How do you solve a Complex Volume problem?

A four-digit number N leaves a remainder of 10 when divided by 21, a remainder of 11 when divided by 23, and a remainder of 12 w

aksik [14]

Answer:

16

Step-by-step explanation:

There may be a formal way to solve simultaneous Diophantine equations, but I don't know what it is. So, I addressed this by solving them 2 at a time.

First of all, we want to find a relation between integers p and q such that ...

21p+10 = 23q +11

21p -23q = 1 . . . . . . subtract 23q+10

Using the extended Euclidean algorithm or trial and error or astute observation, we find that (p, q) = (11, 10) is a solution to this equation. Then we can write p and q as ...

p = 23n+11
q = 21n+10

for some integer n.

Then our original numbers become ...

21p+10 = 21(23n+11)+10 = 483n +241 . . . . for some integer n

For this to be a 4-digit number, we must have n such that

1000 ≤ 483n +241 ≤ 9999

1.6 < n < 20.2

__

Modulo 25, the number (483n+241) is ...

8n +16

and we want that value to be 12:

(8n +16) mod 25 = 12

(8n +4) mod 25 = 0 . . . . . subtract 12

4(2n+1) mod 25 = 0 . . . . factor out 4

For this to be true, we must have 2n+1 be a multiple of 25. The only value of n that is in the required interval [2, 20] is n=12.

__

When n=12, 483n+241 = 6037. The sum of digits of 6037 is 16.

3 0

3 years ago

In a certain lottery, an urn contains balls numbered 1 to 34. from this urn, 6 balls are chosen randomly, without replacement

Anarel [89]

$\displaystyle |\Omega|=\binom{34}{6}=\dfrac{34!}{6!28!}=\dfrac{29\cdot30\cdot31\cdot32\cdot33\cdot34}{2\cdot3\cdot4\cdot5\cdot6}=1344904\\ |A|=1\\\\ P(A)=\dfrac{1}{1344904}$

7 0

3 years ago

Graph the equation y = k x + 1 if it is known that the point M belongs to it:<br> M (2, −7)

Elenna [48]

Answer:y = k x + 1

(1,3)

3 = k(1) + 1

2 = 1k

k = 2

Step-by-step explanation:

7 0

2 years ago

How many garbage cans should be placed in each campground? Explain how you can use addition or multiplication to find the answer

Vikentia [17]

Answer:

Step-by-step explanation:

Multiplication

Ratio of Tables to garbage = 8:3

Small camp ground:

40 picnic tables and let x be the number of garbage cans

$\frac{8}{3}=\frac{40}{x}\\\\\frac{8*5}{3*5}=\frac{40}{x}\\\\\frac{40}{15}=\frac{40}{x}\\\\$

x = 15

Small campground with 40 picnic table needs 15 garbage cans

Large picnic ground:

120 picnic tables and let y be the number of garbage cans

$\frac{8}{3}=\frac{120}{y}\\\\\frac{8*15}{3*15}=\frac{120}{y}\\\\\frac{120}{45}=\frac{120}{y}$

y= 45

Big campground with 120 picnic table needs 45 garbage cans

3 0

2 years ago

In the figure below, ⎯⎯⎯⎯⎯⎯⎯⎯,⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯, and ⎯⎯⎯⎯⎯⎯⎯⎯⎯ are medians of △. If KC = 108 and OC = 2n+10, then n=

Nimfa-mama [501]

The question was incomplete. Below you will find the missing content.

In the figure below, BL, AM, and CK are medians of △ABC. If KC = 108 and OC = 2n + 10, then n=

The picture is attached below.

The value of n is 31.

The median is the line connecting the vertex and its midpoint on the opposite side of the triangle.

The intersection of all 3 medians of the triangle is called the centroid.

As we know centroid divides the median in the ratio of 2:1.

In the given picture,

Medians of triangle △ABC are AM, BL, and CK.

So the centroid of the triangle is O.

Given that KC= 108

As the centroid O divides the line KC in 2:1.

Let OC=2x

KO= X

As KC= KO+OC

⇒ 108= x+2x

⇒ 3x=108

⇒ x=108/3

⇒ x=36

Then OC= 2x= 2*36= 72

As given in the question OC= 2n+10

putting the value of OC in above equation

⇒ 72= 2n+10

⇒ 2n= 72-10

⇒ 2n=62

⇒ n=62/2

⇒n=31

Therefore the value of n is 31.

Learn more about the median of the triangle

here: brainly.com/question/2264495

#SPJ10

5 0

2 years ago