## Saturday, February 8, 2014

### 22: The Five Digit Number

I am thinking of a 5-digit number. Well, I'm not thinking of it right now, but just go with it, okay?

If you put a "1" at the end of the number, you get a 6-digit number that is three times as big as the 6-digit number you'd get if you put a "1" at the beginning of the number.

In other words, the number _ _ _ _ _ 1 is three times as big as 1 _ _ _ _ _

What is the 5-digit number?

1. The five-digit number is 42,857.

Let x represent the five-digit number in question. By sticking a 1 at the end of this number to get a six-digit number, you are effectively multiplying x by 10, then adding 1 (thus getting 10 x + 1).

If you put 1 at the beginning of the number to get a different six-digit number, you are effectively adding 100,000 to x. Three times that is 300,000 + 3x.

So we can say 10x + 1 = 300,000 + 3x. Solving for x, we get the intermediate equation 7x = 299,999, and dividing by 7 on both sides yields x = 42,857.

You can then confirm this solution by noting that 428,571 is indeed three times 142,857.

2. I'm curious to see if anybody has an alternate method of solving this. I have to confess, my solution came about when I was trying a much more complex strategy. I let a, b, c, d, and e represent the five digits, in order from left to right, of the five-digit number. Thus, I said 10^5 a + 10^4 b + 10^3 c + 10^2 d + 10e + 1 = 3(10^5 + 10^4 a + 10^3 b + 10^2 c + 10d + e). I was thinking, "Ugh, I need four more equations, maybe this isn't a good idea" then I noticed that the abcde sum on the left was the same as the one on the right, except each term was multiplied by 10. Then I ran with that idea, and got what you see above.

3. Treat it like a multiplication problem, using the better algorithm, not that lattice crap. In the beginning: 3(1abcde) = abcde1.
3e = _1. The only value that gives a product with a 1 in the units place is 3(7) = 21, so e=7, and carry the 2.

3(d)+2 = _7 ... d=5 and carry the 1.
3(c)+1 = _5 ... c=8 and carry the 2
3(b)+2 = _8 ... b=2, no carry.
3(a)+0 = _2 ... a=4

4. What about the "looks like the decimal expansion of 1/7" aspect to this problem? It looks like other problems could be generated from this one (and perhaps from other interesting repeating decimal expansions for fractions):

1. There is a 4 digit number which, when you put the number "14" at the end, becomes a 6 digit number that is twice as large as if you put the 14 at the beginning of the number. What is the 4 digit number?
2. There is a 2 digit number which, when you put the number "1428" at the end, becomes a 6 digit number that is 4 times as large as if you put the 1428 at the beginning of the number. What is the 2 digit number?
3. There is a 3 digit number which, when you put the number "142" at the end, becomes a 6 digit number that is 6 times as large as if you put 142 at the beginning of the number. What is the number?

I suppose this would work with other fractions:

There is a 14 digit number which, when you put the number "94" at the beginning, becomes 8 times larger than if you had put "94" at the end of the number. What is the 14 digit number?