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Mathematics 18 Online
OpenStudy (goformit100):

Find a number which when divided by 10 leaves remainder 9, when divided by 8 leaves remainder 7, etc. down to where when divided by 2, it leaves remainder 1.

OpenStudy (goformit100):

@bahupepe

OpenStudy (goformit100):

@ganeshie8

OpenStudy (anonymous):

Managed to solve it by trial and error... I don't know if there's a better way. The divided by 2 leaves remainder 1 makes it odd The divided by 10 leaves 9 makes its unit digit 9

OpenStudy (perl):

this is a chinese remainder theorem problem

OpenStudy (anonymous):

Aahh.. clearly out of my league. lol

OpenStudy (unklerhaukus):

work backwards ?

terenzreignz (terenzreignz):

pff... It's A number, and no specifications to it. Try 10! - 1

OpenStudy (goformit100):

The Question I am posting are from International mathematics Olympiad Level-1 sample papers given By My Coach Professor Binay Kumar Sinha

terenzreignz (terenzreignz):

LOL I'm just kidding :)

terenzreignz (terenzreignz):

Or am I?

terenzreignz (terenzreignz):

You're probably looking for the smallest number that satisfies your conditions, though, @goformit100 and 10! - 1 is simply ghastly XD

OpenStudy (perl):

x = 9 mod 10 x = 8 mod 9 x =7 mod 8 x = 6 mod 7 x = 5 mod 6 x = 4 mod 5 x = 3 mod 4 x = 2 mod 3 x = 1 mod 2

OpenStudy (goformit100):

I personally solved this question in this was am i correct ?

terenzreignz (terenzreignz):

Okay.... time to work ... 1 divides 1 1x2 divides 1 and 2 1x2x3 divides 1,2,3,6 1x4x3 divides 1,2,3,4,6 1x4x6 divides 1,2,3,4,6,8 1x3x4x6 divides 1,2,3,4,6,8,9 1x3x4x5x6 divides 1,2,3,4,5,6,8,9,10 1x3x4x5x6x7 divides 1,2,3,4,5,6,7,8,9,10 So 2520 (that product) would be divisible by all from 2 to 10 so take away 1 it will yield the desired remainders 2519 is your answer ;)

OpenStudy (goformit100):

SEE Let the number = N So, N = 10base(a9) + 9 = 9 base(a8) + 8 = 8 base(a7) + 7 = 7 base(a6) + 6 … … = 2abase(1) +1 i.e. N+1 is having 2,3,4,5, ….. 10 as a factor Taking LEM of 2,3,4,5,….. 10 = 2520 is it correct ?

terenzreignz (terenzreignz):

LOL @goformit100 !!!! :)

OpenStudy (goformit100):

Am I correct ?

terenzreignz (terenzreignz):

Yup :)

OpenStudy (goformit100):

Next http://openstudy.com/study#/updates/517bbcaae4b0249598f7a76d

OpenStudy (perl):

what does LEM stand for?

OpenStudy (anonymous):

@terenzreignz Color me impressed!!! :-)

ganeshie8 (ganeshie8):

@terenzreignz wonderful logic !!!

terenzreignz (terenzreignz):

Thanks, guys :) I could summed it all up in what @goformit100 says, though (LEM, which I suspect means something along the lines of "least common multiple") I have a tendency to overdo things :/ But a big thanks to you guys :D

OpenStudy (perl):

im lost, what is the next step

terenzreignz (terenzreignz):

You find the least common multiple of the numbers from 1 to 10, and take away 1 from that least common multiple.

OpenStudy (perl):

x = 10 a_9 + 9 = 9 a_8 + 8 = 8 a_7 + 7 . . . = 2 a_1 + 1

OpenStudy (perl):

why do you take away 1 from the LCM

OpenStudy (perl):

x-1 = 2*a_1 x-2 = 3*a_2 x-3 = 4*a_3

terenzreignz (terenzreignz):

Because, since that number 2520 = 0 (mod <all of them>) if you take away 1, 2519 = -1(mod 10) = 9(mod 10) 2519 = -1(mod 9) = 8(mod 9) 2519 = -1(mod 8) = 7(mod 8) etc...

