## 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!

June 2nd, 2010 at 7:47 pm

24*4=42

as 42 is always the answer

method-brute force

June 2nd, 2010 at 8:06 pm

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?

June 2nd, 2010 at 8:10 pm

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.

June 2nd, 2010 at 8:50 pm

102564 is the least such number

410256 = 4 * 102564

June 3rd, 2010 at 1:45 am

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

June 3rd, 2010 at 3:54 am

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.

June 3rd, 2010 at 10:24 am

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

June 4th, 2010 at 2:26 am

the number is 104 !!

June 4th, 2010 at 6:38 pm

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

June 5th, 2010 at 3:44 am

The answer is 102564. -by Dino1111

June 5th, 2010 at 6:24 pm

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.

June 8th, 2010 at 6:42 am

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

June 9th, 2010 at 12:17 am

Answer is 4

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

June 9th, 2010 at 4:22 am

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

Hi Nimble1. That looks, errr nimble;)

June 10th, 2010 at 8:24 pm

Answer is 102564

Because

102564 is a integer &

102564*4=410256

June 15th, 2010 at 8:09 pm

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

June 24th, 2010 at 6:13 am

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