Hey dave13

First things first, the first list of numbers are of form 3x and 7x, where x takes integer values starting from 1.

Every time you add 2 to 3x, you will get a number of the form 3x + 2.
Similarly, we get 7x + 5 when you add 5 to 7x.

3x + 2 -> 2,5,8,11,14,17,20(7 numbers)
7x + 5 -> 5,12,19(3 numbers)

So, you can extrapolate and find out that there are more numbers of the form 3x + 2 than of the form 7x + 5.

I think the reason we re-write 3a+2 as 3x–1 and 7b+5 as 7b-2 is to explain that they yield a common list.

Here, we need to find the numbers which are present in both the lists. So, you can't equate in this case.

Hope this helps you!

pushpitkc thank you for your nice explanation

one question:

what are these numbers in red (quotients ?) and how do we get them ?... "Since the next number will also belong to both the lists above so it will be 3/6/9/12/15/18/21… away from 5 and it will also be 7/14/21/28/35/42… "

thanks and have a great day

https://gmatclub.com/forum/new-units-di ... 68569.html is no longer valid. Bunuel am I looking at something wrong?

In this post, I would like to focus on a particular type of remainder questions and how to solve them in a particular way. For the type of questions I am going to discuss today, I like to use “Binomial Theorem.” You might be tempted to run away right now and save yourself some precious time if you are not a Math geek but wait! We will just use an application of Binomial which I will explain in very simple language. I am quite certain that you will be comfortable with the method if you just give it a chance.

Question 1: What is the remainder when \(\frac{(3^{84})}{26}\)

(A) 0
(B) 1
(C) 2
(D) 24
(E) 25

Solution: First up, GMAT questions don’t involve any painful calculations. So my thought is that there has to be an obvious link between 3 and 26. 26 is 1 less than the cube of 3. (It helps one to know the squares of first 20 numbers and cubes of first 10 numbers.)

So, \(3^3 = 27\)

But how is it going to help us? Now we come to binomial theorem. Let me start with something you already know.

\((a + b)^2 = a^2 + 2ab + b^2\)

\((a + b)^3 = a^3 + 3ba^2 + 3ab^2 + b^3\)

What about \((a + b)^4\) or \((a + b)^5\) or higher powers? Binomial theorem just tells us how to expand these expressions. It gives you a general formula:

(a + b)^n = a^n + n*a^(n-1)*b + n(n-1)/2*a^(n-2)*b^2 +……..+ n*a*b^(n-1) + b^n

I know the above looks intimidating but our concern is limited to the last term of the expression. Notice that every term above is divisible by ‘a’ except for the last term b^n. Every term but the last has ‘a’ as a factor. That is all you need to understand about Binomial Theorem.

Now for some quick applications:

What is the last term when you expand \((8 + 1)^{20}\)? It is 1^20 (which is just ‘1’).

When you expand \((8 + 1)^{20}\), is every term divisible by 8? Yes, except for the last term, 1, because every term has 8 as a factor except for the last term.

If I divide \((8 + 1)^{20}\) by 8, what will be the remainder? Since every term (except for the last one) in the expansion of \((8 + 1)^{20}\) is divisible by 8, we can say that \((8 + 1)^{20}\) is 1 more than a multiple of 8. Hence the remainder when we divide it by 8 will be 1.

Or I can say that when I divide \(9^{20}\) (which is just \((8 + 1)^{20}\) by 8, the remainder is 1.

Now let’s look at our original question.

\((3^{84}) = (3^3)^{28} = 27^{28} = (26 + 1)^{28}\)

Every term of \((26 + 1)^{28}\) will be divisible by 26 except for the last one. The last term will be 1^28 = 1. Hence, when you divide 27^28 by 26, the remainder will be 1.

Answer (B). This question is discussed HERE.

