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I was randomly thinking and I stumbled upon this question in my mind. I was thinking of numbers, and I observed that any number is greater than or equal to the product of digits in many cases. I was thinking, is this true for all natural numbers or for only a few cases? Please help.

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looks like it is true. $9^n\lt 9\times 10^n$ – Sabyasachi Mar 14 '14 at 19:29
Comments on my answer quite right - withdrawn. – Tom Collinge Mar 14 '14 at 19:38
@TomCollinge That withdrawl is gonna earn you a "Peer Pressure" badge. – Hawk Mar 14 '14 at 19:39
I'm new at this - didn't want to leave my missunderstanding lying around. – Tom Collinge Mar 14 '14 at 20:01
@ Ray Toal k. I got it. :) – mridul Mar 15 '14 at 17:48
up vote 45 down vote accepted

Let $k$ be the number of digits, and let $a$ be the first digit (assume $a>0$). Then the product of the digits is bounded above by $a\times 9^{k-1}$. The value of the number itself is bounded below by the total value of that first digit: $a\times 10^{k-1}$.

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nice, +1, and it immediately proves the analogous result for any base. – mjqxxxx Mar 14 '14 at 20:56
nice answer :) . But how to find product of digit in the case of number less than 10?. – mridul Mar 15 '14 at 6:26
@mridul That is not a special case. For a number $<10$, $k=1$. The product is bound above by $a*9^0=a$ and the value is bound below by $a*10^0=a$, and the relation still holds: $a\geq{a}$ – Jason C Mar 15 '14 at 9:55
I get your logic. But ,to find product we want at-least 2 number.If k=1 , it is single digit number. In the case of 12 product of digit is 1x2 =2. In the case of number 6 product of digit is 6x what? How to find product of digits in single digit number? – mridul Mar 15 '14 at 10:11
I think u compare number system with base 10 and number system with base 9. – mridul Mar 15 '14 at 10:33

$\begin{eqnarray} {\bf Hint} & & a + 10\ b + 10^2 c + 10^3 d\\ &=& a+\color{#c00}{10}(b+\color{#c00}{10}(c+\color{#c00}{10}d))\\ &\ \ge\ & a+\ \color{#c00}a\ (b +\ \color{#c00}b\ (c+\ \color{#c00}c\ d))\quad\ \ {\rm by}\ \ \ \color{#c00}{10 > a,b,c}\\ & =& a+\ a\,\ b + a\ b\ c+ abcd\ \ge\ abcd \end{eqnarray}$

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(Note: This is not a formal, mathematical proof, just a way of thinking about it. IANAM)

Let's look at the number 12: a good, honest number who always pays its taxes on time and helps old ladies cross the street. 12 won't tell you this, because his mother taught him to be polite, but he's made up of two components: 12 = 1*10 + 2

Now let's look at 12's good friend 26. He is also made up of two components: 26 = 2*10 + 6.

As a last example, we'll have a look at their long lost cousin 794, which exhibits the exact same behaviour: 794 = 7*100 + 9*10 + 4.

We can see that the product of the numbers they're made up of is smaller than themselves:

  • 12 > 1*2
  • 26 > 2*6
  • 794 > 7*9*4

The reason for this is revealed when you ignore the advice of 12's mother and look at what they're made of. As we saw, one of 12's components is 1*10. Since there's no such digit as 10, all of the 10+x family (where x is a digit) will be larger than the product of their digits, because the number itself exhibits a multiplication larger than any single digit.

Let's expand this to be more formal. A number in our decimal system looks like this:

$a*10^n + b*10^{(n-1)} + ... + z*10^0$

We can see our digits here, they're a, b, ... z. Looking at their product:

$a * b * ... * z$

And we can see that in the former, they're being multiplied by things far larger than they.

In conclusion, this makes sense because our decimal system works by multiplying digits by powers of 10. No digit can be bigger than that multiplication (and the number is a sum of these multiplications), so it stands to reason that a number is greater or equal to the product of its digits.

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small currection. A number in our decimal system looks like this .......+ z*10^0. – mridul Mar 15 '14 at 13:51
@mirdul Thanks! Corrected. – Zirak Mar 15 '14 at 13:52
which numbers satisfy the condition a number is equal to the product of its digits? – mridul Mar 15 '14 at 13:55
Looking at the logic above, we see that if our number is larger than 10, it contains multipliers greater than a possible digit. So for a number to be equal to the product of its digits, it must not have this behaviour. Which numbers have only digit-possible multiplications? Single digit numbers. – Zirak Mar 15 '14 at 14:09
@mridul usually we define the empty X as something that is an identity on X. In the case of field of real numbers we usually define the empty product as the 1 or the field or just one. This is for consistency and brevity, just like we define the empty union as the empty set and the empty intersection as everything. – Benjamin Gruenbaum Mar 15 '14 at 20:15

Let k be the number of digits, and ak-1 ak-2 .... a0 are the digits of the number.

for example, Let the number is 265 then k=3 , a2 = 2, a1=6, a0 = 5.

Then product of digit of the number is          ak-1 * ak-2 * ... * a0

The actual value of number is                        ak-1 * 10k-1 + ak-2 * 10k-2 +....+a0 * 100

Compare product of digit of the number and 1st part of actual value of the number

         ak-1 * ak-2 * ... * a0

         ak-1 * 10k-1

In product of digit of the number after first digit (ak-1) there are product of k-1 digit such that 0<=digit<10.

In actual value of number after first digit (ak-1) there are product of k-1 10.

10k-1 always grater than product of k-1 number in {0-9}.


If k>1 the number is always grater than product of digit of the number.

And if k=1 we can't find the product of digit of the number because product is binary operator.

Any natural number is greater than or equal to its product of digits.

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why down vote? please add comment . I don't understand the error with my answer.. – mridul Mar 15 '14 at 13:30
I didn't downvote, but I think the most likely reason for the downvotes is that this answer is essentially just repeating Uniwisdom's earlier answer. – Andreas Blass Mar 15 '14 at 13:54
No. in Unwisdom's answer he said that a number is greater or equal to the product of its digits. But in I can't find any number satisfy the condition a number is equal to the product of its digits. And I conclude that Any natural number(number>=10) is greater than its product of digits. – mridul Mar 15 '14 at 14:00
In the first place, the question was about "greater or equal". In the second place, "equal" actually does happen, namely in the case of one-digit numbers, which you artificially excluded by declaring that the product of a single factor is undefined. In fact, the product of a single factor is ordinarily defined as that factor. – Andreas Blass Mar 15 '14 at 14:08
Soory. I think product of a single factor is "undefined". thanks. – mridul Mar 15 '14 at 14:18

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