Basics of Mathematics

[surendar]

Hi,

I know its a lovely subject.. Its been years since i practiced Mathematics..Hence lost touch a bit in basics. And i am attending an exam this week for which we should be strong enough in basics.

I have some doubts in basic questions.. can you help me out,

1) Is there a formula to calculate

 n + n^2 + n^3 + n^4 + ........ + n^x

or

 1 + n + n^2 + n^3 + ..... + n^x

:what


2) Can any one suggest a solution to this problem?

A person went to market with a red and a green bag and bought 10 carrots & 6 radishes. On the way to home he distributed vegetables into the two bags in such a way that no bag is empty. In how many ways can he do this?

It should be a permutation and combinations.. but my mind is going blank somehow.

3)

Given 125w + 25x + 5y + z = 264. Also if w, x, y & z are non-negative integers and less than 5, then the value of w + x + y + z is?

4)

Given a sequence 8, -7, 6, -5, 4, ------------ 80. Find the sum of first 27 numbers?

Just want to confirm whether it's (-5) is answer. When i calculated -5 only came. Correct me if i am wrong.

5)

If the median of a seven number sequence is 2n+2 then find the Arithmetic mean of that sequence?

My mind is going blank with what approach to proceed.

For Question 3 : I proceeded as below but am not sure with the approach i had taken the problem.

Since 125w + 25x + 5y + z = 264, w can be either 1 or 2.

w=1 --> 25x + 5y + z = 139

In this case, even if the x takes atmost value 4, it leaves us, 5y + z = 39 from which it looks like equation doesn't satisfies the condition given ( all are non neg integers < 5 ) to proceed further. Hence w cant be 1.

w=2 --> 25x + 5y + z = 14

For this scenario, x must be 0. Which will take to 5y + z = 14 from which we can substitute y = 2 and z = 4 to satisfy the equation.

=> w=2, x=0, y=2,z=4 ( sum : 8 )

Is this the only approach to proceed with these type of problems? Or any shortcuts available? perhaps, the exam requires to solve this problem in one minute to 90 seconds. :

 

[shravi]

I know its a lovely subject

I beg to differ!

 

[Kshitiz_Indian]

lol, all these are pretty good ones. I'm feeling too tired to solve them all right now.

But what I feel from the back of my mind is that the first one can be solved using binomial theorem, its pretty similar to this!

 

[surendar]

I beg to differ!

hehe.. it does depend on the interest which one puts up in mathematics. This mathematics is going to play a crucial role almost everywhere in our real life.. Even the age of this earth has been estimated till Dec 12,2012 using some graph interpretation. ( Accuracy of this calculation is unknown though! )

lol, all these are pretty good ones. I'm feeling too tired to solve them all right now.

But what I feel from the back of my mind is that the first one can be solved using binomial theorem, its pretty similar to this!

Yeah Kshitiz, i also thought.. but isn't it like application of binomial theorem like (1+x)^n something like that? :what

Problem is, I forgot all those formulas and even have lost my 10th std Mathematic text book which does contain all the standard formulas for basics

 

[Kshitiz_Indian]

I'll get you binomial theorem concept in an hour... just got to find that damn book.

 

[jordox]

Maths is so bad. Oh my god.

 

[usy]

I was like the best in Maths 7 years ago. Like The A Best.

 

[ZoraxDoom]

Maths is awesome.

I'll get to work on those soon...

 

[ZoraxDoom]

1) I learnt something like this. Sequences...
Wait, let me try something out.

n+...n^6

When n=2
2+4+8+16+32+64...
= 126 (= n^7 - n)
So, simplified, n+...n^x = n^(x+1) - n

When n=3
3+9+27+81+243+729
= 1092 (=3^7 - 1095. 1095/3 = 365. 365 = n+...n^5 + 2. Interesting...)
Simplified, = n^(x+1) - n(n+...n^(x-1) + 2)

When n=5
5+25+125+625+3125+15625
= 19530 (=5^7 - 58595. 58595/5 = 11,719. 11,719 = n^6 - n+...n^5 + 1. Hmm. What is going on here...)
Simplified, =n^(x+1) - n(n^x - (n+...n^(x-1) + 1) )

n+...n^7
2+4+8+16+32+64+128
=254 (=n^8 - n. Hmm...)

I'll pick up on this later. I think I see a pattern emerging, but I'm not sure.

 

[Kshitiz_Indian]

Surendar, I think for the first series you HAVE to use Binomial theorem. Have a look. http://en.wikipedia.org/wiki/Binomial_theorem

Using x = 1 we can get your result but in the reverse way, but it doesn't matter in an addition, and since x = 1 any power of x will be 1!

 

[surendar]

@ Zorax and Kshitiz : Thank you for the having a try on the first sum. I was also trying that sum this evening. And let me share, what I found out.

Series is : n+n^2+n^3+n^4+..... n^x.

This can be re-written as,

n(n^0) + n(n^1) + n(n^2) + ..... n(n^x-1) -- eq 1

This can be solved by Geometric progression where a = n and r = n as the GP series can be generally written as ar^0, ar^1, ar^2, ar^3.... ar^n

And the formula would be,

Sn = a(1-r^n+1) / 1-r where n = x-1 in eq 1.

Thank you again for the try on this one. Perhaps, I am still clueless on the other sums though. I tried first one alone and that took time to recognize that it was a GP one.

 

[Highlander999]

Hi,

I know its a lovely subject.. Its been years since i practiced Mathematics..Hence lost touch a bit in basics. And i am attending an exam this week for which we should be strong enough in basics.

This is so obvious it I am shocked you haven't done it. This got me through 5 years of Maths. It is called the sly glance at the paper next to you, or my favourite the old steal the answer sheet before the exam

 

[surendar]

This is so obvious it I am shocked you haven't done it. This got me through 5 years of Maths. It is called the sly glance at the paper next to you, or my favourite the old steal the answer sheet before the exam

hehe.. You cannot do that because, every day and in fact every batch(morning/eve) the questions keep changing to different individuals. And no one knows where the question paper is coming from. It's neither a univ exam nor a school exam.

Basics should be strong enough because, as I mentioned, each question had to be solved in less number of minutes. Time management is one of the major constraints in these kind of exams.

 

[surendar]

Can anyone figure it out whether the below question is proper?

If n is divided by 24, the remainder is 21. Which of the following is the divisor of n?

I somehow feel that question is missing something or wrong :what If it's a proper question, what will be the solution?

The options given were 3,4,6,9,10! I am clueless what to choose :what

 

[Spinksy]

I have no idea, sorry. You guys are teaching me new things to be perfectly honest. My class is just starting to learn directed numbers. Ghay!