All you had to do was to look for a power of 3 which is 1 more or 1 less than 26. We found that the third power of 3 is 1 more than 26. We adjusted the power to make 27 the base and split it into (26 + 1). We got the remainder as 1. Why do we necessarily look for a power 1 more or 1 less? We do that because 1^n is always 1. If we are left with 2^28, we again have a problem since we don’t know what 2^28 is. Let’s use this concept in another problem now:

Question 2: What is the remainder when 2^86 is divided by 9?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 8

I have added a few complications in this question. Let’s tackle them one by one. We start by looking for a power of 2 which is 1 more or 1 less than 9. We know 2^3 = 8 which is 1 less than 9.

Next, let’s adjust the power to make the base 8.

\(2^{86} = 8^?\)

86 is not divisible by 3. The closest integer less than 86 that is divisible by 3 is 84. So, separate out two 2s and work with the rest of the 84 2s as of now.

\(2^{86} = (2^2) * (2^{84}) = (4) * (2^3)^{28} = 4* (8^{28})\)

I am going to forget about the 4 for the time being.

\(8^{28} = (9 – 1)^{28} = [9 + (-1)]^{28}\)

Every term of this expression will be divisible by 9 except for the last term \((-1)^{28}\) which is again equal to 1.

Hence, 8^28 will give a remainder 1 when divided by 9.

I can say that \(8^{28} = 9m + 1\) where m is some positive integer. Now, we need to consider the 4 that we left out in the previous step. Our actual expression is

\(4 * 8^{28} = 4 * (9m + 1) = 4*9m + 4\)

When I divide this by 9, 4*9m is divisible by 9. So, 4*9m + 4 is 4 more than a multiple of 9. Hence the remainder will be 4.

This question is discussed HERE.

A question to ponder on: How will you solve this question if I change it to “What is the remainder of 2^83 is divided by 9?”

Here, we will discuss the question we left you with in the last post. It involves a lot of different concepts – remainder on division by 5, cyclicity and negative remainders. Since we did not get any replies with the solution, we are assuming that it turned out to be a little hard.

It actually is a little harder than your standard GMAT questions but the point is that it can be easily solved using all concepts relevant to GMAT. Hence it certainly makes sense to understand how to solve it.

Question: What is the remainder when 3^(7^11) is divided by 5? (here, 3 is raised to the power (7^11))

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

Solution: As we said last week, this question can easily be solved using cyclicity and negative remainders. What is the remainder when a number is divided by 5? Say, what is the remainder when 2387646 is divided by 5? Are you going to do this division to find the remainder? No! Note that every number ending in 5 or 0 is divisible by 5.

2387646 = 2387645 + 1

i.e. the given number is 1 more than a multiple of 5. Obviously then, when the number is divided by 5, the remainder will be 1. Hence the last digit of a number decides what the remainder is when the number is divided by 5.

On the same lines,

What is the remainder when 36793 is divided by 5? It is 3 (since it is 3 more than 36790 – a multiple of 5).

What is the remainder when 46^8 is divided by 5? It is 1. Why? Because 46 to any power will always end with 6 so it will always be 1 more than a multiple of 5.

On the same lines, if we can find the last digit of 3^(7^11), we will be able to find the remainder when it is divided by 5.

Recall from the discussion in your books, 3 has a cyclicity of 4 i.e. the last digit of 3 to any power takes one of 4 values in succession.

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
3^6 = 729

and so on… The last digits of powers of 3 are 3, 9, 7, 1, 3, 9, 7, 1 … Every time the power is a multiple of 4, the last digit is 1. If it is 1 more than a multiple of 4, the last digit is 3. If it is 2 more than a multiple of 4, the last digit is 9 and if it 3 more than a multiple of 4, the last digit is 7.

What about the power here 7^(11)? Is it a multiple of 4, 1 more than a multiple of 4, 2 more than a multiple of 4 or 3 more than a multiple of 4? We need to find the remainder when 7^(11) is divided by 4 to know that.

Do you remember the binomial theorem concept we discussed many weeks back? If no, check it out here.

