@abakcus Thanks so much, that came right in time!

@abakcus Good to know, as soon as someone asks me to multiply that number that’l come in handy. :)

@abakcus And if you multiply 142857 x 3, just move the first digit to the back! 😁

@abakcus How did you stumble across this?

@abakcus bookmarking for future ref, thx

@abakcus which immediately implies that it is divisible by 99

@abakcus where were you yesterday?!

@abakcus Also, you can multiply by 9 by simply moving the last two digits (the 93) to the front, to get 93103448... . You can multiply by 4 by moving the last 7 digits to the front, to get 413793103448... . Any one digit multiple of this number comes from such a rearrangement.

@abakcus 'shifting' per se

@abakcus Thanks

@abakcus how convenient!

@abakcus @mathladyhazel I don’t need, thanks.

@abakcus Definitely. The digit before the last digit is "zero", consequently, whatever the carryout it is not going to affect 3x1

@abakcus Is is cyclic ? Can we repeat this operation with the result we just got ?

@abakcus 105,263,157,894,736,842 times 2 has the same effect. 105,263,157,894,736,842 * 2 = 210,526,315,789,473,684

@abakcus Why on Earth the one would want to multiply 10344827586206896551724137 with 3?

@abakcus +62

@abakcus 3n=3·10^k+(n-3)/10⇒n=(30·10^k-3)/29 30·10^k-3≡0(mod 29)⇔10^k-3≡0(mod 29)⇔10^k≡3 (mod 29) As 29 is prime, 10^28≡1(mod 29) and 10^27≡3(mod 29) (∵ 3·10≡1(mod 29)) ∴ k=28m+27 n=(30·10^(28m+27))-3)/29 m=0⇒n=1034482758620689655172413793 For other m, n is m copies of that

@abakcus Why @sudarshaniisc sir what is the reason?

@abakcus It works for any large number