You got 1min to do this

- imqwerty

You got 1min to do this

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- jamiebookeater

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- imqwerty

show that all nos whose digits are permutation of 1234567890 are not prime

- imqwerty

u have to prove none of the permutation is prime no

- imqwerty

Lol

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## More answers

- Kainui

All the digits add to 3, so it's divisible by 3.

- imqwerty

yes :D
correct!

- Kainui

Less than a minute even ;P

- imqwerty

lol

- Kainui

Yeah actually this is a really cool thing to prove

- Kainui

No, it's not that bad.

- imqwerty

i'll be doing and asking such ques today cx

- Kainui

Lol I don't even understand the twin paradox.
Proof goes like this, I'll stray away from modular stuff, so let's say we have a 3 digit number a,b,c are the digits, so we can write it like this:
\[a*100+b*10+c\]
However we can write all of these like:
\[a*(99+1)+b*(9+1)+c\]
move the trash to the side like this:
\[a*99+b*9 + a+b+c\]
Since \(a*99+b*9\) is divisible by 3, it doesn't affect that the number is divisible by 3 or not, so w just have to look at if a+b+c is divisible by 3.

- imqwerty

evryday i hear abt new topic from u and i feel so bad

- Kainui

Hahaha are you mocking astrophysics or do you really feel that way? xD

- imqwerty

lol xD

- imqwerty

ok next question-

- imqwerty

we are given any 37positive integers and we have to prove that it is possible to choose any 7 whos sum is divisible by 7.

- ganeshie8

are teh integers different ?

- imqwerty

yes

- ganeshie8

if so, we can try messing with pigeonhole principle

- imqwerty

yes :) it has to be done with pigeonhole

- ganeshie8

Consider the remainders when the integers are divided by 7 :
\[\{0,1,2,3,4,5,6\}\]
7 holes, and 37 pigeons
By pigeonhole principle, one hole must contain at least 6 pigeons.

- Kainui

Is this really true, for any 7 numbers,
\[a+b+c+d+e+f+g \equiv 0 \mod 7\]

- ganeshie8

no,
if we're given 37 different positive integers, then there exists at least one combination such that the sum is divisible by 7

- ganeshie8

its not true for arbitrarily chosen 7 integers, consider an example :
7+14+21+28+35+42+1

- DLS

@ganeshie8 ..Consider a sequence containing a mixture of multiple of 6s and 7s.
Somewhat like..
6, 12, 18, 24, 30, 36...and 7, 14, 21,28..etc
So every element in the set of 37 integers is either of form 7k or 6k so remainders would have a pattern of 0 and 6's . In that case it is impossible for sum to be divisible by 7 I guess.
Correct me if I'm wrong..:|

- ganeshie8

if you take 7 integers of form 7k+1.
the remainders add up to 7, which is divisible by 7. so we're good right ?

- DLS

Yes its true for the form of 7k+1.

- imqwerty

yes :)

- ganeshie8

thats true for any 7 integers of form \(7k+a\)
when \(a\) is fixed

- ganeshie8

because, the remainders then add up to \(7a\)
\(7a\) is divisible by \(7\)

- DLS

Can you prove how is it true if all 37 numbers are multiples of 6 ?

- ganeshie8

why are you considering multiples of 6 ?

- ganeshie8

i mean, any piarticular reason ?

- DLS

Numbers are distinct and +ve, therefore obey the constraints of the question :P
You can consider it as a corner case.

- ganeshie8

it is difficult to analyze a sum of multiples of 6 in modulus 7

- imqwerty

the 37 multiples of 6 are of this form->6m
we choose any seven
the sum will be like this-\[6[m_{1}+m_{2}+m_{3}....m_{7}]\]
again we have to prove that m1,m2,m3.... this sum is divisible by 7
so we jst come back to our original question

- ganeshie8

nice, so we don't get anywhere...

- Kainui

Imagine all the 37 are the same mod 7, then it doesn't matter what the numbers are. In fact, if you only had 2 different types of numbers mod 7 in there, you'd be able to always pick a set of all the same. Actually we can keep doing this and have 5 different types of numbers in there that are all the same mod 7, since 5*7=35, so we could always find some number in there that occurs 7 times.
It's only when we add a 6th number and 7th number variety to the mix that we have to think about it being a problem.

- ganeshie8

Here is the complete proof :
Consider the remainders when the integers are divided by 7 :
\[\{0,1,2,3,4,5,6\}\]
label the holes as \(0,1,2\ldots, 6\).
Case 1: each hole has at least one pigeon.
7 holes, and 37 pigeons
By pigeonhole principle, one hole must contain at least 6 pigeons.
Take 6 numbers form that hole, the remainders add up to 6a, which is same as \(-a\) under modulus 7.
Next, for the 7th number, take it from the hole \(a\).
the reaminders add up to 0.
Case 2 : at least one hole is empty. (assume exactly one hole is empt for definiteness)
6 holes occupied, and 37 pigeons
By pigeonhole principle, one hole must contain at least 7 pigeons.
Take the 7 numbers from that hole.
the remainders add up to \(7a\), which is divisible by \(7\)

- imqwerty

yes correct :)

- imqwerty

ok a quickques-
A quadrilateral ABCD has side lengths AB = 6, BC = 7, CD = 9, DA = 8. Prove that there exists a point P in ABCD such that the perpendiculars from P to the sides of ABCD are equal.

- ganeshie8

6 + 9 = 7 + 8
so a circle can be inscribed in this quadrilateral

- imqwerty

correct :)

- ganeshie8

that ends the proof too if you're familiar with pitot's theorem already :)

- imqwerty

yes (: if u hav any inscribed circle in a quad then all the sides are tangents and then like this-|dw:1447592259222:dw|

- ganeshie8

that is a simple yet cute result :)

- imqwerty

nxt ques
Find the remainder when \[(5 ·15122014^{2014} + 1)(5 · 15122014^{2014 }+ 2)(5 · 15122014^{2014} +3)(5 · 15122014^{2014} + 4) \]is divided by 25

- ikram002p

dot means \times or decimal comma ?

- ganeshie8

that should be easy, 24 right ?

- imqwerty

yea :)
(5n + 1)(5n + 2)(5n + 3)(5n + 4) ≡ 24 (mod 25)
jst expand
its times

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