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There are certain primes whose remainders exhibit a periodicity that correlates with the prime itself. 7 is one of them. This question exposes readers to this property about such primes:Feep wrote:This is the second question I've seen from you that seems highly unlikely to be actually asked on a real GMAT exam. When a number is stated as "a", it is usually implied that the number is a constant value, and yet you imply that "a" is a changing value. Does "b" also change? In the answer choices, can it be known which a or b value is being used?
If you're inventing questions, dtweah (and I'm not saying you are), PLEASE state as such in the thread. You'd be doing many people a great disservice by introducing questions that are unlike those found on the actual exam. What is the source of this question?
It took me several minutes just to figure out what you meant by this question, since as written it doesn't make sense; when you divide a by b, of course you will always get the same remainder, and not a sequence of different remainders. As best I can tell, you don't mean "in the calculation of a/b", but rather "in the calculation of a^n/b, for n = 1, 2, 3, ...". Then if you let a = 3, for example, when you divide 3^1 by 7, the remainder is 3; if you divide 3^2 by 7 the remainder is 2; if you divide 3^3 by 7, the remainder is 6, and so on. That produces the sequence in the question: 3, 2, 6, 4, 5, 1, 3, ... ad infinitum. Of course, a need not be 3 here; it could be any number with a remainder of 3 when divided by 7: a could be 10, 17, 24, etc.dtweah wrote:In the calculation of a/b the following remainder sequence was observed: 3264513.... On the basis of this the following propositions were made:
I. a +b > 12
II 9<a+b<18
III b>a
doesn't accord with the above interpretation at all. The powers of 1 will produce remainders of 1 every time, and the sequence will be 1, 1, 1, 1, ...dtweah wrote: 1/7 has remainder 1326451
There aren't just 'certain' primes that exhibit this periodicity; this is true for all primes. If you take any prime p, you can always find a positive integer x less than p for which x^1, x^2, x^3, ... x^(p-1) produce all possible nonzero remainders when you divide by p. When p = 7, both x = 3 and x = 5 produce all six possible nonzero remainders when raised to powers 1 through 6, for example. When p=11, x=2 will produce all ten nonzero remainders when raised to the powers 1 through 10. One can find similar examples for every prime. This is a famous result of Gauss, and is a fundamental theorem in Algebraic Number Theory.dtweah wrote:
There are certain primes whose remainders exhibit a periodicity that correlates with the prime itself. 7 is one of them.
I don’t understand what you are talking about.Ian Stewart wrote:It took me several minutes just to figure out what you meant by this question, since as written it doesn't make sense; when you divide a by b, of course you will always get the same remainder, and not a sequence of different remainders. As best I can tell, you don't mean "in the calculation of a/b", but rather "in the calculation of a^n/b, for n = 1, 2, 3, ...". Then if you let a = 3, for example, when you divide 3^1 by 7, the remainder is 3; if you divide 3^2 by 7 the remainder is 2; if you divide 3^3 by 7, the remainder is 6, and so on. That produces the sequence in the question: 3, 2, 6, 4, 5, 1, 3, ... ad infinitum. Of course, a need not be 3 here; it could be any number with a remainder of 3 when divided by 7: a could be 10, 17, 24, etc.dtweah wrote:In the calculation of a/b the following remainder sequence was observed: 3264513.... On the basis of this the following propositions were made:
I. a +b > 12
II 9<a+b<18
III b>a
That said, this:
doesn't accord with the above interpretation at all. The powers of 1 will produce remainders of 1 every time, and the sequence will be 1, 1, 1, 1, ...dtweah wrote: 1/7 has remainder 1326451
There aren't just 'certain' primes that exhibit this periodicity; this is true for all primes. If you take any prime p, you can always find a positive integer x less than p for which x^1, x^2, x^3, ... x^(p-1) produce all possible nonzero remainders when you divide by p. When p = 7, both x = 3 and x = 5 produce all six possible nonzero remainders when raised to powers 1 through 6, for example. When p=11, x=2 will produce all ten nonzero remainders when raised to the powers 1 through 10. One can find similar examples for every prime. This is a famous result of Gauss, and is a fundamental theorem in Algebraic Number Theory.dtweah wrote:
There are certain primes whose remainders exhibit a periodicity that correlates with the prime itself. 7 is one of them.
Still, it will never be tested on the GMAT, since no GMAT test taker would ever be expected to know it. It is certainly an interesting property of numbers, I'll agree.
Well at least I finally understand what you meant by the question - by 'remainder sequence', you are referring to the sequence of remainders when a is divided by b, when 10a is divided by b, when 100a is divided by b, when 1000a is divided by b, and so on. Those are precisely the remainders you obtain when you are performing long division. That is, the meaning of the question would have been clear if you had said that the term r_n in the 'remainder sequence of a/b' was equal to the remainder when a*10^(n-1) is divided by b. When you simply talk about the remainder when a is divided by b, there can only be a single numerical answer, of course, which is why the wording of the original question is unclear, and as you can tell from the responses above, none of the respondents understood what you meant.dtweah wrote: I don’t understand what you are talking about.
1/7= 0.1428571428571428571…… 1 will always occur in the 7th place
In doing this division: The remainder sequence is 326451326451326451…… 3 will always occur in the 7th place
From the above:
8/7=1. 1428571428571428571…..
And the remainder sequence is 326451326451326451……
(7n-6)/7 will have the remainder sequence 326451326451326451…… for all n=1,2,……
All primes don't exhibit this periodicity. There are only 7 such primes less than 100 that do.
11 does not exhibit the periodicity above. 11 has a periodicity of in the division of 1/11= .0909090909...... The period here is 2. Different from the case of 7
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