = $25,000 (2nd kid finishes college)
Assuming the rate ( R)at 8%, and total number of years(N) =6
Plugging the uneven cash flows in financial calculator as Cf=0 , CF1= $25,000, CF2=$25,000, CF3= $50,000, CF4=$50,000, CF5=$25,000, CF6 = $25,000
From the financial calculator,
The Present Value (PV) = Net Present Value (NPV) = $153,793.54
Now, to find the annuity for 16 years,
N=16
FV at the of 16 year = $153,793.54
PV=0
R = 8%
From the financial calculator,
Payment (pmt) = $5,071.63 per year.
Well, it is still not one of the answers Robert Frost provided.
|
| Robert Frost |
Posted
on 06-Mar-04 10:44 AM
FV=25,000
T=16
Rate = 8%
First Child,
For T = 1, PV = 23148.15 (1)
For T = 2, PV = 21433.47 (2)
For T = 3, PV = 19845.81 (3)
For T = 4, PV = 18375.75 (4)
Adding all PV = 82803.18
Second Child,
For T = 3, PV = 19845.81 (1)
For T = 4, PV = 18375.75 (2)
For T = 5, PV = 17014.58 (3)
For T = 6, PV = 15754.24 (4)
Adding all PV = 70990.38
Adding both PV's = 153793.56
Now FV = 153793.56
T = 16
R = 8%
Payment = ?
FV = PMT [((1+R)^T-1)/R]
153793.56 = PMT (2.426/.08)
PMT = 153793.56/30.325
Therefore PMT = 5071.51
DAMN !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
|
| acharya |
Posted
on 06-Mar-04 01:37 PM
how many sajhaites does it take to .....?
|
| SITARA |
Posted
on 06-Mar-04 03:06 PM
The answer is B ($5477.36).
Present value of each payment at the end of the 16th year: (No need to bring PV all the way to present, but you could)
$25,000.00
$23,148.15
$42,866.94
$39,691.61
$18,375.75
$17,014.58
Total Present Value at the end of year 16 = $166,097.03
An annuity at 8% (annual compounding) requires a payment of $5477.36 for 16 years so that it will have $166,097.03 in the account by the end of 16th year to match the necessary amount of payments.
***********************************************************
Posted for a Sajhaite friend-- with a mathematical bend, who forgot his password!!!
As for me, I can only be a mathematical geeeeeeeeeeeeeeee! ;)
|
| hum-ba-ba |
Posted
on 06-Mar-04 03:40 PM
Here's my solution - I get 4695.95--
CHILD 1 CHILD 2
YRS INT BALANCE SAVE INT BALANCE SAVE
0 2528.32 2167.63
1 1.08 2730.59 2528.32 1.08 2341.04 2167.63
2 1.08 5679.62 2528.32 1.08 4869.36 2167.63
3 1.08 8864.57 2528.32 1.08 7599.95 2167.63
4 1.08 12304.32 2528.32 1.08 10548.99 2167.63
5 1.08 16019.26 2528.32 1.08 13733.95 2167.63
6 1.08 20031.38 2528.32 1.08 17173.71 2167.63
7 1.08 24364.48 2528.32 1.08 20888.64 2167.63
8 1.08 29044.22 2528.32 1.08 24900.77 2167.63
9 1.08 34098.35 2528.32 1.08 29233.88 2167.63
10 1.08 39556.80 2528.32 1.08 33913.63 2167.63
11 1.08 45451.93 2528.32 1.08 38967.76 2167.63
12 1.08 51818.67 2528.32 1.08 44426.22 2167.63
13 1.08 58694.75 2528.32 1.08 50321.36 2167.63
14 1.08 66120.91 2528.32 1.08 56688.11 2167.63
15 1.08 74141.17 2528.32 1.08 63564.20 2167.63
16 1.08 82803.05 0 1.08 70990.37 0
17 1.08 89427.29 -25000 1.08 76669.60 0
18 1.08 69581.48 -25000 1.08 82803.17 0
19 1.08 48148.00 -25000 1.08 89427.42 -25000
20 1.08 24999.84 -25000 1.08 69581.62 -25000
21 0 1.08 48148.15 -25000
22 1.08 25000.00 -25000
23 0
|
| hum-ba-ba |
Posted
on 06-Mar-04 03:58 PM
oh crap, forgot that html only take one space at the most
Here's my solution - I get 4695.95--
CHILD1 _____________________CHILD2
YRS_INT__BALANCE___SAVE_____INT__BALANCE____SAVE
0___________________2528.32___________________2167.63
1___1.08__2730.59 ____2528.32__1.08 __2341.04_____2167.63
2___1.08__5679.62 ____2528.32__1.08 __4869.36_____2167.63
