Remainder

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Remainder

by dtweah » Fri Apr 24, 2009 4:24 am
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

Which of the above Must be true?

A. I only

B. II only

C. III only

D. I and II

E. I, II and III

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by Feep » Fri Apr 24, 2009 6:53 pm
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?
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by dtweah » Sat Apr 25, 2009 10:48 am
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?
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:

1/7 has remainder 1326451
3/7 has 3264513.
In fact 3/7, 10/7 .... (7n-4)/7 exhibit 3264513.

That is all that is needed to solve the queston or variations on that question, which is the important thing. Question was intended to expose readers to this important property about the Number 7.
If I did them a disservice I apologize!

We can have endless debates about the value and utility of this forum. I probably see it different than you. Readers can decide WHETHER IT IS BEYOND GMAT TO TEST THIS PROPERTY ABOUT THE NUMBER 7. The important thing is they be exposed to the property, not about how a particular question is phrased since they may never meet that identical question on the GMAT. Hope this helps.

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by Feep » Sat Apr 25, 2009 2:35 pm
I'm not doubting that this question has merit. I merely ask, for the sake of everyone on this forum, that you mention in your posts that it is a question of your own design. Your questions, whether intentional or not, do differ in protocol from standard GMAT questions, and I want students to understand that.

It certainly is an interesting property, though...
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by Ian Stewart » Sat Apr 25, 2009 5:16 pm
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
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.

That said, this:
dtweah wrote: 1/7 has remainder 1326451
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:
There are certain primes whose remainders exhibit a periodicity that correlates with the prime itself. 7 is one of them.
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.

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.
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by dtweah » Mon Apr 27, 2009 1:19 pm
Ian Stewart wrote:
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
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.

That said, this:
dtweah wrote: 1/7 has remainder 1326451
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:
There are certain primes whose remainders exhibit a periodicity that correlates with the prime itself. 7 is one of them.
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.

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.
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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by Ian Stewart » Mon Apr 27, 2009 3:17 pm
(For GMAT test-takers, I'd note that my response below is entirely irrelevant to the GMAT test - no need to know any of this for the exam)
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
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.

My best guess was that you were referring to primitive roots mod p, which produce the cyclical patterns that appear in your question, and are ubiquitous in Number Theory, but now I understand you meant something different. The following is a consequence, incidentally, of Gauss' result that I mentioned above: if p is a prime different from 2 or 5, the number of repeating digits in the decimal expansion of 1/p is always a divisor of p-1.

And there are nine primes (not seven) less than 100 which produce what you call a 'remainder sequence' with period p-1; 7, 17, 19, 23, 29, 57, 59, 61, and 97.
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