Math 6 Chapter 1 Lesson 16: Common divisors and common multiples
1. Summary of theory Tóm
1.1. The common divisor of two or more numbers is the divisor of all of them
If \(\left. \begin{array}{l}a \vdots x\\b \vdots x\\c \vdots x\end{array} \right\} \Rightarrow x \in \) OK(a; b;c)
Example 1:
U(8) = {1; 2; 4; 8}
U(12) = {1; 2; 3; 4; twelfth}
UC(8; 12) = {1; 2; 4}
If the set A = U(8) and the set B = U(12) are represented, then UC(8;12)= \(A \cap B = {\rm{\{ }}1;2;4\} \)
Among the common divisors of two or more numbers, the largest number is called the greatest common divisor. UCLN symbol.
How to find the GCC of two or more numbers:
– Step 1: Parse each number into prime factors
– Step 2: Pick out the common prime factors.
– Step 3: Make a product of the selected factors, each factor taken with the smallest exponent.
The product found is the UCLN to find
Example 2: Find the GCC (48; 168; 360).
We have: \(48 = {2^3}.3,\,\,\,168 = {2^3}.3.7,\,\,360\, = {2^3}{.3^2} .5\)
UCLN (48; 168; 360) = \({2^3}.3 = 24\)
1.2. A common multiple of two or more numbers is a multiple of all of them.
\(\left. \begin{array}{l}x \vdots a\\x \vdots b\\x \vdots c\end{array} \right\} \Rightarrow x \in BC(a;b;c) )\)
Example 3:
B(6) = {0; 6; twelfth; 18; 24; 30; 36; 42; 48;…}
B(8) = {0; 8; 16; 24; 32; 40; 48; …}
BC(6 ;8) = {0; 24; 48; 72;…}
\(BC(6;8) = B(6)\,\,\, \cap \,\,B(8)\,\, = \,\,{\rm{\{ }}0;\, \,\,24;\,\,\,48;… {\rm{\} }}\)
Least common multiple (BCNN) of two or more numbers is the smallest nonzero number in the set of multiples of those numbers.
* How to find the balance sheet of two or more numbers:
– Step 1: Factor each factor into prime factors.
– Step 2: Choose common and special factors.
– Step 3: Build the product of the selected factors, each factor taking the largest exponent.
The product found is the BCNN to be found.
Example 4:
\(\begin{array}{l}84 = {2^2}.3.7\\140 = {2^2}.5.7\\360 = {2^3}{.3^2}.5\end{ array}\)
\(BCNN = {2^3}{.3^2}.5.7 = 2520.\)
2. Illustrated exercise
Question 1: Find a fourdigit natural number A such that, when divided by 131, the remainder is 112, when divided by 132, the remainder is 97 but is divisible by 99.
Solution guide:
According to the problem, we have: A = 131p + 112 = 132q + 97
Or 131p = 132q – 15 = 131q + (q – 15)
\( \Rightarrow q – 15\,\,\, \vdots \,\,\,131\,\, \Rightarrow \,\,q = 131x + 15\,\,(x \in \mathbb{N} )\)
where A = 132q + 97 = 132. (131x + 15) = 132 .131x + 1980
Since A has four digits, x = 0 and 1980 : 99 = 20
So the number to find is A = 1980.
Verse 2: Let a = 123456789; b=987654321.
a) Find the LCC of (a; b)
b) Find the remainder in the division of the state balance sheet (a; b) by 11.
Solution guide:
a.
\(a\,\,\, \vdots \,\,\,9\) and \(b\,\,\, \vdots \,\,\,9\) (because the sum of its digits divides end of 9)
On the other hand b – 8a = 9 so if UC(a; b) = d then \(9\,\, \vdots \,\,d\)
So every GCC of a, b is a GCC of 9 or 9 = GCLN (a; b)
b.
Because \(BCNN(a;b) = \frac{{ab}}{{UCLN(a;b)}} = \frac{{ab}}{9} = \frac{a}{9}.b\ )
But \(\frac{a}{9} = 11m + 3;\,\,\frac{b}{9} = 11n\,\, + 5.\)
So BCNN (a;b) = 11p + 4
So the required balance is 4.
Question 3:
a. Find \(a \in {\mathbb{N}^*}\), knowing \(a\,\, \vdots \,\,\,378,\,\,a\,\, \vdots \,\ ,594.\)
b. Find \(b \in {\mathbb{N}^*}\), know \(112\,\,\, \vdots \,\,\,b;\,\,280\,\, \vdots \ ,\,\,b.\)
Solution guide:
a. \(a\,\, \vdots \,\,\,378,\,\,a\,\, \vdots \,\,594\,\, \Rightarrow \,\,a\,\, = BCNN \,\,(378;\,\,594)\)
\(378 = {2.3^3}.7,\,\,594\,\, = \,{2.3^3}.11\)
So a = BCNN(378; 594)
b. \(112\,\, \vdots \,\,b,\,\,280\,\, \vdots \,\,b\,\, \Rightarrow \,\) b = UCLN (112; 280)
\(112 = {2^4}.7,\,\,280\,\, = \,{2^3}.5.7\)
So b = GCLN(112; 280) = \({2^3}.7 = 56.\)
3. Practice
3.1. Essay exercises
Question 1: Find a threedigit natural number A such that, when divided by 13, the remainder is 9, when divided by 15, the remainder is 5 but is divisible by 31.
Verse 2: Let a = 420; b=630.
a) Find the LCC of (a; b)
b) Find the remainder in the division of the state balance sheet (a; b) by 11.
Question 3:
a. Find \(a \in {\mathbb{N}^*}\), know \(a\,\, \vdots \,\,\,12,\,\,a\,\, \vdots \,\ ,15.\)
b. Find \(b \in {\mathbb{N}^*}\), know \(330\,\,\, \vdots \,\,\,b;\,\,451\,\, \vdots \ ,\,\,b.\)
3.2. Multiple choice exercises Bài
Question 1: Find the set of common multiples of 15 and 18 less than 200
A. \(A = \left\{ {0;45;90;120} \right\}\)
B. \(A = \left\{ {0;45;90;120;180} \right\}\)
C. \(A = \left\{ {0;90;80} \right\}\)
D. \(A = \left\{ {0;60;90;120} \right\}\)
Verse 2: The set UC(4, 12) is:
A. \(\left\{ {0;1;2;4} \right\}\)
B. \(\left\{ {1;2;4} \right\}\)
C. \(\left\{ {1;2;3;4} \right\}\)
D. \(\left\{ {1;2;3;4;6} \right\}\)
Question 3: Given a set A consisting of multiples of 8, set B consisting of multiples of 100, and set C having a common multiple of 8 and 100. State the relationship between set C and two sets A and B.
A. \(C \subset A,C \subset B\)
B. \(A \subset C,B \subset C\)
C. \(C \subset A,B \subset C\)
D. \(A \subset C,C \subset B\)
Question 4: Find the intersection of two sets A and B, knowing that
A = {book; pen; ruler; eraser}
B = {book; book; bag; ruler; eraser}
A. C={book; book; eraser}
B. C={book; pen; book; eraser}
C. C={book; ruler; eraser}
D. C={book; book; bag}
Question 5: List the elements of the set A = UC{20; 30}
A. A={1; 2; 4; ten}
B. A={1; 2; 5;10; 15}
C. A={1; 2; 5}
D. A={1; 2; 5;10}
4. Conclusion
Through this lecture on Common Divisors and Common Multiples, students need to complete some of the objectives given by the lesson, such as:

Forming the concept of common divisor, common multiple.

How to find the common divisor, common multiple of two or more numbers.
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