A warped sense of humour

Number Theory Level 5

We will call a positive integer $N$ warped if there exists a multiple of $N$ whose digits are all the same. Find the sum of the first 10 non-warped numbers

The answer is 331.

2 solutions

Xuming Liang
Jun 1, 2014

Note that all numbers with the same digits can be expressed as $a(\frac {10^n-1}{9})$ , where $1\le a\le 9$ . Hence $N$ is a wraped number $\iff$ there exists $a, n$ such that $N\mid a(\frac {10^n-1}{9})$ .

Lemma: If $gcf(N,10)=1$ , then there exists $n$ such that $N\mid \frac {10^n-1}{9}$

Proof: In mod notation: $\frac {10^n-1}{9}\equiv 0\pmod N$

$\Rightarrow 10^n\equiv 1\pmod {9N}$ , since $gcf(9N,10)=1$ , hence by Euler's theorem $n=\phi (N)$ satisfy our condition.

Let $k$ denote a number such that $gcf(k,10)=1$ . By this lemma, numbers that can be expressed as $2^{e_1}k,5^{e_2}k$ where $0\le e_1\le 3, 0\le e_2\le 1$ (don't forget $a$ ) are wraped numbers. On the other hand, numbers that are divisible by any of $10,16,25$ are not wraped numbers. Listing these numbers in increasing order gives:

$10+16+20+25+30+32+40+48+50+60=\boxed {331}$ .

Really nice proof with Euler's theorem; I feel that the crux of the proof is the change of perspective from trying to construct a warped or a non-warped number to trying to construct the multiple. Also, not that this could be solved without Euler's formula - for those who don't know it - by just proving the following specific lemma:

Lemma: Let $r$ be an integer with $gcd(r,10)=1$ . Then there exists an $a$ such that $r | 10^{a}-1$

Proof: since $gcd(10,r)=1$ we can find two integers $x,y$ , $x>y$ , such that

$10^{x}=10^{y}$ (mod $r$ ) $\Rightarrow$ $10^{y}(10^{x-y}-1)=0$ (mod $r$ )

Since we know $gcd(10,r)=1$ we must have $r|10^{x-y}-1$

And then you just progress as you did :)

Daniel Remo - 7 years ago

Nice. But the numbers are wARped, not wRAped. ;)

Jubayer Nirjhor - 6 years, 11 months ago

but 10 * 101 = 1010 where the digits of original number are only used and according to me there can't exist a warped number if you are just talking about the original digits of number being used to form its multiple

akash deep - 6 years, 10 months ago

Did the same but $0$ is a multiple of every number.

Kartik Sharma - 6 years, 2 months ago

Vinay Sipani
Jun 1, 2014

These are some of the points that I have observed.

All 1-digit numbers are warped numbers.
All positive integers ending with 0 are non-warped numbers
All prime numbers (except 2) and their multiples upto 9 are warped numbers.
All numbers of the form $5^{n-2}$ and $2^n$ (where $n>3$ ) and their multiples are non-warped numbers.

Based on these,the first 10 non-warped numbers are: $10,16,20,25,30,32,40,48,50,60$

Therefore,sum= 331

Very nice: can you then prove the following culmination of your observations? All non-warped numbers are of the form $10k, 16k$ or $25k$ for an integer $k$

Daniel Remo - 7 years ago

A warped sense of humour

The answer is 331.

2 solutions

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