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3 years ago

Least common multiples of two- digit numbers. I need two examples. SHOW WORK

Mathematics

1 answer:

Lerok [7]3 years ago

4 0

9514 1404 393

Answer:

LCM(21, 51) = 357
LCM(42, 90) = 630

Step-by-step explanation:

For a pair of numbers, I find it convenient to think in terms of factors unique to the number, and factors shared with the other number.

Consider the numbers 21 and 51, for example.

21 = 3·7

51 = 3·17

The factor 3 is shared by both numbers. The factors 7 and 17 are unique to one number or the other.

If we group these factors like this ...

(factors unique to 1 [ shared factors ) factors unique to 2]

= (7 [3 ) 17]

The numbers in ( ) parentheses are the factors of 21, and the numbers in [ ] brackets are the factors of 51.

The LCM (least common multiple) is the product of the factors in those brackets:

LCM(21, 51) = 7 × 3 × 17 = 357

The unique factors of the numbers don't have to be prime (as in the above example); they just cannot be shared. Here's another example for the numbers 42 and 90

(7 [ 3·2 ) 3·5 ] = (7 [ 6 ) 15] . . . . factors of (42) and [90]

Then the LCM is ...

LCM(42, 90) = 7 × 6 × 15 = 630

_____

Additional comment

What we have called "shared factors" here is the same as the "greatest common divisor"(GCD) or "greatest common factor" (GCF).

If we divide our little diagram so it shows the product of the two numbers:

(42)[90] = (7×6)·[6×15]

we can see that the LCM is the quotient of the product and the GCD.

LCM(A, B) = A·B/GCD(A, B)

This is an occasionally handy relationship, as it is not difficult to find the GCD of a pair of numbers using Euclid's algorithm.

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3 years ago

Let D = {-48, -14, -8, 0, 1, 3, 16, 23, 26, 32, 36} Determine which of the following statements are true and which are false. a)

morpeh [17]

Answer:

$\forall x\in D$ if x is odd then x> 0 is true statement.

$\forall x\in D$ if x is even then x≤0 is false statement.

$\forall x\in D$ If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.

$\forall x\in D$ if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.

Step-by-step explanation:

Consider the provided information.

D = {-48, -14, -8, 0, 1, 3, 16, 23, 26, 32, 36}

Part (A) $\forall x\in D$ if x is odd then x> 0

Here only even numbers are less than 0 that means the statement is true.

$\forall x\in D$ if x is odd then x> 0 is true statement.

Part (B) $\forall x\in D$ if x is less than 0 then x is even.

Here only even numbers are less than 0 that means the statement is true.

$\forall x\in D$ if x is odd then x> 0 is true statement.

Part (C) $\forall x\in D$ if x is even then x≤0

Here we can see that 16, 26, 32, 36 are even number and also greater than 0. Thus the statement is false.

$\forall x\in D$ if x is even then x≤0 is false statement.

Part (D) $\forall x\in D$ If the ones digit of x is 2, then the tens digit is 3 or 4.

There is only one number whose ones digit is 2. i.e. 32 also the tens digit of the number 32 is 3. Which makes the above statement true.

$\forall x\in D$ If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.

Part (E) $\forall x\in D$ if the ones digit of x is 6, then the tens digit is 1 or 2.

Numbers having ones digit 6 are: 16, 26 and 36

Here, the tens digits are 1, 2 and 3 which is contradict to our statement. Hence the provided statement is false.

$\forall x\in D$ if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.

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