Question

i was doing this ques and i find it quiet interesting

Answer 1

Find all 7 digit nos made from 5 and 7 such that they are divisible by 5 and 7 both

Answer 2

I wonder what the first 6 digits are :P

Answer 3

what do u mean made from 5 and 7

Answer 4

yea last digit will be 5 :)

Answer 5

yeah units digit has to be 5

xxx xxx 5

I guess rest of the 6 digits must be found from divisibility rule of 7

Answer 6

i mean like this-
5757575
555557
7 digit nos having only 5 and 7

Answer 7

ah okay :)

Answer 8

oaky so lets start somewhere

Answer 9

\(\sum\limits_{i=0}^{6}a_i10^i \equiv \sum\limits_{i=0}^{6}a_i 3^i \pmod{7} \)

\(a_0 = 5\)

Answer 10

i think like umm lets take an example-
in this case we are considering only 4 digits number made from 5 and 7

so let the number be abcd so
d=5
and
abcd=1000a+100b+10c+5

this is divisible by 7
so\[\frac{ 1000a }{ 7}+\frac{ 100b }{ 7 }+\frac{ 10c }{ 7 }+\frac{ 5}{ 7}=abcd\]
we need to get rid of that 5/7 so
if we can a 30/7 from 1000a/7 +100b/7 +10c/7 then we can get rid of it cause 30/7+5/7=5 :)

so basically we need to find number of total combinations following this logic

Answer 11

the number of possiblities for the 6 digits is the same as the number of 6 digit binary numbers

Answer 12

that wuld be abcd/7 and not abcd

Answer 13

\(\sum\limits_{i=0}^{6}a_i10^i \\~\\ \equiv \sum\limits_{i=0}^{6}a_i 3^i \\~\\ \equiv a_0 + 3a_1+2a_2 -a_3 - 3a_4-2a_5+a_6 \pmod{7} \)


and we know that \(a_0 = 5\)

Answer 14

so you want to choose \(a_1, \ldots, a_6\) such that the expression, \(5 + 3a_1+2a_2 -a_3 - 3a_4-2a_5+a_6\) is divisible by \(7\)

Answer 15

I don't see an easy way to figure out all those 6 coefficients using just one eqn..

Answer 16

I was trying to figure out the second digit this way by looking at this
\[(2n+1)35 \equiv 70n+35\mod 100\]

since our number must be an odd multiple of 35 in order to have the last digit remain 5. The only values of n that will work are when:

\[n \equiv 2 \mod 10\] or \[n \equiv 6 \mod 10\]
then I did a mini fourier transform of two terms to interpolate the cases:

\[ n = 4 +(-1)^k\]

so k=0 when n=2 and k=1 when n=6 and I plugged this in:

\[ 70n+35 \equiv 70(4+(-1)^k)+35 \equiv 350 +(-1)^k 70 +35\mod 100\]
Maybe writing mod 100 is inappropriate, I'm slowly building up the pieces although this is probably a tough and confusing road to walk. lol

Answer 17

nice, but i know the answer, second digit is either 5 or 7 :P

Answer 18

Yeah of course what am I thinking lol.

Answer 19

nice, but i know the answer, second digit is either 5 or 7 :P
neither has 100% probability for sure

Answer 20

In my mind I'm thinking of looking for a way to sift out the numbers from 5555555 to 5777777 that are divisible by 35.

Answer 21

Wow that's dyslexia probably

Answer 22

In my mind I thought 7777775 to 5555555 and thought, oh that's backwards and then I flipped it haha

Answer 23

Ahh I think that might work if we consider binary number system, but we're considering decimal.. so we don't really have any easy hold on how many "decimal" numbers are divisible by 35 that use only the digits 5 and 7

Answer 24

Yeah beats me for right now, kinda just playing around. I am not to good with fermat's little theorem but I immediately thought we could do something clever with divisibility since we know:

\[35=6^2-1\]

I have this sixth sense when it comes to geometric series, I'm always looking for something that's like 1 less than a power xD

Answer 25

Can we start at 5555555 and somehow rule out how to add \(2*10^k\) to it in multiple ways?

Answer 26

It's kinda like an array of booleans or something

Answer 27

Or is this the binary number thing you were talking about earlier that I somehow missed out on :O

Answer 28

i ended up making a tree .-.

Answer 29

ok it worked :)
first solution 7555555

Answer 30

ok all solusions

7555555
7755755
5557755
5755575
7775775
5757775
5577775

Answer 31

ok so i have figured it out
u guys want hint or solution

and ikram u got most of solutions and the second last is wrong

Answer 32

ugh lol

Answer 33

ok waiting ur solution :)

Answer 34

ok :)

we know that the last digit is 5
now we need to find the 6digits
if the number is divisible by 7 then if we replace the all the 7s in number with 0 still the number has to be divisible by 7.
for example-
5577775 is divisible by 7
and 5577775=5500005+77770
we know that 77770 is divisible by 7 so 5500005 has to be divisible by 7
so each number is formed by adding some numbers from the set-
{50, 500,5000, 50000, 500000, 5000000} along with 5 (:

now lets see what remainders we get on dividing the numbers frm the set by 7

the respective remainders are-{1,3,2,6,4,5}

now its sufficient to check the remainders which add up to a number of form 2+7k
since the last digit is already 5.
putting k=0,1.....
we get
the remainders which add up to 2+7k form--->{2}, {3,6} , {4,5}, {2,3,4}, {1,3,5}, {1,2,6}, {2,3,5,6}, {1,4,5,6}, {1,2,3,4,6}
these correspond to these numbers-
7775775
7757575
5577775
7575575
5777555
7755755
5755575
5557755
7555555

:)

Answer 35

nice :) good work

Answer 36

very clean explanation

Answer 37

thanks :)

Answer 38

Brute force check:

Select[Map[FromDigits,
Select[Permutations[{5, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7}, 7],
Length[#] == 7 &]], Divisible[#, 7] && Divisible[#, 5] &]

{5557755, 5577775, 5755575, 5777555, 7555555, 7575575, 7755755, 7757575, 7775775}