A no. has 6 at the end. When it is shifted to front the no. becomes 4 times. find the least such possible no.?

Answer 1

The smallest one is 153,846. (No, Steve, there are an infinite number of solutions.)

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I see PB under me gives a nice algebraic solution. My approach didn't involve algebra at all... I figure that, if the number (call it x) ends in 6, then 4x ends in 4. Since the 6 is being moved to the front of x, it follows that the second to last digit of x has to be the same as the last digit of 4x; therefore x ends in 46. And, since x ends in 46, 4x ends in 84. So, x ends in 846, etc. - continue in this way until you see the numbers start to repeat.

Answer 2

there are an infinite no. of solutions.
to get the smallest, let the no. be x and the no. of digits be n.
form d equation, we get

(x-6)/10 + 6(10^n) = 4x

x-6 + 6(10^n+1) = 40x

x+6(10^n+1 -1) = 40x

39x = 6(10^n+1 -1)

13x = 2(10^n+1 -1)

find the least possible (10 ^n+1 -1) which is diviible by 13.
it comes out to be 999999.
n is 6 and substitute n to get the least value of x, which is 153846

Answer 3

60 +x = 4(10x + 6)
36 = 39x (doesn't work - so no two-digit answer)

600 + x = 4(10x +6)
594 = 39x (doesn't work - so no three-digit answer)

6000 + x = 4(10x +6)
5994 = 399x (doesn't work - so no 4-digit answer)

60000 + x = 4(10x +6)
59994 = 3999x (doesn't work

Keep going, x approaches 15 as a limit, but never makes it since 15 * a number ending in 9 will never equal a number ending in 4

There is no solution.