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Four fun

Posted by Chris on June 2, 2010 – 7:39 pm

What is the smallest integer that ends with a 4 and that when that 4 is moved from the end to the beginning, makes the number 4 times as large.

i.e. I want something like 412 = 4*124 (which obviously doesn’t work).

I bet you think I’ve lost it!


This post is under “Mathemagic” and has 17 respond so far.
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17 Responds so far- Add one»

  1. 1. Knightmare Said:

    24*4=42
    as 42 is always the answer

    method-brute force

  2. 2. Mudkip Said:

    The answer is the number in question itself.

    16 = 4 * 4

    Four is the smallest integer ending in four that is four times larger when multiplied by four. What do you think Chris?

  3. 3. Mudkip Said:

    I think i worded that wrong. Four is the smallest integer ending in four (since itself is only one digit), and when the ending number (which would be the four) is moved to the beginning, the number would become four times as large.

  4. 4. Nathan Said:

    102564 is the least such number

    410256 = 4 * 102564

  5. 5. Mohamed Said:

    4 * 10^n + x = 4(10 x + 4) ——-(1)
    4 * 10^n = 40 x + 16
    39 x = 4 * 10^n – 16
    3 * 13 x = 4 * 10^n – 16

    0 = 4 * 10^n – 16 (mod 13)
    4 * 10^n = 16 (mod 13) ——-(2)

    To use Fermat’s little theorem thereafter, we need something on the form of
    10^(n+i) = 1 (mod 13) ——-(*1)
    Applying Fermat’s little theorem -> n = 12 – i ——(*2)

    And since we seek the smallest integer, we shall minimize n. But from (*2), to minimize n we need to first maximize i.

    To reach (*1), multiply both sides of (2) by 22
    10^(n+1) = 1 (mod 13) ——-(a1)

    To find the values of i that satisfies (*1), we shall proceed on by successively multiplying both sides by 10 mod (13), until we reach a RHS of 1.

    10(n+2) = 10 (mod 13)
    10(n+3) = 22 (mod 13)
    10(n+4) = 25 (mod 13)
    10(n+5) = 16 (mod 13)
    10(n+6) = 4 (mod 13)
    10(n+7) = 1 (mod 13) ——(a2)

    Apparently, the values spans a closed set {10,22,25,16,4,1}. So, the next value of i that satisfies (*1) will be 13. 10(n+13) = 1 (mod 13) (call it a3).. and so on.

    Substituting into (*2),
    a1: n = 12 – 1 -> n = 11
    a2: n = 12 – 7 -> n = 5
    a3: n = 12 – 13 -> n = -1

    Hence, our answer is a2, i.e. n = 5.

    Substituting into (1), and solving for x
    x = 10256

    So, our two numbers are:
    410256
    102564

  6. 6. Chris Said:

    102564 it is. I posted it because Fermat’s little theorem wasn’t applicable.

    I did it as follows. Having got to 39x = 4*10^n – 16,
    we obtain, 4*10^n = 16 (mod 39)
    multiply by 10 => 10^n = 4 (mod 39)
    multiply by 10 => 10^(n+1) = 1 (mod 39)
    I then used methodical trial and error, n = 0,1,2,3,4,…
    until I found n = 5 satisfied the equation.

    I decided to stick with mod 39 as it seemed to retain more information. My instinct told me that working mod 3 or mod 13 may give an incorrect result. Right now there’s loads of things I’d like to try out. I can’t as I have to have moved out of my place by tomorrow afternoon.

    I, in part, posted the problem because I was intrigued that starting from
    4*10^n = 16 (mod 39). Dividing by 4 correctly => 10^n = 4 (mod 39)
    I’m dying to know if that always works if the division is exact.

  7. 7. Karl Sharman Said:

    Well, after the “3″ problem, I felt confident diving into this one with the solution happily tucked under my belt.

    I now can confirm after reading the posts that Chris is correct, it doesn’t calculate for 4. I could have worked it out with brute force maths…..

    I shall now be hunting Chris down for setting me up like that ;-)

  8. 8. M.shosha Said:

    the number is 104 !!

  9. 9. Chris Said:

    104 doesn’t work. 104*4 = 416 not 410.

  10. 10. Dino1111 Said:

    The answer is 102564. -by Dino1111

  11. 11. ulas Said:

    the answer is 102564.

    4abcde = 4 x abcde4

    400.000 + 10.000a + 1.000b + 100c + 10d + e = 400.000 + 40.000b + 4.000c + 400d + 40e + (4×4)

    399.984 = 390.000a + 39.000b + 3.900c + 390d + 39e

    39 x (10256) = 39 x (10.000a + 1.000b + 100c + 10d + e)
    abcde

    then the answer becomes like this : 102564

    the smallest number must be bigger than 100.000 because numbers which are smaller than 100.000, don’t fit the equilibrium of above.

  12. 12. Nimble1 Said:

    This is really a simple excercise requiring little math.

    Start with the number 4. The second right digit is 4 x 4 = (1)6 (carry the 1). Digit 3 is (6 x 4) + 1 = (2)5 (carry the 2). Digit 4 is (5 x 4) + 2 = (2)2 (carry the 2). Digit 5 is 2 x 4) + 2 = (1)0 (carry the 1). Digit 6 is 0(0 x 4) + 1 = 1.
    Since 1 x 4 = 4, you have reacheded your answer: 102564

  13. 13. Ajay Shahi Said:

    Answer is 4

    4 * 4 = 16 i.e., 4 times greater than 4.

  14. 14. Chris Said:

    Hi Ulas. How did you know that it was a 6 digit number up front?

    Hi Nimble1. That looks, errr nimble;)

  15. 15. Muthiah Ganesh Said:

    Answer is 102564

    Because

    102564 is a integer &
    102564*4=410256

  16. 16. Chris Said:

    Hi Muthiah. Although your number is right, how do you know for sure that it’s the lowest number that works?

  17. 17. Chris Said:

    Hi Ajay Shahi. Start with 4, move it to the beginning still gives 4; 4 is not 16.

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