Page 44 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

1.115 There are 95 numbers a 1 ,a 2 , ··· ,a 95 , which can only be +1 or -1. Find the
minimum value of the sum of all products of any two, S = a 1 a 2 + a 1 a 3 + ··· + a 94 a 95 ;

also determine how many (+1)’s and how many (-1)’s in the 95 numbers such that the
minimum S is obtained.

Proof: Assume there are m (+1)’s and n (-1)’s in a 1 ,a 2 , ··· ,a 95 , then m + n = 95 (i).

2
2
a 1 a 2 + a 1 a 3 + ··· + a 94 a 95 = S , multiply it by 2 plus a + a + ··· + a 2 95 = 95 :
1
2
(a 1 + a 2 + ··· + a 95 ) =2S + 95. a 1 + a 2 + ··· + a 95 = m − n , then (m − n) =2S + 95.
2
2
The minimum value of S to make 2S + 95 a perfect square is S min = 13. When S = S min ,
2
2
(m − n) = 11 , that is, m − n = ±11 (ii). (i)(ii) imply that m + n = 95,m − n = 11 or
m + n = 95,m − n = −11, from which we have m = 53,n = 42 or m = 42,n = 53. Hence,
when there are 53 (+1)’s and 42 (-1)’s, or there are 42 (+1)’s and 53 (-1)’s, S = S min = 13.

1.116 Let p = n(n + 1)(n + 2) ···(n + 7), where n is a positive integer, show

√
2
[ p]= n +7n +6.
4
Proof: Let a = n +7n +6, then
2
2
2
p = n(n+7)(n+1)(n+6)(n+2)(n+5)(n+3)(n+4) = (n +7n)(n +7n+6)(n +7n+.
2
2
4
2
4
10)(n +7n+12) = (a−6)a(a+4)(a+6) = a +4a(a −9a−36) = a +4a(a+3)(a−12)
4
When n ≥ 1, a> 12, then a <p . On the other hand,
2
2
3
3
2
4
2
4
4
(a+1) −p = a +4a +1+4a +2a +4a−a −4a +36a +144a = 42a +148a+1 >
4
3
2
2
2
4
3
4
2
(a+1) −p
0 ⇒ p< (a + 1) 4 = a +4a +1+4a +2a +4a−a −4a +36a +144a = 42a +148a+1 >
√
√
4
2
4
0 ⇒ p< (a + 1) . Hence, a <p < (a + 1) ⇒ a< 4 p <a +1 ⇒ [ p]= a = n +7n +6.
4
4
1.117 The real numbers a,b,c,d, e satisfy
2
2
a + b + c + d + e =8,a + b + c + d + e = 16, find the maximum value of e .
2
2
2

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