## . . . 6

This puzzle is from the 1962 International Mathematics Olympiad. We have a number (integer) with 6 as the last (right-most) digit (936 for example). If we erase the 6 and put it on the left end of the number (693 in our example), then we have a number four times our original number. We see that 936 doesn't work. What is the smallest number that does fit the above conditions? Try it, before looking at the following solution.

Solution:

We can solve this by trial and error. But first we can deduce that the left-most digit must be a 1 (6.../4 = 1...). Then a little trial and error shows that the answer must have at least four digits. There are a lot of numbers to try, with four or more digits. So trial and error is beginning to seem a little too time-consuming.

First solution: Maybe we can solve it algebraically. If our number is x, then 4x=6m+(x-6)/10, where m is some power of 10. 40x=6n+x-6, where n is the next power of 10. 39x=6n-6 or 13x=2(n-1) or x=2(n-1)/13. So n-1 (one less than a power of 10, in other words, a string of nines) would seem to be a multiple of 13. We can now test the first few strings or nines, to see if any of them is a multiple of 13. 999999 turns out to be a multiple of 13 (13 x 76923). So x=2 times 76923 = 153846. And multiplying this by 4, to see if it solves our problem, we get 615384. So it is the solution.

We can do this first solution by writing out x as a sum of powers of ten (10^(s-1)a+10^(s-2)b+...+10p+6). The algebra is similar to the above.

Second solution: Here is a clever way to solve it. 1...6x4=61... So the last digit of the product must be 4: 1...46x4=61...4. So the next to last digit of the solution must be 8 (do you see why?): 1...846x4=61...84. So the third to last digit must be 3: 1...3846x4=61...384. So the next digit must be a 5: 1...53846x4=61...5384. The next digit must be a 1, which is what we were looking for. The smallest such number is 153846, which should work. But we must multiply it out to make sure we haven't made a mistake: 153846x4=615384. It works.

Third solution: This is my own solution. We can do something similar to the second solution, but work from the left end. 1...6x4=61... 61.../4 is between 1525... and 154999... So our second digit is 5: 15...6x4=615... 615.../4 is between 15375... and 153999... So our third digit is 3: 153...6x4=6153... 6153.../4 is between 153825... and 15384999... So our fourth digit is 8: 1538...6x4=61538... 61538.../4 is between 153845... and 1538474999... So the fifth digit is 4. We've been looking for a 4, as the right-most digit of our product must be a 4. So 153846x4 should be 615384. Multiplying it out shows that it is our answer.

One might ask what the next such number is. You might want to try that. Using any of the above methods, the answer this problem is 153,846,153,846. Noticing a pattern there, maybe you can find the third smallest such integer. You probably do not need to do any calculations, except to confirm your deduction by multiplying by 4.