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Topic: Deleting the first digit (Read 343 times) |
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fatball
Senior Riddler
Can anyone help me think outside the box please?
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Deleting the first digit
« on: Jan 22nd, 2006, 9:02pm » |
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Find all powers of 2 such that, after deleting the first digit, another power of 2 remains. (For example, 25 = 32. On deleting the initial 3, we are left with 2 = 21.) Numbers are written in standard decimal notation, with no leading zeroes.
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Eigenray
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Re: Deleting the first digit
« Reply #1 on: Jan 23rd, 2006, 2:50am » |
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| hidden: | Note that if the first digit k is even, then dividing the number by 2 gives another solution. So let's assume the first digit k is odd. Say
2x = k10y + 2z.
Since x>z, the largest power of 2 dividing k10y=2x-2z is y=z. Thus
2x-z = k5z + 1.
Now, if we're not allowing the second digit to be 0, that is, 2x has just one more digit than 2z, then 0<x-z<7, and we find only the solution k=3, z=1. Thus the only such number with odd first digit is 32, and the only other number is then 64.
But we can handle the more general case. As explained before, 2x=2z mod 5z implies that x=z mod 4*5z-1, so that x-z > 4*5z-1. But it should not take much convincing that
24*5^(z-1) < 10 * 5z + 1
is only possible for z=1. |
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