Problem for you #4 : A number ends with 7. Find out the smallest number such that this number becomes 5 times if 7 is shifted to the 1st position. Good Luck!!

Question

anonymous · Answer

How do you define shifting?

ABC7 -> 7ABC ?

anonymous · Answer

Yeah.

anonymous · Answer

and 7ABC =5 * ABC7, right?

apoorvk · Answer

Well that means the no. (original) ends with '57'.

anonymous · Answer

@FFM Yeah. Excpet that you dont know if its 4 digit. But otherwise yes.

experimentx · Answer

it's my wild guess that first digit must be 1

apoorvk · Answer

Hmm. So so so...

The first digit of the no. has got to be '1'.

So, the whole thing is like, suppose there are 'n' digits, then..

apoorvk · Answer

Yeah @experimentX your wild guess ain't so wild!

experimentx · Answer

lol ... second digit must be 4 or 5 ...

apoorvk · Answer

$$5 \times (10^n + ...... + 5 \times 10^2  + 7) = 7 \times 10^n  + 10^{n-1} + ......+ 5$$

anonymous · Answer

* LHS last is 5 * 10

apoorvk · Answer

something like this^.

Yeah and now I guess, it's "find the next digit" game.


Yes SORRY. - LHS's second last term would be 5x10 and NOT 5x10^2 - thanks @siddhantsharan

experimentx · Answer

mos likely the second digit is 4 ... and the third digit it 2 or 3

anonymous · Answer

Apoorv. You got the problem with your way?

experimentx · Answer

http://www.wolframalpha.com/input/?i=142857*5 ??

anonymous · Answer

@experimentX That is correct. Method?

experimentx · Answer

it's hard to explain 

2*5 = 10 ... so first digit is cannot be greater than 1
5*1 = 5
we we need 2 carry ... which means 4 and 5

apoorvk · Answer

yeah, that explanation for '1' is alright to me.

experimentx · Answer

4 or 5 !!
it's difficult to get 5 9*5 = 45 <-- so i choose 4

anonymous · Answer

How'd you know the number of digits involved.
Or wait. You just proceeded till you got a 7?

experimentx · Answer

something like this and that ... more of it .. hit and trial
and a bit of luck

anonymous · Answer

Haha. You always do get there. Nice going man!! :D:D

experimentx · Answer

lol ... not really ... these types of question you usually ask never favors me ... sometimes it get lucky that's all

experimentx · Answer

any way thanks for the question too

anonymous · Answer

Do you know the answer?

anonymous · Answer

Well I have solved it.

anonymous · Answer

Not a bad problem, I have solved something like this 2 years back.

anonymous · Answer

HINT: $ \frac 1 7 $

anonymous · Answer

142857

anonymous · Answer

Precisely. Knew you would. How did you do it?

anonymous · Answer

How'd you do ittttt??

anonymous · Answer

I knew this property. I had solved this problem before PE-52 http://projecteuler.net/problem=52