3___1.08__8864.57____ 2528.32__1.08 __7599.95_____2167.63
4___1.08__12304.32___2528.32__1.08 __10548.99____2167.63
5___1.08 __16019.26__ 2528.32__1.08 __13733.95___2167.63
6___1.08 __20031.38__ 2528.32__1.08 __17173.71___2167.63
7___1.08 __24364.48__ 2528.32__1.08 __20888.64___2167.63
8___1.08 __29044.22__ 2528.32__1.08 __24900.77___2167.63
9___1.08 __34098.35__ 2528.32__1.08 __29233.88___2167.63
10__1.08 __39556.80__ 2528.32__1.08 __33913.63___2167.63
11__1.08 __45451.93__2528.32___1.08__38967.76___2167.63
12__1.08 __51818.67__2528.32___1.08__44426.22___2167.63
13__1.08 __58694.75__2528.32___1.08__50321.36___2167.63
14__1.08 __66120.91__2528.32___1.08__56688.11___2167.63
15__1.08 __74141.17__2528.32___1.08__63564.20___2167.63
16__1.08 __82803.05__0________1.08__70990.37___0
17__1.08 __89427.29__-25000___1.08__76669.60___0
18__1.08 __69581.48__-25000___1.08__82803.17___0
19__1.08 __48148.00__-25000___1.08__89427.42__ -25000
20__1.08 __24999.84__-25000___1.08__69581.62__-25000
21_______0__________________1.08__48148.15___-25000
22__________________________1.08__25000.00___-25000
23__________________________________________0
|
| thaag |
Posted
on 06-Mar-04 04:12 PM
I guess me and Humbaba calculated same way, we both have the same answer but with different approach :)
|
| nepali_angel |
Posted
on 06-Mar-04 04:43 PM
From the choices Robert Frost gave, it seems that B is the correct answer. Here is how I did it:
For 5477.36:
End of 1st year: 5477.36
end of 2 years: 11392.9088
end of 3 years: 17781.701504
end of 4 years: 24681.59762432
end of 5 years: 32133.4854342656
end of 6 years: 40181.5242690069
end of 7 years: 48873.4062105274
end of 8 years: 58260.6387073696
end of 9 years: 68398.8498039592
end of 10 years: 79348.1177882759
end of 11 years: 91173.327211338
end of 12 years: 103944.553388245
end of 13 years: 117737.477659305
end of 14 years: 132633.835872049
end of 15 years: 148721.902741813
end of 16 years: 166097.014961158
end of 18 years: 137575.558250695
end of 20 years: 48148.1311436101
end of 22 years:-0.0198340931777784
That is, the value is almost 0 at the end of 22 years if we start out with 5477.36. Now let's compare with other choices:
End of 1st year: 9612.1
end of 2 years: 19993.168
end of 3 years: 31204.72144
end of 4 years: 43313.1991552
end of 5 years: 56390.355087616
end of 6 years: 70513.6834946253
end of 7 years: 85766.8781741953
end of 8 years: 102240.328428131
end of 9 years: 120031.654702381
end of 10 years: 139246.287078572
end of 11 years: 159998.090044858
end of 12 years: 182410.037248446
end of 13 years: 206614.940228322
end of 14 years: 232756.235446588
end of 15 years: 260988.834282315
end of 16 years: 291480.0410249
end of 18 years: 283822.319851443
end of 20 years: 218730.353874723
end of 22 years:198967.084759477
Clearly there is no way that this is the answer.
Now with 12507.29, we have:
End of 1st year: 12507.29
end of 2 years: 26015.1632
end of 3 years: 40603.666256
end of 4 years: 56359.24955648
end of 5 years: 73375.2795209984
end of 6 years: 91752.5918826783
end of 7 years: 111600.089233293
end of 8 years: 133035.386371956
end of 9 years: 156185.507281712
end of 10 years: 181187.637864249
end of 11 years: 208189.938893389
end of 12 years: 237352.424004861
end of 13 years: 268847.907925249
end of 14 years: 302863.030559269
end of 15 years: 339599.363004011
end of 16 years: 379274.602044332
end of 18 years: 386225.895824508
end of 20 years: 338173.884889707
end of 22 years:338286.019335354
Too much of a surplus huh?
