imqwerty
  • imqwerty
You got 1min to do this
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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imqwerty
  • imqwerty
show that all nos whose digits are permutation of 1234567890 are not prime
imqwerty
  • imqwerty
u have to prove none of the permutation is prime no
imqwerty
  • imqwerty
Lol

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Kainui
  • Kainui
All the digits add to 3, so it's divisible by 3.
imqwerty
  • imqwerty
yes :D correct!
Kainui
  • Kainui
Less than a minute even ;P
imqwerty
  • imqwerty
lol
Kainui
  • Kainui
Yeah actually this is a really cool thing to prove
Kainui
  • Kainui
No, it's not that bad.
imqwerty
  • imqwerty
i'll be doing and asking such ques today cx
Kainui
  • 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
  • imqwerty
evryday i hear abt new topic from u and i feel so bad
Kainui
  • Kainui
Hahaha are you mocking astrophysics or do you really feel that way? xD
imqwerty
  • imqwerty
lol xD
imqwerty
  • imqwerty
ok next question-
imqwerty
  • 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
  • ganeshie8
are teh integers different ?
imqwerty
  • imqwerty
yes
ganeshie8
  • ganeshie8
if so, we can try messing with pigeonhole principle
imqwerty
  • imqwerty
yes :) it has to be done with pigeonhole
ganeshie8
  • 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
  • Kainui
Is this really true, for any 7 numbers, \[a+b+c+d+e+f+g \equiv 0 \mod 7\]
ganeshie8
  • 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
  • ganeshie8
its not true for arbitrarily chosen 7 integers, consider an example : 7+14+21+28+35+42+1
DLS
  • 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
  • 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
  • DLS
Yes its true for the form of 7k+1.
imqwerty
  • imqwerty
yes :)
ganeshie8
  • ganeshie8
thats true for any 7 integers of form \(7k+a\) when \(a\) is fixed
ganeshie8
  • ganeshie8
because, the remainders then add up to \(7a\) \(7a\) is divisible by \(7\)
DLS
  • DLS
Can you prove how is it true if all 37 numbers are multiples of 6 ?
ganeshie8
  • ganeshie8
why are you considering multiples of 6 ?
ganeshie8
  • ganeshie8
i mean, any piarticular reason ?
DLS
  • DLS
Numbers are distinct and +ve, therefore obey the constraints of the question :P You can consider it as a corner case.
ganeshie8
  • ganeshie8
it is difficult to analyze a sum of multiples of 6 in modulus 7
imqwerty
  • 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
  • ganeshie8
nice, so we don't get anywhere...
Kainui
  • 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
  • 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
  • imqwerty
yes correct :)
imqwerty
  • 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
  • ganeshie8
6 + 9 = 7 + 8 so a circle can be inscribed in this quadrilateral
imqwerty
  • imqwerty
correct :)
ganeshie8
  • ganeshie8
that ends the proof too if you're familiar with pitot's theorem already :)
imqwerty
  • 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
  • ganeshie8
that is a simple yet cute result :)
imqwerty
  • 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
  • ikram002p
dot means \times or decimal comma ?
ganeshie8
  • ganeshie8
that should be easy, 24 right ?
imqwerty
  • imqwerty
yea :) (5n + 1)(5n + 2)(5n + 3)(5n + 4) ≡ 24 (mod 25) jst expand its times

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