79 (mod 10) = 73(3) (mod 10).Quote:
Originally Posted by desiresjab
Using Google sheets, 73 = 7*7*7 = 343 and 343 (mod 10) = 3.
Using your example,
73(3) (mod 10) = 33 (mod 10) = 27 (mod 10) = 7
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79 (mod 10) = 73(3) (mod 10).Quote:
Originally Posted by desiresjab
Using Google sheets, 73 = 7*7*7 = 343 and 343 (mod 10) = 3.
Using your example,
73(3) (mod 10) = 33 (mod 10) = 27 (mod 10) = 7
Good deal, old boy, you are right there with me. That one may have been trickier than the first several examples. This type of manipulation is fundamental to doing number theory. Many other fundamental ideas and techniques are indispensable. Every one that is learned and studied adds a speck to one's basic understanding of our number system and numbers in general.Quote:
Originally Posted by YesNo
Many people either forgot or never quite realized that something as routine as $867 is nothing more than an algebraic equation with the unknowns filled in as 10.
867=8(10)2+6(10)1+7(10)0
Banned again for what I am not doing. I don't get it. And I have forgotten how to extract myself from this mess.
Hmmmm....Which math notation did I use that it did not like? Could it be as simple as parentheses where it did not understand them?
Will it take a letter to a numerical exponent?
a6
Will it take aa and (a)a and a(a)?
Okay, I am stumped.
I will submit this post piece by piece until I find what is preventing it from passing. It will be done when I put QED at the bottom.
I may as well keep going with simple but intriguing ideas found within everday numbers. The next technique is one I would not expect anyone without experience in number theory to find, so I will go ahead and show the techniqe in a solution first, then explain its general workings. I know I can count on Yes/No, I hope others will follow the reasoning, as well, since it seems to be the way God thinks about some things, to mix metaphor and mathematics.
Prove that for every prime number p, there is a multiple of p whose every digit is 9. To me, at least, this law is not intuitively apparent without a studied understanding of numbers first. I suppose a Newton or a Gauss might look at it for the first time and see it immediately, but not most of us. Perhaps I am elevating my own dimness by suggesting it would take a Gauss or Newton. I still think it would require a massively bright person to work this out on their own without former experience in related problems.
A concrete example first, using 7 as our prime. We notice that N, 1/7=.142857... This is part of the technique, in case we are not Newton or Gauss and did not think of this approach. The three dots signify that the numbers after the decimal point comprise the repetend, and keep repeating.
When we multiply by(10)6, we get the digits of the repetend out front of the decimal point, but the repetend behind the decimal is unchaged, like this:
N=.142857.
(.142857...)(10)6=142857.142857...=1000,000N.
We know this is a hundred thousand of this repetend. If we subtract one of these repetend, like so:
(142857.142857...)-.142857...=142857=999,999N.
In other words, 999999/142857=7 and of course, 7·142857=999999, showing that seven has such a multiple.
Okay, everything got through except the general description at the last. I will try to fix that and post it next.
QED
In general: N ·10R where R is the length of the repetend of the prime p and N the repetend itself, the above technique must yield a similar result of a string of 9's of length R.
R will always be a divisor of p-1, or it will be p-1 itself.
There was a proof of the general case to go along with this, but since I have screwed up my own home document trying to fix this problem, I will now have to fix that.
That's all. Hopefully, it will be understood without the proof of the general case.
Here is the rest of it, not so much a proof as a clarification.
N10R-N=[9(10)R+9(10)R-1...+9(10)0]. This is indeed a string of 9's. Now,
Z=999999N, then Z/999999=N=1/p. Therefore p=999999/Z, and p·Z=999999.
Nice result.Quote:
Originally Posted by desiresjab
I've seen the technique used as a way of showing that a repeating decimal can be represented as a ratio of integers. This shows that such numbers are "rational". However, I did not realize that this implied that any prime has a multiple where the product is all nines. Now that you point it out, it makes sense that this should be the case.
I will give the forum three days to work on a problem that took me longer than that to solve. If more time is requested, I will give it gladly.
* * * * *
If I summed all the digits of the decimal representation of 44444444, and called that number C, then summed all the digits of that number, what would the resulting number D be, exactly.
It looks like this: AA=B, the sum of the digits of B=C, and the sum of the digits of C=D. Find D.
Can anyone do this problem? Give it a try.
I wouldn't be able to get beyond the neat way you present the problem, desiresjab.Quote:
Originally Posted by desiresjab
Well, this problem is no slouch. I believe I found it in an old math Olympiad exam. It was a bonus question for the brightest teenagers on earth. That is kinda disgusting, isn't it? Although I have had possession of the problem for years, I did not look at it much until I began to collect some tools. Then it took a long time of laying it down for long spells and forgetting all about it. A problem I want to solve but have no way to approach can drive me crazy if it stays in my mind. I just loved the shape of this problem and what it was asking for looked so impossible.Quote:
Originally Posted by Dreamwoven
What solving stuff like this comes down to is having the tools. There are always properties, laws and techniques which are related somehow and can be used to excavate the answer. If you have no idea about the existence of some property which will lead you right to the answer you are in for a long haul, probably an impossible one. Only cats like Leibniz could solve such a monster unprepared, independently rediscovering such laws and properties as are needed along the way.
I will show how this one is done soon. Right now I have personal obligations begging to be resolved around the house.
Sum the digits of the decimal representation of 44444444, then sum the digits of the new number. The form of the problem looks like this:
AA=B, the sum of B's digits equals C, the sum of C's digits equals D. Find D, exactly.
