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

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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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: $a\times 10^{k-1}$.

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

(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.

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}.

Therefore,

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.