from "101 problems in algebra" Log Out | Topics | Search
Moderators | Register | Edit Profile

Ask NRICH » Archive 2010-2011 » Onwards and Upwards » from "101 problems in algebra" « Previous Next »

Author Message
johncena
Frequent poster

Post Number: 173
Posted on Monday, 29 August, 2011 - 06:12 pm:   

Let x1,x2,x3,...,xn be the sequence of integers such that

1) -1 <= xi <= 2 for i=1,2,3,...,n
2) x1 + x2 + x3 + ... + xn = 19
3) (x1)^2 + (x2)^2 + ... + (xn)^2 = 99

Determine the minimum and maximum possible values of (x1)^3 + (x2)^3 + ... + (xn)^3

Problem is that i can't understand the solution given in the book....
they take a= number of -1's in the sequence b=1's and c=2's

and obtain -a+b+2c = 19 and a+b+4c = 99

I can't understand this step....pls help .....

thanks in advance.
Graeme McRae
Veteran poster

Post Number: 2917
Posted on Monday, 29 August, 2011 - 06:51 pm:   

Johncena,

Let me just explain one of those two equations, and then I think you'll understand the other one as well.

If there are "a" -1's, then in your equation 3), the number of terms equal to (-1)^2 is "a", and since each of the terms equal to (-1)^2 has a value of 1, these terms combined contribute the value of "a" to the total.

If there are "b" 1's, then in your equation 3), the number of terms equal to (1)^2 is "b", and since each of the terms equal to (1)^2 has a value of 1, these terms combined contribute the value of "b" to the total.

If there are "c" 2's, then in your equation 3), the number of terms equal to (2)^2 is "c", and since each of the terms equal to (2)^2 has a value of 4, these terms combined contribute the value of "4c" to the total.

So then, reviewing the last three paragraphs I wrote, the value of the terms of equation 3) is "a" plus "b" plus "4c", and equation 3 tells us this sum is 99. This explains the second equation in the solution given in the book.

Did you follow that? If not, write back, and perhaps someone else will explain it differently.
johncena
Frequent poster

Post Number: 174
Posted on Tuesday, 30 August, 2011 - 04:40 pm:   

I got it now ....


thank u very much sir....
Graeme McRae
Veteran poster

Post Number: 2919
Posted on Tuesday, 30 August, 2011 - 06:13 pm:   

You're quite welcome. Do you need help with the next few steps in the solution? For example, given that -a+b+2c=19 and a+b+4c=99, can you express b and c in terms of a?
johncena
Frequent poster

Post Number: 175
Posted on Wednesday, 31 August, 2011 - 06:58 am:   

-a+b+2c = 19...(1)

a+b+4c = 99...(2)

=> a+19+a+2c = 99
=> a=40-c or, c=40-a

So , (1)=> -3a+80 +b = 19
or, b = 3a-61
Graeme McRae
Veteran poster

Post Number: 2920
Posted on Wednesday, 31 August, 2011 - 02:16 pm:   

Good! Do you completely understand the solution of this problem now?

Add Your Message Here
Posting is currently disabled in this topic. Contact your discussion moderator for more information.

Topics | Last Day | Last Week | Tree View | Search | Help/Instructions | Program Credits Administration