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Find the first positive number that has remainder of 3 when divided by 4, a remainder of 4 when divided by 5, a remainder of 5 when divided by 6, and a remainder of 6 when divided by 7! Show how you approached this problem--guess and check, other methods!

By Serene (Serene) on Thursday, August 14, 2003 - 01:31 pm: Edit

LCM(4,5,6,7) = 35*12=420
420-1=419.

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 01:32 pm: Edit

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By Sup4 (Sup4) on Thursday, August 14, 2003 - 01:33 pm: Edit

ans = 15*9*11*13

By Serene (Serene) on Thursday, August 14, 2003 - 01:35 pm: Edit

15*9*11*3 will give remainder 0 when divided by 5.

sorry fairy haha i couldn't do multiplication =)

By Zerg_Vvins (Zerg_Vvins) on Thursday, August 14, 2003 - 01:35 pm: Edit

serene how did u approach this problem?

By Sup4 (Sup4) on Thursday, August 14, 2003 - 01:36 pm: Edit

sorry made an error

By Serene (Serene) on Thursday, August 14, 2003 - 01:37 pm: Edit

If you add 1 to the number, it gives 0 remainder for 4, 5, 6, and 7.

By Zerg_Vvins (Zerg_Vvins) on Thursday, August 14, 2003 - 01:38 pm: Edit

damn I am stupid, I see it now.

I approached it wrong, I found the lcm and then tried to add 6 to it...

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 01:41 pm: Edit

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By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 01:41 pm: Edit

Slightly harder. Find the 1st positive number having remainder 3 dividing 8, remainder 6 dividing 7, and remainder 2 dividing 11! =)

By Serene (Serene) on Thursday, August 14, 2003 - 01:54 pm: Edit

2*56+3*385+6*176 = 2323
2323=475(mod 616)

ans = 475

great problem fairy. got me confused for a while. =)

By Zerg_Vvins (Zerg_Vvins) on Thursday, August 14, 2003 - 02:01 pm: Edit

positively confused, y the number? y 2323 explain plz

By Serene (Serene) on Thursday, August 14, 2003 - 02:02 pm: Edit

zerg: I didn't write out the whole process on purpose so you can take your time thinking about it... =)

By Serene (Serene) on Thursday, August 14, 2003 - 02:10 pm: Edit

btw zerg, a similar math problem appears in JinYong's novel Shediao Yingxiong Zhuan (Legend of the Condor Heroes). That problem was incorporated into a quest in this online game I used to play (you couldn't advance unless you knew the code, which was given to you in remainders! =D), so that game was where I first looked at this type of problem. =)

By Zerg_Vvins (Zerg_Vvins) on Thursday, August 14, 2003 - 02:11 pm: Edit

grrr. at work now...
oh god this is gonna hurt my brain

By Zerg_Vvins (Zerg_Vvins) on Thursday, August 14, 2003 - 02:12 pm: Edit

does it appear in the novel or the game??

By Serene (Serene) on Thursday, August 14, 2003 - 02:12 pm: Edit

Both. =) The novel even included a poem that gave the particular solution for remainders when divided by 3, 5, and 7.

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 02:14 pm: Edit

And you wonder why it's called the CCCCCCCCrt. ;) hmm maybe i said too much

By Serene (Serene) on Thursday, August 14, 2003 - 02:15 pm: Edit

crt?

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 02:18 pm: Edit

Yea

By Serene (Serene) on Thursday, August 14, 2003 - 02:21 pm: Edit

what does crt stand for/mean?

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 02:21 pm: Edit

It's a theorem

By Soulofheaven8 (Soulofheaven8) on Thursday, August 14, 2003 - 02:22 pm: Edit

Pretty esoteric thread. =)

By Serene (Serene) on Thursday, August 14, 2003 - 02:23 pm: Edit

lol... I see. Except it's ok to say the name, I'm looking at the formula and still can't understand it completely
I just think of the problem as "preserving the remainder"... but of course, it took me some trial and error to get the multiplying factor

By Xiggi (Xiggi) on Thursday, August 14, 2003 - 02:28 pm: Edit

1.3 seconds is the answer

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 02:29 pm: Edit

You had this 2*56+3*385+6*176 = 2323

from:

2*56*1 + 3*77*5 + 6*88*2 right?

