The professor who's teaching us the ropes in weekly sessions is a great guy.

Here's one problem I did, from the promo poster:

Find the smallest natural number n which has the following properties:

(i) its decimal representation has a 6 as its last digit, and

(ii) if the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as large as the original number n.

If n starts with 6, and n = 4k, then k must start with a 1. So:

n = 61 ..,

k = 1 ... 6

Now the cool part: some recursive long multiplication.

Multiplying 6*4 gives you 24, so 4 is the last digit of n. This means that 4 is also the second-to last digit of k. So 4*4 +2 = 18. 8 is the second-to-last digit of n, and third-to-last digit of k. We know that 1 is the second digit of n, so eventually, we have to get to a 1.

2

1....6

x4

61...4

1

1...46

x4

61..84

3

1..846

x4

61.384

25

153846

x4

615384

This is true. So I think that 153846 is the smallest n that ends in 6 where shifting the digits right gives you 4n.

I love this stuff.

## 1 comment:

Yeah, that's the same answer I got...here's how I did it:

let 10x+6 be the number. Then this relation holds true:

4(10x+6) = 6*10^m + x

for some m.

Reducing, we get:

13x = 2*10^m - 8

so therefore

2*10^m = 8 (mod 13)

10^m = 4 (mod 13)

The lowest value of m for which that works is m=5, so now we can solve for x:

13x = 2*10^5 - 8

x = 15384

Therefore the number is 10x+6 = 153846

This is definitely the smallest value as well, so props.

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