Find a number such that when you rotate the number to the left,
the new number is 1.5 times the original number.
To rotate the number left, move the first digit to the last place
of the number like 75837 rotated to the left is 58377)
January 26, 2011
IQ puzzle
Find a number such that when you rotate the number to the left,
the new number is 1.5 times the original number.
To rotate the number left, move the first digit to the last place
of the number like 75837 rotated to the left is 58377)
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8 comments:
for (int i = 10; i < int.MaxValue; i++)
{
int original = i;
int shifted = int.Parse(i.ToString().Substring(1) + i.ToString()[0]);
if ((original * 1.5F) == shifted)
{
Console.WriteLine("Original: " + original + "\nShifted: " + shifted);
break;
}
}
That's how I discovered
Original: 352941176
Shifted: 529411763
@Andre
Original * 1.5 = Shifted
but 352941176 * 1.5 = 529411764
not 529411763
Just missed !!! :)
What about 13, 1 rotated to left is 31. which is exactly one whole and a half times the original number. It didn't say it had to be a huge number, correct?
Sorry Candace but 1.5 * 13 = 19.5
What is the answer?
Zero should be a good answer. It does not ask us to come up with a non-zero number.
Has it got something to do with decimal places? complex numbers? Or is it just an integer.
One of the Numbers iss:
3529411764705882
^^
6 :)
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