[Guest]

A curious whole number, whose last digit is 7, has the curious property that in order to multiply it by 7, all you need to do is to move the seven from the right end and place it at the beginning.

What is the number?

[Guest]

Good riddle, i can remember seeing this some where before, but for the life of me can't recall at the moment, so for the time being, without aid of computer program, i am trying trial and error without success! I just need to recall where... Oh well nearly finshed work for the year and this will keep me occupied until it's time to go.

All the best for 2009 everyone in the Den

nb edit spelling

Edited December 31, 2008 by underground_dan

[HoustonHokie]

1,014,492,753,623,188,405,797 x 7 = 7,101,449,275,362,318,840,579

[Guest]

Top banana HoustonHokie, that would have taken me to next new year!!!

NB edit spelling

Edited December 31, 2008 by underground_dan

[HoustonHokie]

Top banana HoustonHokie, that would have taken me to next new year!!!

NB edit spelling

I've never had to automate multiplication digit by digit before, but my spreadsheet gave out at 15 digits. It was an interesting process to set up formulas to multiply two numbers, add the "carryover" number, figure out the 0 and 10 place, and report those in various locations. Then it was a matter of carrying it out place after place and looking for a spot where the product began with "71". First time for everything...

[Guest]

I could have never, ever figured that out. Good job, man!

[Guest]

A curious whole number, whose last digit is 7, has the curious property that in order to multiply it by 7, all you need to do is to move the seven from the right end and place it at the beginning.

What is the number?

1014492753623188405797....but would be interested to know ur way of solvn it...

[Guest]

I had already heard this somewhere and thought i would post it up. Btw, Houston how did you get it?

[Jiminy Cricket]

I've never had to automate multiplication digit by digit before, but my spreadsheet gave out at 15 digits. It was an interesting process to set up formulas to multiply two numbers, add the "carryover" number, figure out the 0 and 10 place, and report those in various locations. Then it was a matter of carrying it out place after place and looking for a spot where the product began with "71". First time for everything...

Impressive! Good job!

[Guest]

I've never had to automate multiplication digit by digit before, but my spreadsheet gave out at 15 digits. It was an interesting process to set up formulas to multiply two numbers, add the "carryover" number, figure out the 0 and 10 place, and report those in various locations. Then it was a matter of carrying it out place after place and looking for a spot where the product began with "71". First time for everything...

wow

[HoustonHokie]

I had already heard this somewhere and thought i would post it up. Btw, Houston how did you get it?

Very carefully...

Basically I performed multiplication one digit at a time and kept expanding the candidate number based on the results of my previous multiplications. Because that sentence makes almost no sense to me, here's a better look at what I did:

7
x 7
---
49

Now, I moved the 4 and copied the 9 like so (I moved the 4 up so it would be
added to the product of 7 & 9 - just the way I learned to do it in grade school!):

4
97
x 7
---
679

Moving the 6 and copying the 7 like so:

64
797
x 7
----
5579

and on and on and on. Here's the way my calculation looked at the end:

013361532412165204564
1014492753623188405797
x 7
----------------------
7101449275362318840579

I knew that the last digit of the number was 7, so I started with that:

Actually, I put together a little spreadsheet to help me make sure that I multiplied, added, and copied numbers correctly, but the work above was my thought process.

I knew that I had to be looking for a 10 at the beginning so that when I took the product of 7 x 0 it would have no carryover number and then the product of 7 x 1 would be 7. When I saw that, I stopped!

Edited December 31, 2008 by HoustonHokie

[Guest]

Very carefully...

Basically I performed multiplication one digit at a time and kept expanding the candidate number based on the results of my previous multiplications. Because that sentence makes almost no sense to me, here's a better look at what I did:

7
x 7
---
49

Now, I moved the 4 and copied the 9 like so (I moved the 4 up so it would be
added to the product of 7 & 9 - just the way I learned to do it in grade school!):

4
97
x 7
---
679

Moving the 6 and copying the 7 like so:

64
797
x 7
----
5579

and on and on and on. Here's the way my calculation looked at the end:

013361532412165204564
1014492753623188405797
x 7
----------------------
7101449275362318840579

I knew that the last digit of the number was 7, so I started with that:

Actually, I put together a little spreadsheet to help me make sure that I multiplied, added, and copied numbers correctly, but the work above was my thought process.

I knew that I had to be looking for a 10 at the beginning so that when I took the product of 7 x 0 it would have no carryover number and then the product of 7 x 1 would be 7. When I saw that, I stopped!

:o