Now with 5329.45:
End of 1st year: 5329.45
end of 2 years: 11085.256
end of 3 years: 17301.52648
end of 4 years: 24015.0985984
end of 5 years: 31265.756486272
end of 6 years: 39096.4670051738
end of 7 years: 47553.6343655877
end of 8 years: 56687.3751148347
end of 9 years: 66551.8151240214
end of 10 years: 77205.4103339431
end of 11 years: 88711.2931606586
end of 12 years: 101137.646613511
end of 13 years: 114558.108342592
end of 14 years: 129052.20701
end of 15 years: 144705.833570799
end of 16 years: 161611.750256463
end of 18 years: 132343.945499139
end of 20 years: 42045.9780301957
end of 22 years:-7117.5712255797
Missed by quite a bit.
Now, let's see with 4944:
End of 1st year: 4944.84
end of 2 years: 10285.2672
end of 3 years: 16052.928576
end of 4 years: 22282.00286208
end of 5 years: 29009.4030910464
end of 6 years: 36274.9953383301
end of 7 years: 44121.8349653965
end of 8 years: 52596.4217626282
end of 9 years: 61748.9755036385
end of 10 years: 71633.7335439296
end of 11 years: 82309.2722274439
end of 12 years: 93838.8540056394
end of 13 years: 106290.802326091
end of 14 years: 119738.906512178
end of 15 years: 134262.859033152
end of 16 years: 149948.727755804
end of 18 years: 118740.19605437
end of 20 years: 26178.5646778172
end of 22 years:-25625.322159794
This can't be it eithers.
So, the answer is choice B.
|
| john doe |
Posted
on 06-Mar-04 06:01 PM
The last annuity payment coincides with the first tuition payment. We were all missing this crucial point until Sitara showed up with her mathematical genius, albeit borrowed. :)
|
| Robert Frost |
Posted
on 06-Mar-04 11:12 PM
Alright, there you go..!! beautiful, Sitara..!
Who forgot the password, btw??? Mine is still woods hai.....;)
Nepali_angel, clap clap..!!
|
| Robert Frost |
Posted
on 06-Mar-04 11:32 PM
Here is another one:
Bob is 20 years old today and is starting to save money, so that he can get is MBA. He is interested in a 1-year MBA program. Tuition and expenses are currently $20,000 per year, and they are expected to increase by 5 percent per year. Bob plans to begin his MBA when he is 26 years old, and since all tuition and expenses are due at the beginning of the school year, Bob will make his one single payment 6 years from today.
Right now, Bob has $25,000 in a brokerage account, and he plans to contribute a fixed amount to the account at the end of each of the next six years (t=1,2,3,4,5, and 6). The account is expected to earn an annual return of 10 percent each year. Bob plans to withdraw $15,000 from the account two year from today (t=2) to purchase a used car, but he plans to make no other withdrawals from the account until he starts the MBA program. How much does Bob need to put in the account at the end of each of the next six years to have enough money to pay for his MBA?
|
| Prem Charo |
Posted
on 07-Mar-04 08:24 AM
Frost ji, What is your major? tapaiko major k ho? Hope not Finance or Accounting. If so, then you better switch your major. K ho assignment yahi shajhabasi harobatai sakne bichar cha ki k ho?;)
|
| hum-ba-ba |
Posted
on 07-Mar-04 10:48 AM
thaaq, i hadn't read your solution, and in fact i solved the problem almost the same way you mention. i just illustrated it in the table form so that it would be easier to read. mr. frost, don't you think this answer is more appealing -- you can blow an extra 700 every month and also get your kids through college..
|
| CRUSADER |
Posted
on 07-Mar-04 06:45 PM
$579.91 every year ?
