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## Easy as 123451

Posted by Zorglub on December 19, 2014 – 3:08 pm

A function f takes a positive integer and returns another one by moving the leftmost digit to the right.

For example f(12345) = 23451

What is the smallest strictly positive integer n such that f(n) = 1.5 n

This post is under “Tom” and has 5 respond so far.

### 5 Responds so far- Add one»

1. 1. Wizard of Oz Said：

n = 352,941,176,470,588

Let n = (10^r)*d + A, then f(n) = 10A + d = 1.5[(10^r)*d + A)

So 17A = d[3*(10^r) – 2) where d < 10

So 17 must divide into one of 28, 298, 2998, . . .

After a minute or two on the calculator we eventually get A/d = 176,470,588,235,294

2. 2. Chris Said：

Ans: 1,176,470,588,235,294

Let N be an n digit string and D be a single digit. We want to find ND = 1.5*DN
So 10*N + D = 1.5 * (D*10ⁿ + N)
20*N + 2*D = 3*D*10ⁿ + 3*N
17*N = D(3*10ⁿ – 2)

As 1 ≤ D ≤ 9, we must have D ≠ 0 (mod 17), so 3*10ⁿ = 2 (mod 17)
Multiplying by 9 => 27*10ⁿ = 18 (mod 17) => 10^(n+1) = 1 (mod 17)
Fermat’s little theorem says n+1 = 16 = 2^4 satisfies that equation. A quick check shows that n+1 = 8 fails to satisfy it, so n+1 = 16 is the smallest power that satisfies the equation. So n = 15.

Substituting that into 17*N = D(3*10ⁿ – 2) =>
17*N = 2,999,999,999,999,998*D =>
N = 176,470,588,235,294*D
We also need ND = 1.5*DN => (2/3)*ND = DN
i.e. (2/3)*1,764,705,882,352,94D = D,176,470,588,235,294
But (2/3)*176,470,588,235,294 = 117,647,058,823,529.333….
so D = 1. Check: 1,764,705,882,352,941 = 1.5*1,176,470,588,235,294

So the required number is 1,176,470,588,235,294 and either Wiz or I has had too much to drink.

3. 3. Wizard of Oz Said：

Hi Chris, I plead guilty to the disorderly part of being drunk and disorderly. I failed to notice in checking my answer that 1.5*n was out by just 1 in the last digit for my version of n. So, as usual, you’ve confounded me again and shown up my carelessness.

Must do better is my new years resolution. Have a happy new year. You too Zorglub – nice problem.

4. 4. Zorglub Said：

Good job. And nice touch with Fermat’s little Theorem.

5. 5. Chris Said：

Hi Zorglub. This problem is very similar to the very first one that I did on ToM (it used a factor of 2 rather than 1.5). I loved it because of the ridiculously large number that resulted. It caused me to learn about modulo arithmetic. I was a bit naughty in not stating that because 17 is prime, that gcd(D,17) = 1 was assumed. Thanks for posting it.

Hi Wiz. I’m relieved that it wasn’t me who’d goofed.

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