number problem

your one post was enough to make it "tooo obvious" [3]

you dont really need conguencies to solve this one though it can be handy if you know

HEy, where did congruence come from? We aren't talking about triangles.

hey philip u r very right

I used ≡, not =. Look up congruences. Loosely speaking, the remainder is 3 upon division by 7. But here remainder need not lie between 0 and 6

10 = 3 l7l
= 3x7
= 21?
[7]

55555..555(n times) = 5(1+10+10²+...+10^n-1)

Now 10≡3 mod 7,

So (1+10+10²+...+10^n-1) ≡ 1+3+3²+...+3^n-1 mod 7

checking for n=1,2,3 we get for n = 6, 1+3+3²2²+..+3⁵ =364 ≡ 0 mod 7

Hence the number 555555 is the first such number.

Its obvious that the next such number can only be 555...555 (12 5s)

How's this tooooo obvious? [7]
pls tell me someone [7] [2]

[12][12][12][12][12]

[7][7][7][7][7]

[2][2][2][2][2]

OBVIOUS? [11]

thats tooo obvious

thats tooo obvious

How did you know these were the numbers?

since n1 and n2 are not necessarily the first such nos the general ans would be

79365 x 10^6k where k={1,2,3,4......}

hope i am not blabbering [4]

n1=79365
n2=79365079365

|n1-n2| = 79365000000
[1]