0
$\begingroup$

if I take 5688+6984=12672 then sum the result 1+2+6+7+2=18 then sum that result 1+8=9. vs. this. same digits from above. 5+6+8+8+6+9+8+4=54 then sum that result 5+4=9. using this method where the original two are always the same length, four digits in this example, and always re-summing down to one digit, the result is always the same. Is this significant at all??? than you.

$\endgroup$
  • $\begingroup$ This is wellknown, as the repeated digit-sum is $9$ if the given positive integer is a multiple of $9$ (and it is the remainder from dividing by $9$ otherwise). $\endgroup$ – Hagen von Eitzen Oct 30 '14 at 15:58
  • $\begingroup$ Thank you for the feedback, I looked that up just now. Very helpful. $\endgroup$ – shawn Oct 30 '14 at 16:10
2
$\begingroup$

Hint : A number is divisible by $9$ if and only if the sum of its digits is divisible by $9$.

$\endgroup$
  • 1
    $\begingroup$ I see, cool thanks for the feedback. $\endgroup$ – shawn Oct 30 '14 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.