3 digit no.

Is there any 3 digit no. with distinct digits which when reversed divides the original no. perfectly? Can there be such a number for any n digits?

#NumberTheory #MathProblem #Math

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

According to my calculations there are no such 3-digit numbers.

For this problem we need to find $a$ , $b$ and $c$ , so that $n*(100a+10b+c)=100*c+10b+a$ .

To start off we look at the last and first digits:

$n*c \equiv a$ (mod $10$ ) and $n*a \lt 10$ . We make a multiplication table and cross out all impossible combinations.

We are left with;

$n=1$

Here $a=c$ , so we don't have distinct digits.

$n=2$

Here $a=2, c=6$ or $a=4, c=7$ .

$n=3$

Here $a=2, c=4$ or $a=1, c=7$

$n=4$

Here $a=2, c=3$ or $a=2, c=8$

$n=7$

Here $a=1,c=3$

$n=9$

Here $a=1, c=9$ .

Substituting these 8 possibilities in the equation from the second line gives us 8 equations in $b$ . None of these equations has a solution where $0 \le b \le 9$ .

Ton de Moree - 7 years, 8 months ago

The only numbers that work are 510, 540, 810.

Ryan Soedjak - 7 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$