Results 1 to 4 of 4

Math Help - Confusing Math Puzzle

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    1

    Smile 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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by sigmaxi View Post
    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 = 4102564. To see how to get the solution, see the similar problem here.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,683
    Thanks
    615
    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

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2008
    Posts
    1
    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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Confusing Vector Math Problem
    Posted in the Geometry Forum
    Replies: 0
    Last Post: June 29th 2010, 10:46 PM
  2. Math Induction Confusing
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 29th 2010, 04:44 AM
  3. five 3s math puzzle please help
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: November 18th 2007, 05:04 AM
  4. Help me with this Math puzzle...
    Posted in the Math Challenge Problems Forum
    Replies: 6
    Last Post: October 11th 2007, 07:24 PM
  5. confusing math on conics
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 6th 2006, 07:49 AM

Search Tags


/mathhelpforum @mathhelpforum