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Move the 6, Quadruple the Number (Posted on 2003-09-04) Difficulty: 4 of 5
In Move the 2, Double the Number, we found a number that ended in two, for which moving the two to the beginning of the number doubled its value.

For this problem, we have a number with 6 as the last (right-most) digit.
If we erase the 6 and put it on the left end of the number (for example, 936 would become 693), then we have a number four times our original number (we see that 936 doesn't work, of course).

What is the smallest number that fits this condition?

What is the second smallest number that works?

What is the tenth smallest number that works?

See The Solution Submitted by DJ    
Rating: 4.2500 (12 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Not for SIlver Knight's Eyes | Comment 3 of 6 |
define the number we are looking for as x

the ones digit of x is six (specified by the problem)
The tens digit of x will be the ones digit in 4x so it must be equla to the ones digit of the four times the ones digit of x (six as noted above) so the tens digit must be 4
The one hundreds digit of x is the tens digit of 4x and so must be teh ones digit of four times the tens digit of x plus the tens digit of four times the ones digit of x (to accout for the "carried" tens) and is therefore 8 {(4*4)+2 = 18}

Using this technique we continue until we find a value of 4x with a leftmost digit that is 6.
the values we come up with are
x = 153846
4x = 615384

this is the smallest value of x
the second smallest value is this string repeated twice (153846153846)
and teh 10th amllest value would be the string repeated 10 times.

  Posted by FatBoy on 2003-09-05 07:44:25
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