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Topic: Centigrade to Fahreheit (Read 651 times) |
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ThudnBlunder
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Centigrade to Fahreheit
« on: Oct 5th, 2003, 3:18am » |
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Converting 275oC to Fahrenheit we get 527oF. In fact, we could have just moved the 5 to the front.
What is the next largest example where moving the last digit to the front gives the right answer?
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| « Last Edit: Oct 5th, 2003, 4:19am by ThudnBlunder » |
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Sir Col
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Re: Centigrade to Fahreheit
« Reply #1 on: Oct 5th, 2003, 4:21am » |
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Once again, a lovely puzzle, T&B.
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It seems that 275oC is the only example!
Let us consider 3-digit examples, C=100x+10y+z.
F=9C/5+32=180x+18y+9z/5+32, and we are trying to find this equal to 100z+10x+y.
Equating and tidying up, 85(10x+y)+160=491z.
So z must be divisible by 5; but this was obvious from the fact that F=9C/5+32, and C must divide by 5 to obtain integer F. However, we also know that z[ne]0, as this would produce a 2-digit number, by placing the zero at the front. Hence z=5.
Therefore, 85(10x+y)=2295, giving 17(10x+y)=459[equiv]0 mod 9. Hence, 10x+y[equiv]0 mod 9, so, x+y[equiv]0 mod 9.
That is, we're looking for, C, a 3-digit number of the form xy5, where x+y[equiv]0 mod 9. A quick test of possibilities fails in each case: 365, 455, 545. However, C[ge]545 produces F with 4-digits.
My logical is a little faulty from here, but I will be bold enough to claim that no more solutions exist. It is faulty, because there remains a possibility that, F, having more digits than C, may end in zero, in which case it would produce a number with the same number of digits as C.
Can anyone finish off my proof?
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ThudnBlunder
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Re: Centigrade to Fahreheit
« Reply #2 on: Oct 5th, 2003, 4:42am » |
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For the 3-digit case, I get F = 100z + [(C-z)/10] and we know that z must equal 5.
Quote:| Can anyone finish off my proof? |
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No. 
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| « Last Edit: Oct 5th, 2003, 7:55am by ThudnBlunder » |
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towr
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Re: Centigrade to Fahreheit
« Reply #3 on: Oct 5th, 2003, 9:22am » |
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I've tried solving it a little differently
using (10*a+b )*9/5+32= 10^i *b + a (where i is the number of digits in a)
Since it's obvious b=5, we get
18*a+9+32= 10^i *5 + a
so 17*a = 5 * 10^i - 41
which is easy enough to try and find for different i's.
I haven't found any other number in the range of C-integers (2^32)
But I don't yet see any fundamental reason why 5 * 10^i - 41 wouldn't be divisable by 17 for i's over 2.
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Sir Col
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Re: Centigrade to Fahreheit
« Reply #4 on: Oct 5th, 2003, 12:59pm » |
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Towr, what a clever approach...
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The problem reduces to finding 5x10i–41[equiv]0 mod 17, or 5x10i[equiv]7 mod 17.
Using the good old Windows calculator again, I found 5x1018[equiv]7 mod 17. Therefore a=(5x1018–41)/17=294117647058823527.
Hence the next example is, 29411764705882352750C.
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towr
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Re: Centigrade to Fahreheit
« Reply #5 on: Oct 5th, 2003, 1:57pm » |
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hmm.. I should have been able to find that.. If I had programmed more cleverly..
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ThudnBlunder
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Re: Centigrade to Fahreheit
« Reply #6 on: Oct 6th, 2003, 3:31am » |
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Quote:| hmm.. I should have been able to find that.. If I had programmed more cleverly.. |
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: All the solutions are given by C = 5*(1016m+3 - 65)/17 where m = 0,1,2...
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| « Last Edit: Oct 21st, 2003, 2:56am by ThudnBlunder » |
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Sir Col
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Re: Centigrade to Fahreheit
« Reply #7 on: Oct 6th, 2003, 3:40am » |
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Very clever, T&B, but can you prove that your formula works for all values of m and provides the complete solution set? 
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ThudnBlunder
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Re: Centigrade to Fahreheit
« Reply #8 on: Oct 6th, 2003, 6:36am » |
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Quote:| Very clever, T&B, but can you prove that your formula works for all values of m and provides the complete solution set? |
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Yes, I can.
Let C = xn-1*10n-1 + ... + x1*10 + x0
Then F = x0*10n-1 + (C - x0)/10.
F = (9C/5) + 32 => x0 = 5
Hence (9C/5) + 32 = 5*10n-1 + [(C - 5)/10]
This gives C = 5*(10n - 65)/17
As 10 is a primitive root modulo 17, it follows that C is an integer iff n is of the form 16m+3.
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| « Last Edit: Oct 6th, 2003, 12:05pm by ThudnBlunder » |
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Sir Col
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Re: Centigrade to Fahreheit
« Reply #9 on: Oct 6th, 2003, 10:18am » |
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Nice, but could you please explain the last past?
Quote:| As 10 is a primitive root modulo 17, it follows that C is an integer iff n is of the form 16m+3. |
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ThudnBlunder
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Re: Centigrade to Fahreheit
« Reply #10 on: Oct 6th, 2003, 11:28am » |
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Quote:Nice, but could you please explain the last past?
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You asked for a proof. Fair enough, I gave one. But now you are expecting me to understand it??
OK, here is my primitive attempt:
We have C = 5*(10n - 65)/17
Hence we need 10n = 65 (mod 17)
= 14 (mod 17)
By Fermat's Little Theorem,
1016 = 1 (mod 17)
1016m = 1 (mod 17)
1016m+3 = 1000 (mod 17) = 14 (mod 17)
Because 10 is a primitive root mod 17, 1016m+i will run through all possibilities mod 17
Here, i = 3 gives us our 14 (mod 17)
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| « Last Edit: Oct 6th, 2003, 11:45am by ThudnBlunder » |
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