8. Find lcm(143, 227), lcm(306, 657), and lcm(272, 1479).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
2.4 #8
![8:26
K 41
Your Algebra answer is re... ^ V
2.4 #8
b) gcd(24, 138) = 24x + 138y.
© gcd(119, 272) = 119x + 272y.
d) gcd(1769, 2378) = 1769x + 2378y.
3 Prove that if d is a common divisor of a and b, then d = gcd(a, b)
gcd(a/d,b/d) = 1.
[Hint: Use Theorem 2.7.]
4. Assuming that gcd(a, b) = 1, prove the following:
(a) gcd(a + b, a – b) = 1 or 2.
[Hint: Let d = gcd(a + b, a – b) and show that d|2a, d | 2b,
d < gcd(2a, 2b) = 2 gcd(a, b).]
(b) gcd(2a + b, a + 2b) = 1 or 3.
(c) gcd(a + b, a² + b²) = 1 or 2.
[Hint: a² + b² = (a + b)(a – b) + 2b².]
(d) gcd(a + b, a² – ab + b²) = 1 or 3.
[Hint: a? – ab + b² = (a + b)? – 3ab.]
5. For n > 1, and positive integers a, b, show the following:
(a) If gcd(a, b) = 1, then gcd(a", b") = 1.
[Hint: See Problem 20(a), Section 2.2.]
(b) The relation a" | b" implies that a | b.
[Hint: Put d = gcd(a, b) and write a
gcd(r", s") = 1. Show that r =
6. Prove that if gcd(a, b) = 1, then gcd(a + b, ab) = 1.
7. For nonzero integers a and b, verify that the following conditions are
(a) a |b.
(b) gcd(a, b) = |a |.
(c) lcm(a, b) = |b|.
8. Find lcm(143, 227), lcm(306, 657), and lcm(272, 1479).
9.) Prove that the greatest common divisor of two positive integers divides
multiple.
10. Given nonzero integers a and b, establish the following facts concer
(a) gcd(a, b) = lcm(a, b) if and only if a = ±b.
(b) If k > 0, then lcm(ka, kb) = k lcm(a, b).
(c) If m is any common multiple of a and b, then lcm(a, b)| m.
[Hint: Put t =
0<r < t. Show that r is a common multiple of a and b.]
11. Let a, b, c be integers, no two of which are zero, and d
%3D
-
-
rd,b = sd, where gcd(r, s
1, whence a = d.]
%3D
lcm(a, b) and use the Division Algorithm to write
gcd(a, b,
d = gcd(gcd(a, b), c) = gcd(a, gcd(b, c)) = gcd(gcd(@
VIEW ANSWER](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fadf70025-15e9-4d93-968e-9d86049b61bd%2F8301cc62-0701-46c9-b0ea-b97d7feb77cf%2Fbjsbzuc_processed.png&w=3840&q=75)
Transcribed Image Text:8:26
K 41
Your Algebra answer is re... ^ V
2.4 #8
b) gcd(24, 138) = 24x + 138y.
© gcd(119, 272) = 119x + 272y.
d) gcd(1769, 2378) = 1769x + 2378y.
3 Prove that if d is a common divisor of a and b, then d = gcd(a, b)
gcd(a/d,b/d) = 1.
[Hint: Use Theorem 2.7.]
4. Assuming that gcd(a, b) = 1, prove the following:
(a) gcd(a + b, a – b) = 1 or 2.
[Hint: Let d = gcd(a + b, a – b) and show that d|2a, d | 2b,
d < gcd(2a, 2b) = 2 gcd(a, b).]
(b) gcd(2a + b, a + 2b) = 1 or 3.
(c) gcd(a + b, a² + b²) = 1 or 2.
[Hint: a² + b² = (a + b)(a – b) + 2b².]
(d) gcd(a + b, a² – ab + b²) = 1 or 3.
[Hint: a? – ab + b² = (a + b)? – 3ab.]
5. For n > 1, and positive integers a, b, show the following:
(a) If gcd(a, b) = 1, then gcd(a", b") = 1.
[Hint: See Problem 20(a), Section 2.2.]
(b) The relation a" | b" implies that a | b.
[Hint: Put d = gcd(a, b) and write a
gcd(r", s") = 1. Show that r =
6. Prove that if gcd(a, b) = 1, then gcd(a + b, ab) = 1.
7. For nonzero integers a and b, verify that the following conditions are
(a) a |b.
(b) gcd(a, b) = |a |.
(c) lcm(a, b) = |b|.
8. Find lcm(143, 227), lcm(306, 657), and lcm(272, 1479).
9.) Prove that the greatest common divisor of two positive integers divides
multiple.
10. Given nonzero integers a and b, establish the following facts concer
(a) gcd(a, b) = lcm(a, b) if and only if a = ±b.
(b) If k > 0, then lcm(ka, kb) = k lcm(a, b).
(c) If m is any common multiple of a and b, then lcm(a, b)| m.
[Hint: Put t =
0<r < t. Show that r is a common multiple of a and b.]
11. Let a, b, c be integers, no two of which are zero, and d
%3D
-
-
rd,b = sd, where gcd(r, s
1, whence a = d.]
%3D
lcm(a, b) and use the Division Algorithm to write
gcd(a, b,
d = gcd(gcd(a, b), c) = gcd(a, gcd(b, c)) = gcd(gcd(@
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