Question

You got 1min to do this

Answer 1

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

Answer 2

u have to prove none of the permutation is prime no

Answer 3

Lol

Answer 4

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

Answer 5

yes :D
correct!

Answer 6

Less than a minute even ;P

Answer 7

lol

Answer 8

Yeah actually this is a really cool thing to prove

Answer 9

No, it's not that bad.

Answer 10

i'll be doing and asking such ques today cx

Answer 11

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.

Answer 12

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

Answer 13

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

Answer 14

lol xD

Answer 15

ok next question-

Answer 16

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.

Answer 17

are teh integers different ?

Answer 18

yes

Answer 19

if so, we can try messing with pigeonhole principle

Answer 20

yes :) it has to be done with pigeonhole

Answer 21

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.

Answer 22

Is this really true, for any 7 numbers,

\[a+b+c+d+e+f+g \equiv 0 \mod 7\]

Answer 23

no,

if we're given 37 different positive integers, then there exists at least one combination such that the sum is divisible by 7

Answer 24

its not true for arbitrarily chosen 7 integers, consider an example :

7+14+21+28+35+42+1

Answer 25

@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..:|

Answer 26

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 ?

Answer 27

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

Answer 28

yes :)

Answer 29

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

Answer 30

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

\(7a\) is divisible by \(7\)

Answer 31

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

Answer 32

why are you considering multiples of 6 ?

Answer 33

i mean, any piarticular reason ?

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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.

Answer 39

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 \(7\)

Answer 40

yes correct :)

Answer 41

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.

Answer 42

6 + 9 = 7 + 8

so a circle can be inscribed in this quadrilateral

Answer 43

correct :)

Answer 44

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

Answer 45

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

Answer 46

that is a simple yet cute result :)

Answer 47

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

Answer 48

dot means \times or decimal comma ?

Answer 49

that should be easy, 24 right ?

Answer 50

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