By Xiggi (Xiggi) on Thursday, August 14, 2003 - 02:30 pm: Edit

1.3 seconds is the answer to:

How long will it take for Serene to run CCCCCCCCrt though www.google.com and end up reading a Swiss site

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 02:31 pm: Edit

Yea it is much faster to solve 56x=1 (mod 11), 77x=1 (mod 8), and 88x=1 (mod 7) by trial and error then to do the reverse euclidean algorithm.

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 02:33 pm: Edit

Zerg? Have you figured it out yet?

By Serene (Serene) on Thursday, August 14, 2003 - 02:39 pm: Edit

Fairy: yeah that's where I got it from. =)

lol xiggi =)

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 02:43 pm: Edit

Xiggi that swiss site actually has some really cool ascii movies =)

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 02:52 pm: Edit

Can anyone see the stereograms in these two:

http://www.romanm.ch/ascii-3d/ascii-3D_bild1.htm
http://www.romanm.ch/ascii-3d/ascii-3D_bild2.htm

First one is a heart like shape with three prongs. Second one is a smiley face!

By Serene (Serene) on Thursday, August 14, 2003 - 02:56 pm: Edit

Neat!!!!! =)

By Sup4 (Sup4) on Thursday, August 14, 2003 - 03:16 pm: Edit

how do you do this

(105)^-1 (mod 2) = 1

By Zerg_Vvins (Zerg_Vvins) on Thursday, August 14, 2003 - 03:26 pm: Edit

I figured out the answer to it pretty fast by trial and error, I just used multiples of 11 then took away 9 and got the answer, but I was trying to figure out an equation...

By Zerg_Vvins (Zerg_Vvins) on Thursday, August 14, 2003 - 03:31 pm: Edit

2*56*1 + 3*77*5 + 6*88*2

<-- I got to 56, 77, 88 and got stuck

where do u get teh 2,1 3,5 and 6,2?

By Serene (Serene) on Thursday, August 14, 2003 - 03:44 pm: Edit

2, 3, 6 are remainders
1, 5, 2 from trial and error

By Zerg_Vvins (Zerg_Vvins) on Thursday, August 14, 2003 - 04:02 pm: Edit

so in a manner of speaking, u did trial and error as well?

I did 11x-9 trial and error and eventually got 484-9 which worked...

By Serene (Serene) on Thursday, August 14, 2003 - 04:06 pm: Edit

---

Quote:

I just think of the problem as "preserving the remainder"... but of course, it took me some trial and error to get the multiplying factor

---

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 04:13 pm: Edit

Zerg, trial and error isn't necessary but is certainly much faster:

For example, 56x=1 (mod 11), solving the normal way.

56x+11y=1 has a solution since (56,11)=1. 56=5*11+1, so 1=1*56-5*11. So x=1.

Or a longer example 88x=1 (mod 7):

88x+7y=1, 88=12*7+4, 7=1*4+3, 4=1*3+1, 1=4-(1*3)=4-(7-1*4)=2*4-7=2(88-12*7)-7=2*88-25*7, so x=2 there.

By Serene (Serene) on Thursday, August 14, 2003 - 04:44 pm: Edit

Fairy: Wow! That is sooooooooooooo cool =)
PS. If you get a negative number through the 'normal way', you just add more multiples to make it positive right?

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 04:47 pm: Edit

If x0 and y0 are a particular solution to ax+by=c, all solutions are given by x=x0+(b/d)t, y=y0-(a/d)t, where d=(a,b), and t is some integer.

By Zerg_Vvins (Zerg_Vvins) on Thursday, August 14, 2003 - 04:49 pm: Edit

wow fairy, thnx for the explaination...

Just 1 quick question, how do you remember all these math? I mean don't u get formulas or equations confused sometimes?

By Zerg_Vvins (Zerg_Vvins) on Thursday, August 14, 2003 - 04:53 pm: Edit

fairy: btw did you happen to be on that TV show a couple of years back, the child genius one?

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 05:20 pm: Edit

i wish

By Billiam2 (Billiam2) on Thursday, August 14, 2003 - 05:32 pm: Edit

By Zerg_Vvins (Zerg_Vvins) on Thursday, August 14, 2003 - 05:44 pm: Edit

why not? were you not a studious student back then?

By Fairyofwind (Fairyofwind) on Thursday, August 14, 2003 - 05:51 pm: Edit

Much knowledge is really only a matter of seeing more. And any math I know was learned in the past year. For example, I didn't even make the AIME last year as a freshman.