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 #5
avatar+91 
-1

\(\mathrm{Write\:the\:problem\:in\:long\:division\:format}:2381\overline{|5666}\)

\(\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space2\)

\(\mathrm{Divide}\:5666\:\mathrm{by}\:2381\:\mathrm{to\:get}\:2:381\overline{|5666}\space\)

\(\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space2\)

\(\mathrm{Multiply\:the\:quotient\:digit}\:\left(2\right)\:\mathrm{by\:the\:divisor}\:2381:2381\overline{|5666}\)

\(\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\underline{4762}\)

\(\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space2\)

\(\mathrm{Subtract}\:4762\:\mathrm{from}\:5666:2381\overline{|5666}\space\)

\(\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\underline{4762}\)

\(\mathrm{The\:solution\:for\:Long\:Division\:of}\:\frac{5666}{2381}\:\mathrm{is}\:2\:\mathrm{with\:remainder\:of}\:904\)

.
25.01.2021
 #1
avatar+91 
-1

Before we start, note that in the problem 125 divided by 8, the numbers are defined as follows:

125 = dividend
8 = divisor

 

Step 1:
Start by setting it up with the divisor 8 on the left side and the dividend 125 on the right side like this:

       

  8⟌125


Step 2:
The divisor (8) goes into the first digit of the dividend (1), 0 time(s). Therefore, put 0 on top:
 

    0  

  8⟌125


Step 3:
Multiply the divisor by the result in the previous step (8 x 0 = 0) and write that answer below the dividend.
 

    0  

  8⟌125

   0  


Step 4:
Subtract the result in the previous step from the first digit of the dividend (1 - 0 = 1) and write the answer below.
 

    0  

  8⟌125

   -0  

    1  


Step 5:
Move down the 2nd digit of the dividend (2) like this:
 

    0  

  8⟌125

   -0  

    12 


Step 6:
The divisor (8) goes into the bottom number (12), 1 time(s). Therefore, put 1 on top:
 

    01 

  8⟌125

   -0  

    12 


Step 7:
Multiply the divisor by the result in the previous step (8 x 1 = 8) and write that answer at the bottom:
 

    01 

  8⟌125

   -0  

    12 

     8 


Step 8:
Subtract the result in the previous step from the number written above it. (12 - 8 = 4) and write the answer at the bottom.
 

    01 

  8⟌125

   -0  

    12 

   - 8 

     4 


Step 9:
Move down the last digit of the dividend (5) like this:
 

    01 

  8⟌125

   -0  

    12 

   - 8 

     45


Step 10:
The divisor (8) goes into the bottom number (45), 5 time(s). Therefore put 5 on top:
 

    015

  8⟌125

   -0  

    12 

   - 8 

     45


Step 11:
Multiply the divisor by the result in the previous step (8 x 5 = 40) and write the answer at the bottom:
 

    015

  8⟌125

   -0  

    12 

   - 8 

     45

     40


Step 12:
Subtract the result in the previous step from the number written above it. (45 - 40 = 5) and write the answer at the bottom.
 

    015

  8⟌125

   -0  

    12 

   - 8 

     45

   - 40

      5


You are done, because there are no more digits to move down from the dividend.

The answer is the top number and the remainder is the bottom number.

Therefore, the answer to 125 divided by 8 calculated using Long Division is:

15
5 Remainder

30.12.2020
 #1
avatar+91 
+2

How many 4-digit positive integers exist that satisfy the following conditions:

(A) Each of the first two digits must be 1, 4, or 5, and

(B) the last two digits cannot be the same digit, and

(C) each of the last two digits must be 5, 7, or 8?

 

(A)
Each of the first two digits must be 1, 4, or 5 ?

\(\begin{array}{|lllc|c|} \hline & x&x &xx \\ \hline &1 \cdots &\begin{Bmatrix} 1\\4\\5 \end{Bmatrix}\cdots &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} & 3\times 100 \\\\ &4 \cdots &\begin{Bmatrix} 1\\4\\5 \end{Bmatrix}\cdots &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} & +3\times 100 \\\\ &5 \cdots &\begin{Bmatrix} 1\\4\\5 \end{Bmatrix}\cdots &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} & +3\times 100 \\\\ \hline & & & & = 3\times 3\times 100 = {\color{red}900} \\ \hline \end{array}\)

 

There are 900 4-digit positive integers that satisfy the condition.

 

(B)
The last two digits cannot be the same digit ?

\(\begin{array}{|lcc|c|} \hline & xx &xx \\ \hline &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 00 & 90 \\\\ &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 11 & 90 \\\\ &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 22 & 90 \\\\ & \cdots \\ \\ &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 99 & 90 \\\\ \hline & & & = 90\times 10 = 900 \\ \hline \end{array}\)

 

1000 until 9999: There are 9000  four-digit integers.

 

9000 - 900 = 8100

 

There are 8100 4-digit positive integers that satisfy the condition.

 

(C)
Each of the last two digits must be 5, 7, or 8 ?

\(\begin{array}{|lclc|c|} \hline & x&x &xx \\ \hline &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} \cdots &\begin{Bmatrix} 5\\7\\8 \end{Bmatrix}\cdots & 5& 100\times 3 \\\\ &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} \cdots &\begin{Bmatrix} 5\\7\\8 \end{Bmatrix}\cdots & 7& +100\times 3 \\\\ &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} \cdots &\begin{Bmatrix} 5\\7\\8 \end{Bmatrix}\cdots & 8& +100\times 3 \\\\ \hline & & & & = 3\times 100\times 3 = {\color{red}900} \\ \hline \end{array}\)

 

There are 900 4-digit positive integers that satisfy the condition.

29.12.2020