Question

many ques
1)A boy writes three numbers, 1, 2 and 3, on the sheet.
Now, every girl take turns to the sheet and erase one number,
and then replace it by the sum of the two numbers left. After some turns,
is it possible to have the numbers: 6^2012, 7^2013, 8^2014 on the sheet at
the same time?
(:
Give Proof

Answer 1

maby

Answer 2

but since you tell me to pay you $100 to anwser a question do i have to pay you $100 to solve

Answer 3

i know that answer :P

Answer 4

its a question for u

Answer 5

fine but it may take a while to figre it out

Answer 6

ok

Answer 7

No, not possible cause the largest number has to be the sum of the other 2 numbers and
\[8^{2014} \ne 6^{2012}+7^{2013}\]

Answer 8

hmm vry nice i never thought this way
ur correct :)
i'll show my solution

Answer 9

how ever you do it, it will be 2 odd and 1 even?

Answer 10

i cant figure it out

Answer 11

initially we hav 2 odd nos and 1 even no on the sheet.
If an odd no is chosen, the sum of the remaining two nos will be odd.
If an even no is chosen, the sum of the remaining two nos will be even. Hence the sum of the numbers on the sheet will always be even.
6^2012 + 7^2013 + 8^2014 ≡ 1 (mod 2).
Hence it is impossible to have the nos : 6^2012, 7^2013, 8^2014 on the sheet at the same time.

Answer 12

@baru That's cool I like that

Answer 13

yes :)

Answer 14

yipee!

Answer 15

Fun wait let's see if we can't find some other way to solve this now this was interesting. I really kinda wanted to work out some kind of formula for any permutation of picking the 3 numbers though... Or we can do another question if you got it haha

Answer 16

yes i got many cx but they are all messed up
ima search for some nice ques and post it :)

Answer 17

nxt ques-
Find the smallest prime number which does not divide any five digit no whose digits are in strictly increasing order.

Answer 18

Hmmmm am I allowed to use programming? lol

Answer 19

no lol

Answer 20

I strongly suspect it's 11 but not sure.

Answer 21

yea its 11

Answer 22

I basically reasoned out all our numbers will end in 5, 6, 7, 8, or 9 and that immediately means we can't have 2, 3, 5, 7. Multiplying by 11 seems to make numbers create like a pyramid sorta thing so you won't have that strictly decreasing effect from it, but yeah that's not a rigorous proof or anything lol.

Answer 23

nice one

Answer 24

thanks :)
shuld i give a lil hint?

Answer 25

by ascending order do you mean smallest in the units place? or the other way round?

Answer 26

once you have a strong feeling \(11\) is the answer, below might help in proving it

\(10^n \equiv (-1)^n\pmod{11}\)

Answer 27

that means that number srts frm small digit and then bigger and bigger :)

Answer 28

that is a good question @baru

i think he means the digits are incerasing from left to right
but yes the question is ambiguous w/o that info

Answer 29

i think it has to do with the divisibility test for 11

abcde are the digits

then (a+c+e) - (b+d) should be divisible by 11 is the rule i think

Answer 30

yes :)

Answer 31

i'm guessing the number you get from doing that ^, is always less than 11?

Answer 32

Awesome ! that should do...

Answer 33

yes that is the actual approach :)

Answer 34

we should think of the case where (a+c+e) is the largest and (b+d) is the smallest?

Answer 35

try taking a general form :)
like assume c to be x

Answer 36

:(

Answer 37

:)

Answer 38

everybody left?? i still don't know how to finish it :(

Answer 39

ok so
we take c to be x
now the numbers are consecutive so a,b,c,d,e will be-->
a->x-2
b->x-1
c->x
d->x+1
e->x+2

now jst try to apply divisibility test :)

Answer 40

why not
a: x-3
b x-2
c: x
d: x+1
e x+3

or some other random thingy and still ascending order

Answer 41

the numbers are consecutive
so the difference between the numbers should be 1

here u took b=x-2 and c=x
the difference c-b=2 so they are not positive so u can't do this

Answer 42

u didnt say consecutive....

Answer 43

o-o oh no
this was my solution and i read the ques wrong wait a sec ima work on another solution

Answer 44

lets say the number is abcde

then a<b<c<d<e

(a+c+e)-(b+d)
a+(c-b)+(e-d)
c>b so c-b>0
e>d so e-d>0

so clearly
(a+c+e)-(b+d)>a

we can also see it this way-(a+c+e)-(b+d)
e-(d-c)-(b-a)<e

so a<(a+c+e)-(b+d)<e
a and e lie in this range->[1,9] so surely this number is not divisible by 11

Answer 45

we can also conclude that the number need not be like a<b<c<d<e
it can also be like a>b>c>d>e

Answer 46

u make me read all comments to realize it's a work sheet not only one question :-\

Answer 47

nice though @imqwerty

Answer 48

(: thanks

Answer 49

cool!!