Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My child got a question in school (grade) that is:

Find biggest and smallest 3 digits number, which has sum of it's digits equal to product of those digits.

Help please since I cannot explain my child this question.

Example would be:

$1+2+3=1\cdot 2\cdot 3$

$3+2+1=3\cdot 2\cdot 1$

Numbers 123 and 321

But this is just example of this, how to solve it as a problem.

share|improve this question
add comment

3 Answers

up vote 1 down vote accepted

So, we need find digits $a,b,c$ such that $a+b+c=abc$

Clearly, $abc \ne 0$, if $a=0,b+c=0\implies b=c=0$

So, $$\frac{a+b}{ab-1}=c$$ which is integer.

(1)If $ a=3m+1,b=3n+1$ or $a=3m-1,b=3n-1$

$3\mid (ab-1)$, but $3 ∤ (a+b)\implies (ab-1) ∤(a+b)$

(2)If $ a=3m\le 9,b=3n\le 9\implies 1\le m,n\le 3$

$$ \frac{a+b}{ab-1}=\frac{3(m+n)}{9mn-1}$$

$\implies (9mn-1)\mid (m+n)$ as $(9mn-1,3)=1$

But $(m+n)\le 3+3=6$

$\implies (9mn-1)\le 6$ which is clearly impossible as $m,n \ge 1$.

(3)If $ a=3m+1\le 9 ,b=3n-1\le 9\implies 0\le m\le 2, 1\le n\le 3 $,

$$ \frac{a+b}{ab-1}=\frac{3(m+n)}{(3m+1)(3n-1)-1}$$

$\implies (3m+1)(3n-1)-1\mid (m+n)$ as $((3m+1)(3n-1)-1,3)=(3(3mn-m+n)-2,3)=1$

$ (3m+1)(3n-1)-1\le 5\implies (3m+1)(3n-1) \le 6\implies m\le1$ and $n\le 2$

If $m=1\implies 3m+1=4\implies 3n-1\le 1\implies n<1$, but $1\le n\le 3 $

If $m=0$, $$\frac{3(m+n)}{(3m+1)(3n-1)-1} \ becomes \frac{3n}{3n-2}$$

$\implies (3n-2)\mid 3n\implies (3n-2)\mid n$ as $(3n-2,3)=1$

So, $3n-2\le n\implies n\le 1$ ,but $1\le n\le 3$ so, $n=1$

So, $a=1,b=2\implies c=\frac{a+b}{ab-1}=3$

As $a+b+c=abc$ is symmetric $a=3,b=1$ and $a=3,b=2$ will also be solutions corresponding to cases (4) $a=3m,b=3n+1$ and (5) $a=3m,b=3n-1$ respectively.

So, the only solution is $1,2,3$.

Clearly, $321$ is the largest and $123$ is the smallest.

share|improve this answer
add comment

Clearly $0$ cannot be one of the digits: if it were, the product would be $0$, and the sum would not. We can’t have two $1$’s: if the remaining digit is $c$, the product is $c$, but the sum is $c+2$. This means that the smallest two digits must be at least $1$ and $2$.

It’s easy to check that the digits cannot all be the same: the only solutions of the equation $a^3=3a$ are $a=0$ and $=\pm\sqrt3$, none of which is possible in this problem. Thus, if the digits are $a,b$, and $c$, with $a\le b\le c$, we must have $a<c$.

We already know that $b\ge 2$; suppose that $b\ge 3$; then $abc\ge bc\ge 3c>a+b+c$, since we know that $a<c$. Thus, there is no solution with $b\ge 3$. There is also no solution with $a=b=2$: that would require $4c=4+c$, $3c=4$ and $c=4/3$, which is impossible. Thus, every solution must have $a=1$ and $b=2$. This means that $2c=3+c$, whose only solution is $c=3$.

Thus, the three digits must be $1,2$, and $3$. The smallest number that can be formed from them is $123$, and the largest is $321$.

share|improve this answer
add comment

Notice that $1+2+3=1\cdot 2\cdot 3$ $\ldots$

share|improve this answer
    
yes, but this it is not solution. That is the smallest 3 digit number what is the biggest? –  Dolphin Sep 26 '12 at 7:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.