|
| CRUSADER |
Posted
on 07-Mar-04 07:12 PM
Since tuition will increase 5 percent annually in 6 years it will become
20000.00 * (1.05)^6 = 26801.91
For the first 2 years
PV= $25000.00
(time) n = 2 years
(interest) IY = 10%
Compute FV = $30250.00
Out of this $30250.00 he withdraws $15000.00 which leaves him with $15250.00
PV =15250.00
(time) n = 4 years
(interest) I/Y = 10%
Compute FV = 22327.52
He needs $ 26801.91 for his MBA but he only has $ 22327.52. This makes him short $ 4474.37
FV= 4474.37
PV= 0.00
(interest)I/Y = 10
(time) n = 6 years
Compute PMT = 579.91
|
| tabasco |
Posted
on 07-Mar-04 09:57 PM
. crusader, with my not so mathematical mind, i almost got the same answer with the same method except i used simple interest than compond for the fees. but later i realised that the method is little wrong as you are suppose to deposit money at the end of every year. the calculation at the end of two year wud be different as the total amount would be 25000 + interest + the money he deposited at the end of each year which is not equal to 30250 and has to be more than that. same applies for other 3,4,5,6 years too. this was more complicated than i thought.
;)
|
| CRUSADER |
Posted
on 08-Mar-04 06:06 PM
Tobasco,
Those calculations were done to find out what his $25000.00 would yield by the end of the 2nd, and then after deducting $15000.00 what the remainder would yield by the 6th year (Without including the payments because that is what we are trying to find!).
His 25000.00 would yield $ 22327.52, if no payments were made, which would leave him short $ 4474.37. This shortage helped me figure out the payment of
***************
$ 579.91 per year
***************
I though the question was ambiguous when it didn’t specify how the increments in tuition worked. It does not state if it increases by 5 %, of the base year’s tuition, each year ( 5 % of 20,000.00) or if it increases continuously (compound) . I assumed that it increases continuously for the next 6 years to come up with my answer.
***********************************************************************
However, if assumed that it increases by 5 %, of the base year's tuition, each year (Simple interest) for the next 6 years then the tuition payment on the 6th year will become
20000.00 + (20000 * .05 * 6)
= $ 26000.00
If he needs $ 26000.00 for the tuition payment, he is still short ( $26000.00 - $ 22327.52 ) $3672.48
FV = 3672.48
PV= 0.00
I/Y = 10.00 %
N = 6 yrs
********************
Compute PMT = 475.9805
*******************
Robert, did i BOMB this one ? What is the correct answer ?
|
| u_day |
Posted
on 08-Mar-04 06:57 PM
I remember doing that kinda problem back in Nepal when I was in I.Com. 1st year. Now don't ask me what I.Com. is......lol
|
| Robert Frost |
Posted
on 08-Mar-04 08:14 PM
What is I-COM????;)
Crusader, The correct answer is $580, which you had already bombed a long time back.
|
| Prem Charo |
Posted
on 08-Mar-04 10:16 PM
Robert Frost le aafno homework jati sabai yahi baata garne bichar garyaa ho ki kya ho??
Good Idea !! he he
PC :)
|
| Prem Charo |
Posted
on 08-Mar-04 10:28 PM
IF
If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments means (or =)
(a + b)2 = a2 + b2 + 2ab
Question :-- If as many numbers as we please beginning from a unit be set out continuously in double proportion, until the sum of all becomes prime, and if the sum multiplied into the last make some number, the product will be perfect = ????
PC :)
|
| u_day |
Posted
on 09-Mar-04 11:39 PM
square
|
| Rusty |
Posted
on 10-Mar-04 12:18 PM
Charo, what is the answer?
|
| kILLERbUZz |
Posted
on 10-Mar-04 12:48 PM
I WILL SQUEEZE MY BALLS UNTIL I DIE INSTEAD OF TRYING TO SOLVE THIS QUESTION .... THE QUESTION IS MORE CONFUSING THAN THE EXPECTED ANSWER ... APPRECIATE THE EFFORT THOUGH .....
|
| Prem Charo |
Posted
on 10-Mar-04 06:05 PM
Friends,
No any mathematics guru in Sajha, who can answer my above question?? Come on guys !
|
| Prem Charo |
Posted
on 29-Mar-04 11:10 PM
Killer buzz'
Are you still still squzzing your balls or you already dead ? anyway don't squeeze your balls.
To all loser sajhabasi_ _ _ Where is your answer??
Robert frost, and all _ _ _ _
PC :)
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