1. ## Confusing Math Puzzle

Hello everybody I have a math puzzle from "the numerology of Dr. Matrix" It is an old book I think from the 60s I got this puzzle from a magazine that copied it.

Find the smallest positive integer such that, if you place a 4 in front of it, you get a number that is exactly four times as large as you get if you place a 4 at the end of the number.

I have spent hours just trying to understand what is being said. I loathe the convoluted language of some math problems. I think it is unnecessary. Anyway I am not sure if the puzzle is saying

4integer multiplied by 4= 4integer4

which I think is impossible because 4integer multiplied by 4 will give a number that starts with 1 because 4 multiplied by 4 is 16.

Or if the question is saying
4integer multiplied by 4= integer4,
if so the question is poorly written because it says in the last line, "the number" not the "positive integer"
Okay I hope I have made everything clear. I like play on numbers puzzles I dislike the annoying language sometimes used. Its like jeez English is very important.
I welcome any feedback

2. Originally Posted by sigmaxi
Find the smallest positive integer such that, if you place a 4 in front of it, you get a number that is exactly four times as large as you get if you place a 4 at the end of the number.
Suppose that the integer is 123. If you place the digit 4 in front of it, you get the number 4123. If you place the 4 at the end of it, you get 1234. But 4123 is not equal to 4 times 1234. So 123 is not the correct solution.

In fact, the smallest solution is 10256, and 410256 = 4×102564. To see how to get the solution, see the similar problem here.

3. Hello, sigmaxi!

Find the smallest positive integer such that, if you place a 4 in front of it,
you get a number that is exactly four times as large as you get
if you place a 4 at the end of the number.
Suppose the integer is $ABCDE$
We are told that: . $4ABCDE \:=\:4 \times ABCDE4$

We have: . $\begin{array}{cccccccccc}A & B & C & D & E & 4 \\ \times & & & & & 4 \\ \hline 4 & A & B & C & D & E \end{array}$

From the rightmost multiplication, we see that $E = 6$

We have: . $\begin{array}{cccccccccc}A & B & C & D & 6 & 4 \\ \times & & & & & 4 \\ \hline 4 & A & B & C & D & 6 \end{array}$

From the next multiplication, we see that $D = 5$

We have: . $\begin{array}{cccccccccc}A & B & C & 5 & 6 & 4 \\ \times & & & & & 4 \\ \hline 4 & A & B & C & 5 & 6 \end{array}$

From the next multiplicaton, we see that $C = 2$

We have: . $\begin{array}{cccccccccc}A & B & 2& 5 & 6 & 4 \\ \times & & & & & 4 \\ \hline 4 & A & B & 2 & 5 & 6 \end{array}$

From the next multiplicaton, we see that $B = 0$

We have: . $\begin{array}{cccccccccc}A & 0 & 2 & 5 & 6 & 4 \\ \times & & & & & 4 \\ \hline 4 & A & 0 & 2 & 5 & 6 \end{array}$

From the last multiplication, we see that $A = 1$

We have: . $\begin{array}{cccccccccc}1& 0 & 2 & 5 & 6 & 4 \\ \times & & & & & 4 \\ \hline 4 & 1 & 0 & 2 & 5 & 6 \end{array}$

Therefore, the integer is (as Opalg promised): . $ABCDE \:=\:10256$

4. I remember seeing this on the last issue of The Bent. Is that where you saw it, sigmaxi?

I like Soroban's approach to the solving the problem, but I solved it in what I think is a more efficient way.

You see, the main problem I found was the fact that we don't know how many digits long this integer is. If we solve the problem using the approach aforementioned, you'd have to try for A, AB, ABC, ABCD, ABCDE... this can be a little tedious.

My approach uses simple algebra.
Let x be the original number:

If x is a 1-digit number, then 40 + x would be that number with a 4 in front. Similarly, writing a 4 behind is the same as multiplying by 10 and adding 4.

If we think about what is going to happen next, we can notice two things: 1) To write a 4 behind a number, it does not matter how many digits long the original number is. This number is always going to be 10x + 4. 2) To write the 4 in front, we need to know how large the number is. Therefore, we need to try it out for different situations.

The original problem states 4integer = 4 times integer4

So if x is 1 digit long: 40 + x = 4(10x + 4)
And if x is 2 digits long: 400 + x = 4(10x + 4)
Notice the right side of the equation never changes...

The pattern emerges and the final answer could be written as

x = (1/39)[4(10^d) - 16]

where is an integer and d > 0, which represents the number of digits in x.

For d = 5 we obtain our answer. You still have to try d for 1 through 4, but it's still easier to do it this way.

It literally took me 30 seconds to derive this expression and had the answer in no time.

What if the question was asking for the next smallest number that satisfies this condition? Yeah... try doing it the other way. For that situation d = 11, so you'd have to do it many time. That number is 10,256,410,256.

See a pattern?