Page 41 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

If a, b, c are not divisible by 5, then the last digits of a ,b ,c can only be 1,4,6,9. Thus
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2
the last digits of a − b ,b − c ,c − a should have 0 or ±5, that is, at least one of
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2
2
2
2
2
a − b ,b − c ,c − a is divisible by 5. Since 2 and 5 are coprime, thus at least one of
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2

a b − ab = ab(a − b ),b c − bc = bc(b − c ),c a − ca = ca(c − a )a b − ab = ab(a − b ),b c − bc = bc(b − c ),c a − ca = ca(c − a ) is divisible by 10.
3 3
3 3
3 3
2 2
2 2
2 2
2 2
3 3
2 2
2 2
3 3
3 3
1.108 Let a, b, c are positive integers and follow a geometric sequence, and b − a
is a perfect square, log a + log b + log c =6, find the value of a + b + c .
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6
Solution: log a + log b + log c =6 ⇒ log abc =6 ⇒ abc =6 . In addition, b = ac , then
6
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b =6 = 36, ac = 36 . In order to make 36 − a a perfect square, a can only be 11,27,32,35, and
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a is a divisor of 36 , thus a = 27, then c = 48. Therefore, a + b + c = 27 + 36 + 48 = 111.
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1.109 The real numbers a, b, c, d satisfy a + b = c + d, a + b = c + d , show
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a 2011 + b 2011 = c 2011 + d 2011 .
Proof: If a + b = c + d =0, then the conclusion is obviously true.
If a + b = c + d =0, then
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2
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2
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2
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a + b = c + d ⇒ (a + b)(a − ab + b )=(c + d)(c − cd + d ) ⇒ a − ab + b = 2
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2
2 b = c + d ⇒ (a + b)(a − ab + b )=(c + d)(c − cd + d ) ⇒ a − ab + b =
2
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2
2 2
2
2 2
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a + 3
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2
c − cd + d ⇒ (a + b) − 3ab =(c + d) − 3cd ⇒ ab = cd ⇒ (a + b) − 4ab = b =
a + b = c + d ⇒ (a + b)(a − ab + b )=(c + d)(c − cd + d ) ⇒ a − ab +
2
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2
2cd + d ⇒ (a + b) − 3ab =(c + d) − 3cd ⇒ ab = cd ⇒ (a + b) − 4ab =
2
c −
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2
2
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c − cd + d ⇒ (a + b) − 3ab =(c + d) − 3cd
(c + d) − 4cd ⇒ (a − b) =(c − d) ⇒ a − b = ±(c − d) ⇒ ab = cd ⇒ (a + b) − 4ab =
2
2
2
(c + d) − 4cd ⇒ (a − b) =(c − d) ⇒ a − b = ±(c − d)
(c + d) − 4cd ⇒ (a − b) =(c − d) ⇒ a − b = ±(c − d). Hence,
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2
2
a − b = c − d (i)
a + b = c + d (ii)
or
a − b = d − c (iii)
a + b = c + d (iv)
(i)+(ii): a = c ; (i)-(ii): b = d .
(iii)+(iv): a = d ; (iii)-(iv): b = c .
For either case, we have a 2011 + b 2011 = c 2011 + d 2011 .
1.110 The real numbers a, b, c, d satisfy a + b + c + d =0 , show
a + b + c + d = 3(abc + bcd + cda + dab).
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