Division problem, I need help please?

Hello guys, I’ve got another problem.

I’m dividing 575 by 23 using the long division method.

25

23/575

46

115

575

-460

So, I’ll tell you how I worked this out in case it was a problem with my working out.

I asked myself the following, how many times does 23 go into 5 it does not, therefore how many times does 23 go into 57, it goes in 2 times. So, I put the 2 on top of the 7.

Now, I’ve got to ask myself, 2 x 23 = 46 and then I put the 46 under the 23 and subtract 57 – 46 = 11, so I put 11 under the 46.

**GO RUNNING**

Then I bring the 5 down next to the 11 making 115.

I then ask myself how many times 23 goes into 115 well it goes in exactly 5 times. So I put the 5 on the answer column next to the 2.

I then do 25 x 23 = 575 and now the point where am stuck, I’ve got 115 – 575 = -460.

So the answer is 25 remainder -460?

Please can somebody help me out here?

Re: Division problem, I need help please?

Quote:

Originally Posted by

**Subliminalmessage** ...

**I then do 25 x 23 = 575 and now the point where am stuck, I’ve got 115 – 575 = -460.**

...

You only need to multiply 5 x 23, not the entire running quotient, so 5 x 23 = 115

http://i.imgur.com/b69Ov.png

Re: Division problem, I need help please?

Hello, Subliminalmessage!

You have invented a new (and incorrect) method of division.

This reminds me of an old Abbott and Costello routine.

Lou Costello insists that: $\displaystyle 5 \times 14 \,=\,25.$

Bud Abbott tries to correct him, but Lou "proves" it.

On a blackboard, he writes: .$\displaystyle \begin{array}{ccc}1&4 \\ \times & 5 \\ \hline \end{array}$

He says, "Five time four is twenty": . $\displaystyle \begin{array}{ccc}1&4 \\ \times & 5 \\ \hline 2 & 0 \end{array}$

Then "Five times one is five": .$\displaystyle \begin{array}{ccc}1&4 \\ \times &5 \\ \hline 2&0 \\ & 5 \end{array}$

And before Bud can object, Lou draws the line and adds: /$\displaystyle \begin{array}{cc}1&4 \\ \times & 5 \\ \hline 2 & 0 \\ & 5 \\ \hline 2 & 5 \end{array}$

"See?"

Bud says, "You can't multiply like that! .What if you add five 14's?"

Lou writes:. . $\displaystyle \begin{array}{ccc}&1&4 \\ &1&4 \\ &1&4 \\ &1&4 \\+ &1&4 \\ \hline \end{array}$

Bud takes the chalk and adds *up* the right column: "4, 8, 12, 16, 20 ..."

Lou pushes him aside and adds the *down* the left column: "21, 22, 23, 24, 25 !"

And has: .$\displaystyle \begin{array}{ccc}&1&4 \\ &1&4\\&1&4\\&1&4\\+&1&4\\ \hline & 2&5 \end{array}$

Bud complains again and asks "Okay, what is 25 divided by 5?"

Lou write: .$\displaystyle \begin{array}{cccc}&& - & - \\ 5 & ) & 2 & 5 \end{array}$

He says, "5 doesn't go into 2, but 5 goes into 5 once": .$\displaystyle \begin{array}{cccc}&&& 1 \\ && - & - \\ 5 & ) & 2 & 5 \end{array}$

"One times five is five": .$\displaystyle \begin{array}{cccc}&&&1 \\ && - & - \\ 5 & ) & 2 & 5 \\ &&&5 \end{array}$

"25 minus 5 is 20": .$\displaystyle \begin{array}{cccc}&&&1 \\ && - & - \\ 5 & ) & 2 & 5 \\ &&& 5 \\ &&- & - \\ && 2 & 0 \end{array}$

"And 5 goes into 20 four times": .$\displaystyle \begin{array}{ccccc}&&&1 & 4 \\ && - & - & - \\ 5 & ) & 2 & 5 \\ &&& 5 \\ & & - & - \\ & & 2 & 0 \\ && 2 & 0 \\ && - & - \\ \end{array}$

And Bud Abbott gives up . . .

Re: Division problem, I need help please?

Quote:

Originally Posted by

**daigo**

Ha ha, damn I can be stupid sometimes. Thank you very much for a nice and simple answer, it helps when you draw it out like that.

Re: Division problem, I need help please?

Quote:

Originally Posted by

**Soroban** Hello, Subliminalmessage!

You have invented a new (and incorrect) method of division.

This reminds me of an old Abbott and Costello routine.

Lou Costello insists that: $\displaystyle 5 \times 14 \,=\,25.$

Bud Abbott tries to correct him, but Lou "proves" it.

On a blackboard, he writes: .$\displaystyle \begin{array}{ccc}1&4 \\ \times & 5 \\ \hline \end{array}$

He says, "Five time four is twenty": . $\displaystyle \begin{array}{ccc}1&4 \\ \times & 5 \\ \hline 2 & 0 \end{array}$

Then "Five times one is five": .$\displaystyle \begin{array}{ccc}1&4 \\ \times &5 \\ \hline 2&0 \\ & 5 \end{array}$

And before Bud can object, Lou draws the line and adds: /$\displaystyle \begin{array}{cc}1&4 \\ \times & 5 \\ \hline 2 & 0 \\ & 5 \\ \hline 2 & 5 \end{array}$

"See?"

Bud says, "You can't multiply like that! .What if you add five 14's?"

Lou writes:. . $\displaystyle \begin{array}{ccc}&1&4 \\ &1&4 \\ &1&4 \\ &1&4 \\+ &1&4 \\ \hline \end{array}$

Bud takes the chalk and adds *up* the right column: "4, 8, 12, 16, 20 ..."

Lou pushes him aside and adds the *down* the left column: "21, 22, 23, 24, 25 !"

And has: .$\displaystyle \begin{array}{ccc}&1&4 \\ &1&4\\&1&4\\&1&4\\+&1&4\\ \hline & 2&5 \end{array}$

Bud complains again and asks "Okay, what is 25 divided by 5?"

Lou write: .$\displaystyle \begin{array}{cccc}&& - & - \\ 5 & ) & 2 & 5 \end{array}$

He says, "5 doesn't go into 2, but 5 goes into 5 once": .$\displaystyle \begin{array}{cccc}&&& 1 \\ && - & - \\ 5 & ) & 2 & 5 \end{array}$

"One times five is five": .$\displaystyle \begin{array}{cccc}&&&1 \\ && - & - \\ 5 & ) & 2 & 5 \\ &&&5 \end{array}$

"25 minus 5 is 20": .$\displaystyle \begin{array}{cccc}&&&1 \\ && - & - \\ 5 & ) & 2 & 5 \\ &&& 5 \\ &&- & - \\ && 2 & 0 \end{array}$

"And 5 goes into 20 four times": .$\displaystyle \begin{array}{ccccc}&&&1 & 4 \\ && - & - & - \\ 5 & ) & 2 & 5 \\ &&& 5 \\ & & - & - \\ & & 2 & 0 \\ && 2 & 0 \\ && - & - \\ \end{array}$

And Bud Abbott gives up . . .

First time I heard about Abbot & Costello and they're funny! Yes what I just did does resemble one of their routines!

play tennis

Thanks

Re: Division problem, I need help please?

This did remind me of something called "partial quotient division," but I don't want you to get confused so just take a look at it for fun at your leisure