OpenStudy (perl):

ok , LCM {2,3,4,5,6,7,8,9,10} = 2520

ganeshie8 (ganeshie8):

would it be the least number with that property

terenzreignz (terenzreignz):

It would be the least.... I didn't even know what I was doing (finding the LCM) until goformit pointed it out :D

ganeshie8 (ganeshie8):

im not so sure i feel some numbers may exist with these properties before we reach LCM

OpenStudy (perl):

2520 mod 10 = 0 mod 10 , then you subtract both sides by 1 ? 2519 mod 10 = -1 mod 10

OpenStudy (perl):

and -1 mod 10 = 9 mod 10

OpenStudy (perl):

2520 mod 9 = 0 mod 9 2519 mod 9 = -1 mod 9 = 8 mod 9 = 8

OpenStudy (perl):

In general is it true that x mod a = y mod a iff (x + - c) mod a = (y + - c ) mod a ?

terenzreignz (terenzreignz):

Well, @ganeshie8 suppose N is the least positive integer satisfying those remainder properties... then N+1 is a positive integer that is divisible by all from 2 to 10 Then N+1 is a common multiple of 2 to 10 Now, for N to be the "least positive integer satisfying said properties" N+1 must be the least common multiple of those from 2 to 10

terenzreignz (terenzreignz):

Seems legit, @perl

OpenStudy (perl):

ok i think i see what you did now Given x = 9 mod 10 x = 8 mod 9 x =7 mod 8 x = 6 mod 7 x = 5 mod 6 x = 4 mod 5 x = 3 mod 4 x = 2 mod 3 x = 1 mod 2 Then x +1 = 10 mod 10 = 0 x +1= 9 mod 9 = 0 x +1=8 mod 8 = 0 x +1= 7 mod 7 = 0 x +1 = 6 mod 6 = 0 x +1= 5 mod 5 = 0 x +1= 4 mod 4 = 0 x +1= 3 mod 3 = 0 x +1 = 2 mod 2 = 0

OpenStudy (perl):

so x+1 = LCM { 2,3,4,5,6,7,8,9,10} , then x = LCM { 2,3,4,5,6,7,8,9,10} -1

ganeshie8 (ganeshie8):

hmm i got u. but the question doesnt require it to be divisible by 9 or 7 or 5 or 3

OpenStudy (perl):

The question should probably say, find the smallest number?

terenzreignz (terenzreignz):

It doesn't require it to be the least integer that satisfies, neither, @ganeshie8 LOL But I <facedesked> :) Let's use that logic again, but only for those mentioned... LCM{2,4,6,8,10}=120 119 then.

terenzreignz (terenzreignz):

Better tell this to @goformit100

OpenStudy (goformit100):

ya say

terenzreignz (terenzreignz):

Well, as @ganeshie8 pointed out, @goformit100 your question doesn't involve any division by the odd integers from 2-10 {3,5,7,9} So... 119 seems to be our best bet, following similar lines of thinking.

OpenStudy (perl):

isnt this the question ? find the smallest positive integer x such that x = 9 mod 10 x = 8 mod 9 x =7 mod 8 x = 6 mod 7 x = 5 mod 6 x = 4 mod 5 x = 3 mod 4 x = 2 mod 3 x = 1 mod 2

terenzreignz (terenzreignz):

There's no division by 9, 7, etc. @perl

OpenStudy (perl):

ohh

OpenStudy (perl):

ok well we can create a second follow up question :)

terenzreignz (terenzreignz):

But similar methods work. Just altering the arguments...

OpenStudy (perl):

x = 9 mod 10 x =7 mod 8 x = 5 mod 6 x = 3 mod 4 x = 1 mod 2

OpenStudy (goformit100):

ok guys Have we met with the conclusion, whoes answer is correct ?

terenzreignz (terenzreignz):

119. Thank @ganeshie8 kindly :)

OpenStudy (perl):

x = LCM { 2,4,6,8,10} -1 = 119

OpenStudy (perl):

i dont deserve medals for this problem -hangs head in shame-

ganeshie8 (ganeshie8):

finding next integer seems tricky

OpenStudy (perl):

whats the next question?

ganeshie8 (ganeshie8):

i mean, second least integer with given props

terenzreignz (terenzreignz):

Next integer would be 2<LCM> - 1

OpenStudy (perl):

right

ganeshie8 (ganeshie8):

ohh ya :)

OpenStudy (perl):

also we might want to prove that LCM is the least integer such that the respective mods produce zero.

OpenStudy (perl):

anybody has any questions, i see a lot of people observing

terenzreignz (terenzreignz):

You mean the least integer such that if you take away 1, you get the respective moduli?

OpenStudy (goformit100):

Is my next question http://openstudy.com/study#/updates/517bbcaae4b0249598f7a76d

OpenStudy (perl):

the question just asks 'find a number' , you could even use 2*3*6*8*10 -1

OpenStudy (perl):

it should say 'least positive integer'

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