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# remainder - shortcut method

Author Message
Director
Joined: 17 Dec 2005
Posts: 547
Location: Germany
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06 Mar 2006, 14:35
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hello,

if we want to get the remainder of a expression like:

(11*12*13*14)/5

is there a shortcut method?

thanks
Manager
Joined: 20 Mar 2005
Posts: 201
Location: Colombia, South America
Followers: 1

Kudos [?]: 16 [0], given: 0

Re: remainder - shortcut method [#permalink]

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06 Mar 2006, 18:21
Hello,

if we want to get the remainder of a expression like:

(11*12*13*14)/5

is there a shortcut method?

thanks

all I can think of is that a multiple of 5 must have a last digit that is 0 or 5
and its biggest possible mod is 4

if you take the last digits of 11*12*13*14

that would be 1*2*3*4 = 24
so the mod is 4

another example

like 11*12*13*18

1*2*3*8 = 48

mod is 3

I don't have a rigorous proof for this, but it works
SVP
Joined: 24 Sep 2005
Posts: 1885
Followers: 22

Kudos [?]: 321 [0], given: 0

Re: remainder - shortcut method [#permalink]

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06 Mar 2006, 21:11
conocieur wrote:
Hello,

if we want to get the remainder of a expression like:

(11*12*13*14)/5

is there a shortcut method?

thanks

all I can think of is that a multiple of 5 must have a last digit that is 0 or 5
and its biggest possible mod is 4

if you take the last digits of 11*12*13*14

that would be 1*2*3*4 = 24
so the mod is 4

another example

like 11*12*13*18

1*2*3*8 = 48

mod is 3

I don't have a rigorous proof for this, but it works

It's indeed a very good approach and i don't think there're still shorter ways
Director
Joined: 17 Dec 2005
Posts: 547
Location: Germany
Followers: 1

Kudos [?]: 40 [0], given: 0

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07 Mar 2006, 01:49
conocieur

this approach works very well, thank you.
07 Mar 2006, 01:49
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