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Is it possible for a product of digits in a number to equal $650$?

I tried representing 650 in a product of prime numbers; from that representation, I couldn't. ($13$ ruins the story.) Is it still possible to have a product equal to 650?

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    $\begingroup$ Not in the usual base-ten decimal system, no. Factors of $10,13$ ruin that. However, it could be possible in other numerical bases, depending on how you want to frame the problem. Perhaps a fun exercise to look into. $\endgroup$ – Eevee Trainer Mar 17 '20 at 18:02
  • $\begingroup$ @EeveeTrainer Only $13$ ruins that, not $10$. $\endgroup$ – Don Thousand Mar 17 '20 at 18:20
  • $\begingroup$ Yeah, you're right, my bad. $\endgroup$ – Eevee Trainer Mar 17 '20 at 18:22
  • $\begingroup$ Euclid's lemma says that $13$ must divide some digit. It doesn't matter if that digit is prime or not. And if $13$ divides that number that number must be at least $13$ (well, technically it could be negative but lets stick with natural numbers). And if it's at least $13$ it's larger than $9$ and can't be a base $10$ digit... Could do it in any base $14$ or larger though. $\endgroup$ – fleablood Mar 18 '20 at 0:03
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650 has prime factorization $5\times5\times2\times13$. Because 13 is a 2 digit number and you're only allowing single digit numbers in your factorization, it is impossible to write 650 as a product of single digit numbers (in base 10).

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I should point out the Euclid's Lemma points out that as $13$ if you have a product of terms that equals $650$ so $13|650$ then $13$ must divide one of the terms. You can't "regroup the prime factors" around and make the prime factor $13$ go away.

So if one of the terms is divisible by $13$ it must be at least thirteen (assuming we are only considering positive integers) and ....

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This problem depends on the representation of the number — as opposed to the majority of math problems. As you did not mention the number system, let's use base 12.

To that base, 267B has product of digits $$650_{12}=2\cdot6\cdot7\cdot11_{10}=924_{10}$$

The base $B$ cannot be a prime and must be $B\geqslant7$ at least. The representation is possible if $6B+5$ is ($B-1$)-smooth.

The smallest base that satisfies these constraints is 12, the next is 19. In particular, there is no solution for $B=10$.

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