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A Remainders Shortcut for the GMAT

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A Remainders Shortcut for the GMAT  [#permalink]

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New post 25 Mar 2015, 04:52
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FROM Veritas Prep Blog: A Remainders Shortcut for the GMAT
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We firmly believe that teaching someone is a most productive learning for oneself and every now and then, something happens that strengthens this belief of ours. It’s the questions people ask – knowingly or unknowingly – that connect strings in our mind such that we feel we have gained more from the discussion than even our students!

The other day, we came across this common GMAT question on remainders and many people had solved it the way we would expect them to solve. One person, perhaps erroneously, used a shortcut which upon reflection made perfect sense. Let me give you that question and the shortcut and the problem with the shortcut. We would like you to reflect on why the shortcut actually does make sense and is worth noting down in your log book.

Question: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?

(A) 3

(B) 12

(C) 18

(D) 22

(E) 28

Solution: We are assuming you know how people do the question usually:

The logic it uses is discussed here and the solution is given below as Method I.

Method I:

Positive integer n leaves a remainder of 4 after division by 6. So n = 6a + 4

n can take various values depending on the values of a (which can be any non negative integer).

Some values n can take are: 4, 10, 16, 22, 28, …

Positive integer n leaves a remainder of 3 after division by 5. So n = 5b + 3

n can take various values depending on the values of a (which can be any non negative integer).

Some values n can take are: 3, 8, 13, 18, 23, 28, …

The first common value is 28. So n = 30k + 28

Hence remainder when positive integer n is divided by 30 is 28.

Answer: E.

Perfect! But one fine gentleman came up with the following solution wondering whether he had made a mistake since it seemed to be “super simple Math”.

Method II:

Given in question: “n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5.”

Divide the options by 6 and 5. The one that gives a remainder of 4 and 3 respectively will be correct.

(A) 3 / 6 gives Remainder = 3 -> INCORRECT

(B) 12 / 6 gives Remainder = 0 -> INCORRECT

(C) 18 / 6 gives Remainder = 0 -> INCORRECT

(D) 22 / 6 gives Remainder = 4 but 22 / 5 gives  Remainder = 2 -> INCORRECT

(E) 28 / 6 gives Remainder = 4 and 28 / 5 gives Remainder = 3 ->  CORRECT

Now let us point out that the options are not the values of n; they are the values of remainder that is leftover after you divide n by 30. The question says that n must give a remainder of 4 upon division by 6 and a remainder of 3 upon division by 5. This solution divided the options (which are not the values of n) by 6 and 5 and got the remainder as 4 and 3 respectively. So the premise that when you divide the correct option by 6 and 5, you should get a remainder of 4 and 3 respectively is faulty, right?

This is where we want you to take a moment and think: Is this premise actually faulty?

The fun part is that method II is perfectly correct too. Method I seems a little complicated when compared with Method II, doesn’t it? Let us give you the logic of why method II is correct:

Recall that division is nothing but grouping. When you divide n by 30, you make complete groups of 30 each. The number of groups you get is called the quotient (not relevant here) and the leftover is called the remainder. If this is not clear, check this post first.

When n is divided by 30, groups of 30 are made. Whatever is leftover is given in the options. 30 is completely divisible by 6 and by 5 hence the groups of 30 can be evenly divided into groups of 6 as well as groups of 5. Now, whatever is leftover (given in the options) after division by 30, we need to split that into further groups of 6 and 5. When we split it into groups of 6 (i.e. divide the option by 6), we must have remainder 4 since n leaves remainder 4. When we split it into groups of 5 (i.e. divide the option by 5), we must have remainder 3 since n leaves remainder 3. And, that is the reason we can divide the options by 6 and 5, check their remainders and get the answer!

Now, isn’t that neat!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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A Remainders Shortcut for the GMAT  [#permalink]

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New post 02 Apr 2015, 00:24
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Check this useful concepts for Remainder Theorem.
FERMAT’s REMAINDER THEOREM
Let p be a prime number and a be a number that is co-prime to p. Then, the remainder obtained when ap-1 is divided by p is 1.

For Example:
Find the remainder when 2100 + 2 is divided by 101?
Solution:
101 is a prime number and 2 is co-prime to 101.
2100 = 2101-1 = This when divided by 101 gives remainder 1 according to Fermat’s Theorem.
Also, 2 when divided by 101 give 2 as remainder.
Hence, Final Remainder = 1 + 2 = 3

WILSON’s REMAINDER THEOREM
If p is a prime number, then (p-1)! + 1 is divisible by p.
For Example:
Find the remainder when 16! Is divided by 17?
Solution:
17 is a prime number.
16! = (16! + 1) – 1 = [(17-1!) +1] – 17+16
According to Wilson’s Theorem [(17-1!) +1] is divisible by 17.
Hence, Final remainder = 16

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Re: A Remainders Shortcut for the GMAT  [#permalink]

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New post 19 Jun 2018, 11:38
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Re: A Remainders Shortcut for the GMAT &nbs [#permalink] 19 Jun 2018, 11:38
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