Find the smallest number such that if its rightmost digit is placed at its left end, the new number so formed is precisely 50% larger than the original number.
Let M denote the number satisfying the conditions of the problem. We are required to evaluate the smallest value of M.
Let us assume that M is a nonnegative integer. Then we observe that the smallest value of M is zero, as writing M = 00....00( k zeros) we can trivially observe that it satisfies all the conditions of the problem.
Let us now assume that M is a positive integer.
Then, writing M as 10x+y, we must have:
3(10x+y)/2 = (10^n)*y+x
=> 2*(10^n)*y+2x=30x+y
=> 28x= y*(2*(10^n )-3)
=> 7*4*x = y*(2(10^n)-3) ....(#)
Now, 2*10^n-3 is obviously odd, and so:
gcd(2*10^n-3, 4)=1
This is only possible if y is a multiple of 4.
Now, 7 must divide 2*10^n-3
Therefore, 2*10^n = 3 (mod 7) = 10 (mod 7)
=> 10^n = 5(mod 7)
Now, 10=3(mod 7) => 10^2 = 2(mod 7) => 10^3 = 6(mod 7)
=> 10^4= 4 ( mod 7) => 10^5= 5(mod 7)
Accordingly, from (#), we have:
(2*10^5-3)*y= 7*4*x
=> (199997/7)*y= 4*x
=> 28571*y= 4*x
Now, we recall that we have shown that y must be a multiple of 4. But since y is a single digit, we must have y=4,8
y=4 gives, x= 28571
y=8 gives, x= 57142
Recalling that M =10*x+y, we observe that:
M can assume the values 285714 or, 571428.
Consequently, the required smallest number that satisfies al, the conditions of the puzzle under reference is 285714.
As a check, we can easily verify that: (3/2)*(285714) = 428571, which is perfectly in consonance with the given conditions,
Edited on December 26, 2021, 7:33 am