Divisors, (Extended) Euclidean algorithm 1. Express the numbers 12! = 479,001,600 and the binomial coefficient (2) = 2,704,156, each as products of their prime factors. Do this without using your calculator in any way. Use this to calculate the greatest common divisor, and least common multiple of the two numbers. Confirm how these relate to the product of the numbers. 2. I purchased a number of items that cost $31.70 mch and some other items that met $116.90 each. The total cost of my purchases was $ XXXXXXXXXXHow many items of mch type did I purchase? 3. Let g be the greatest common factor of a = 2,637,110,736 and 6 = 2,961,110,016. (a) Find g using Euclid's algorithm% and (b) using one of the techniques shown in lectures, Sod suitable integers x and y to satisfy the equation g = 2,961,110,016 x+2,637,110,7369. (c) Check whether 6768 is a multiple of g. (It should be!) Find integers s and t mob that 527,760 = 2,961,110,016s + 2,637,110,736 t. Don't just give a single s and t, but find all pairs of integers (s, t) that satisfy this equation. (d) As part of your answer to part (c) you should check that the numbers you have found do indeed satisfy the given equation. If you haven't done so, then do it here, as if a separate task. Explain any difficulties you encountered while doing mob a check, and how you overcame them.

First downloaded: 21/10/2017 at 20:51,17 tl'ou may use appmpriate software when answering parts of this exercise, provided you explain what programs you have ..a, for what purpose, and how you actually made use of them.

First downloaded: 21/10/2017 at 20:51,17 tl'ou may use appmpriate software when answering parts of this exercise, provided you explain what programs you have ..a, for what purpose, and how you actually made use of them.

Answered Same DayDec 27, 2021

1)

12! = 12(11)(10)…1 = 479,001,600

i.e. 3(2

2

)(11)(2)(5)(3

2

)(2

3

)(7)(3)(2)(5)(2

2

)(3)(2)(1)

= 2

10

(3

5

)(5

2

)(7)(11)

24C12 =

=

( )( )( ) ( )

( ) ( )

=2704156

i.e. 2704156 =

(13)(2)(7)(3)(5)(2

4

)(17)(2)(3

2

)(19)(2

2

)(5)(3)(7)(2)(11)(23)(2

3

)(3)/12!

= 2

12

(3

5

)(5

2

)(7

2

)(11)(13)(17)(19)(23)/ 2

10

(3

5

)(5

2

)(7)(11)

= 2

2

(13)(17)(19)(23)(7)

Thus we expressed both 12! And 24

C

12 as a product of prime factors.

Greatest common factor = factors appearing in both

= 2

2

(7)=28.

Least common multiple= Highest factor appearing in either

= 2

10

(3

5

)(5

2

)(7)(11)(13)(17)(19)(23)

LCM = 46260537523200

Product of HCF and LCM = = 2

2

(7)(13)(17)(19)(23)* 2

2

(7)

=2

4

(7

2

)(13)(17) )(19)(23)= 1295295050649600

Product of two numbers = = 1295295050649600

Hence we find that product of two numbers = Product of hcf and lcm

*

Let x be the number of items purchased for 31.70 and y for 116.90

Total cost = 31.70x +...

12! = 12(11)(10)…1 = 479,001,600

i.e. 3(2

2

)(11)(2)(5)(3

2

)(2

3

)(7)(3)(2)(5)(2

2

)(3)(2)(1)

= 2

10

(3

5

)(5

2

)(7)(11)

24C12 =

=

( )( )( ) ( )

( ) ( )

=2704156

i.e. 2704156 =

(13)(2)(7)(3)(5)(2

4

)(17)(2)(3

2

)(19)(2

2

)(5)(3)(7)(2)(11)(23)(2

3

)(3)/12!

= 2

12

(3

5

)(5

2

)(7

2

)(11)(13)(17)(19)(23)/ 2

10

(3

5

)(5

2

)(7)(11)

= 2

2

(13)(17)(19)(23)(7)

Thus we expressed both 12! And 24

C

12 as a product of prime factors.

Greatest common factor = factors appearing in both

= 2

2

(7)=28.

Least common multiple= Highest factor appearing in either

= 2

10

(3

5

)(5

2

)(7)(11)(13)(17)(19)(23)

LCM = 46260537523200

Product of HCF and LCM = = 2

2

(7)(13)(17)(19)(23)* 2

2

(7)

=2

4

(7

2

)(13)(17) )(19)(23)= 1295295050649600

Product of two numbers = = 1295295050649600

Hence we find that product of two numbers = Product of hcf and lcm

*

Let x be the number of items purchased for 31.70 and y for 116.90

Total cost = 31.70x +...

SOLUTION.PDF## Answer To This Question Is Available To Download

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