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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=40c or, c=40a So , (1)=> 3a+80 +b = 19 or, b = 3a61 
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? 
