0
$\begingroup$

How many different ways three two-digit numbers can be chosen so that their product ends in four or more zeros.

I could'n do more: three two-digit numbers - $XX$, $YY$, $ZZ$

  1. $XX$, $YY$, $ZZ$ are ending in zero -- $X0.Y0.Z0$

$X=1-9$, $Y=2,4,6,8$, $Z=5$

$9.4.1=36$

There are much more ways and I can't get to the solution.

Can someone give me an idea?

$\endgroup$
  • 1
    $\begingroup$ Hint: A (positive) number ends with four or more zeroes iff in the number's prime factorization there are at least four factors of $2$ as well as there are at least four factors of $5$. $\endgroup$ – JMoravitz Aug 18 '18 at 16:48
  • 1
    $\begingroup$ Extended hint: Since we are interested only in using two-digit numbers to accomplish this, consider first how the $5$'s are distributed. Note that $5^3=125>99$ so each term may have at most two factors of five each. As you have four $5$'s to distribute among the three locations, at least one must have two factors of five (and so is either $25,50,$ or $75$). Continue... $\endgroup$ – JMoravitz Aug 18 '18 at 16:54

Your Answer

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

Browse other questions tagged or ask your own question.