To find the number of digits in the decimal representation of 44444444, in other words how long the number is, the algebraic technique is to take the log of and add 1, while chopping whatever remains behind the decimal, leaving us with a whole number:
AA, like so:
↓{1+log AA}↓=↓{1+A log A}↓=16211, in the case of
44444444.
The actual calculations are as follows:
4444(log 4444)=16210.707879.... After we chop the decimal, which is irrational and goes on forever without pattern, and add 1 we know the huge number AA has 16211 digits. This number is far larger than the number of atomic particles in the universe, which is only in the neighborhood of 10120 particles, tops.
How can we sum the digits of B when we do not even know the number? We detour creatively, by supposing every one of those 16211 digits to be a 9, so that when we sum them they will equal 9(16211)=145899.
145899, then, is the upper limit for C, it can be no larger. But we are allowed to pretend again. We pretend that all six digits of C are equal to 9.
6X9=54. Ah, do you know what that is? It is an upper limit for D, the number we are after.
Suddenly, we are somewhere. One can almost smell solutions, but how do we get there? The answer lies again in observation and technique, in that order.
When any huge numbers are multipled, it is always easy for us to ascertain what the last digit is. In the case of 4444 times itself, no matter how many times successively we perform the operation, the last digit is always 4 or 6. Even powers of 4444 end in 6, and odd powers end in 4. Of course 4444 as an exponent is an even power, so 44444444 ends in a 6, and when divided by 10 would leave a remainder of 6.
We must detour now for some pretty facts, lest we arrive at our final destination with our route still shrouded in mystery.
* * * * *
Any time you sum the digits of a number X, that sum S will remain congruent to X (mod 9) through successive summing operations. This only happens with 9 because we use base 10. In base 8, 7 would have this same property through successive summing operations. Any base. (The above preservation property is also true for 3 in base 10).
Notice that once we ascertained the number of digits of 44444444, summing the digits successively is the only operation we have performed.
Successive powers under a modulus (any particular divisor ) always bend back and reapeat themselves in a cycle. This is called a power residue cycle, a cyclotomic number, to throw in a fancy term dangerously. They are certainly cyclical, but I do not know if that makes them cyclotomic. The host confesses.
A Modulus does not allow any number in its system to be as large as itself. The modulus is king. Larger numbers are bent back by division until only a remainder is left. Any whole number, when divided by 10, for instance, will leave one of ten remainders: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. That's it. The modulus can systematically reduce any giant number to one smaller than itself. And as far as the king is concerned, any larger number than itself which it can divide is equal to zero. 30=0 (mod 10).
Let's look at the power residue set of 4444, (mod 9). It is okay to begin with the zero power of 4444, which we know is always 1 for any number, and 1 will always be the first number in the power resudue set of any number.
44440=1 (mod 9)
44441=16=7 (mod 9)
44442=19749136=40=4 (mod 9)
44443=87765160384=55=1 (mod 9)
...
...
...
44444444=1 (mod 3)
9(493)=4437, means 4444=7 (mod 9)
44444444=7 (mod 9)
At its 4444th power, 4444 is congruent to 7 (mod 9) and 1 (mod 3).
In fact, the power residue set for this number (mod 3) is
{1, 1, 1, 1, 1....}, and on any power our huge exponential number only equals 1 (mod 3).
The pattern listed above vertically for (mod 9) power residues has produced 1 again at power 3, so we know the complete cycle, which goes 1, 7, 4, repeating every three powers until the highest power is reached. Also notice that to take the congruency of a number with respect to 9 in our base 10 system, we only need to sum its digits then take the remainder when divided by 9.
The more convenient way for calculation is to begin the pattern on the first power so that the power residue set cycles like this from power 1:
{7, 4, 1, 7, 4, 1...}
* * * * *
Gathering what we know so far:
Our target number is less than or equal to 54 and is congruent to 7 (mod 9).
That becomes a small set.
{7, 16, 25, 34, 43, 52}
We have gone from fifty-four possiblities to six.
Now, as was the case with 9, the value, the magnitude, of a number is the same as the sum of its digits (mod 3), so we are further able to say that the correct answer must be congruent to 1 (mod 3). Only three numbers now qualify.
{7, 16, 34}
Now the judging becomes more demanding, more appropriately the search for a tool or a property to distinguish Miss America out of the three.
We know that B ends in a 6. In symbols B=6 (mod 10). But how does this help us? 10 cannot play the same trick that 3 and 9 did, for the preservation of its congruency does not happen. (It would happen on 10 only for base 11, where the factor 5 would also exhibit the property). Ah, but maybe 6 can play the trick. Actually, it cannot help us that way, either.
The simple observation that works is that any number which leaves a remainder of 1 when when divided by 3 and a remainder of 7 when divided by 9, has to leave a reaminder of 4 when divided by 6. But the English of that is so long and messy. Check this out.
If D=1 (mod 3) and= 7 (mod 9), then D=4 (mod 6).
Only one number from the last set qualifies:
{16} is the answer. D=16
One can only assume several of the interior digital places of C are 0 and 1.
Since 16 is smaller than expected, let me recheck the steps. The mistake, if any, is probably in following that (7, 4, 1) cycle correctly to the last power.
I may be back.
I think I can proclaim it correct, since I validated my stepping on the power residue set of 7, 4, 1. At the 4443rd power the cycle is on 1, which puts it on 7 for the 4444th power.
Since we know C is less than 145899 but still a six digit number, I envision a number built along the following lines:
100456. It makes perfect sense now.