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I was exploring digit patterns and wondered whether there are positive integers whose sum of digits is exactly equal to their product of digits.

For example, for 123:

  • Digit sum = 1 + 2 + 3 = 6
  • Digit product = 1 × 2 × 3 = 6

So 123 works.

My question:

Are there infinitely many such numbers? And is there a general pattern or characterization for them?

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Jake T is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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    $\begingroup$ $+1 \ $ Welcome to PSE (Puzzling Stack Exchange)! By the way, 22 also works. $\endgroup$ Commented 9 hours ago
  • $\begingroup$ This question has received two low-quality answers by new users. I have thus protected it to avoid more new users posting answers. If you (OP) wish, I can reverse this. $\endgroup$ Commented 1 hour ago

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Yes.

Take any number with a digit product greater than its digit sum. Adding ones will increase the sum by 1 by not affect the product. Adding enough ones will make them equal.

252 -> 11111111111252

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  • $\begingroup$ That is not a general pattern though. Excluding reveal spoilernumbers already containing a 1, you also have reveal spoilerall other digits, then 22… and I guess that must be it? reveal spoilerAny other number not containing a 1 or 0 must have its digit product greater than its digit sum, and can thus be extended as described. $\endgroup$ Commented 3 hours ago
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Yes, such numbers are not uncommon and typically fall into two categories: Single-digit numbers: For any single-digit number (1-9), the sum of its digits is exactly equal to the product of its digits. For example, take the number 5: its digit sum is 5, and its digit product is also 5.

Multi-digit numbers: These require the combination of digits to balance the sum and product. Examples include 22 (sum = 2+2=4, product = 2×2=4) and 123 (sum = 1+2+3=6, product = 1×2×3=6). Note that 1122 is a counterexample (sum = 1+1+2+2=6, product = 1×1×2×2=4), while 132 is a valid one (both its digit sum and product equal 6).

The core rule is: Avoid excessive "1"s in the digit combination (since 1 increases the sum but not the product), and ensure the digit combination balances the calculation results of the sum and product.

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yousanusb is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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    $\begingroup$ Welcome to Puzzling Stack Exchange. If this answer was AI-generated, it must be attributed. In addition, it does not appear to actually answer the question, which was whether there are infinite numbers with this property. As we are a Q&A site, answers which do not directly address the question will be deleted. Please read How to Answer for more information. $\endgroup$ Commented 1 hour ago
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Well any number multiplied by 1 is equal to itself so you have infinite. For example:

  • 4*4=16
  • 4+4=8

but if you do

  • 4*1*1*1*1*1*1*1*1=16
  • 4+4+1+1+1+1+1+1+1+1=16

so there's infinite as long as you first check the product then add as many 1s as you want

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Aaron Mendoza is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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  • $\begingroup$ This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review $\endgroup$ Commented 2 hours ago
  • $\begingroup$ Despite being too low-quality, I find this answer quite reasonable. This answer is essentially same as ralphmerridew’s, describing that reveal spoileradding many 1s as you want can make all of the numbers satisfy the condition. $\endgroup$ Commented 1 hour ago
  • $\begingroup$ But for the OP of this answer, spoiler tags are (strongly) recommended in answering the puzzle. You can use it by writing “>! “ before the sentence. $\endgroup$ Commented 1 hour ago

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