7^(11) = (8 – 1)^(11)

When this is divided by 4, the remainder will be the last term of this expansion which will be (-1)^11. A remainder of -1 means a positive remainder of 3 (if you are not sure why this is so, check last week’s post here). Mind you, you are not to mark the answer as (D) here and move on! The solution is not complete yet. 3 is just the remainder when 7^(11) is divided by 4.

So 7^(11) is 3 more than a multiple of 4.

Review what we just discussed above: If the power of 3 is 3 more than a multiple of 4, the last digit of 3^(power) will be 7.

So the last digit of 3^(7^11) is 7.

If the last digit of a number is 7, when it is divided by 5, the remainder will be 2. Now we got the answer!

Answer (C)

Interesting question, isn’t it?

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. This question is discussed HERE.

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!

If a and b are positive integers such that a/b = 97.16, which of the following cannot be the remainder when a is divided by b?

(A) 4
(B) 12
(C) 22
(D) 28
(E) 96

Some of my most brilliant students have asked me something similar to this: “When I divide 11 by 4, I get 2.75. What do you mean by, “What is the remainder?’ Where is it?’” So if it is bothering you as well, don’t worry; I will address this issue first.

Say, I tell you the following: Divide 11 by 4. What do you get?

You could answer me with one of the following:

Case 1: You could say, “I get 2.75”

Case 2: You could say, “I get 2 as the quotient and 3 as the remainder.”

Either ways, you are correct. 11/4 = (2 ¾)

When you use the decimal form, you get a .75 which you add to 2 to give you 2.75. This .75 is nothing but the way you express the remainder 3. When you divide 11 by 4, 4 goes into 11 two times and then 3 is left over. When 4 goes into 3, you get 0.75 which is ¾. That is the reason why you can write 11/4 as (2 ¾) in mixed fractions.

Do the following calculations and see what you get. Express the answer in both the forms – Decimal form and quotient-remainder form.

1. Divide 22 by 8.
2. Divide 55 by 20.
3. Divide 275 by 100.

Let’s see what we get in each case.

1. Divide 22 by 8.

Case 1: In decimal form we get 2.75

Case 2: In quotient-remainder form we get 2 as quotient and 6 as remainder (Remember, the divisor is 8 here). We can say 22/8 = 2 (6/8) in mixed fractions.

1. Divide 55 by 20.

Case 1: In decimal form we get 2.75

Case 2: In quotient-remainder form we get 2 as quotient and 15 as remainder (Remember, the divisor is 20 here). We can say 55/20 = 2 (15/20) in mixed fractions.

1. Divide 275 by 100.

Case 1: In decimal form we get 2.75

Case 2: In quotient-remainder form we get 2 as quotient and 75 as remainder (Remember, the divisor is 100 here). We can say 275/100 = 2 (75/100) in mixed fractions.

Notice that the remainder is different in each case above. As the divisor changes, the remainder changes even though in decimal form the answer is the same. This is so because in each of the above cases, 6/8 = 0.75, 15/20 = 0.75 and 75/100 = 0.75. Each of these fractions is equal to ¾. Now if I flip the question and say, “When I divide a by b, I get 2.75. Which of the following cannot be the remainder when a is divided by b: 2, 3, 6, 12, 15?”

We saw above that 2.75 = 2 (75/100) = 2 ¾ in the lowest form. So the remainder will be 3 if the divisor is 4. The remainder will be 6 if the divisor is 8. The remainder will be 15 if the divisor is 20. On the same lines, the remainder will be 12 if the divisor is 16 because 12/16 = ¾. Can the remainder be 2?

3/4 = 2/?

We cannot have an integer value in place of the ‘?’. Hence we will never get 2 as the remainder. The remainder will always be a positive multiple of 3.

Let’s go back to the original question now: If a and b are positive integers such that a/b = 97.16, which of the following cannot be the remainder when a is divided by b?

a/b = 97.16 = 97 (16/100) in mixed fraction = 97 (4/25) in the lowest form.

The remainder must be a positive multiple of 4. 22 is not a multiple of 4 hence you can never have 22 as the remainder

4/25 = 22/?

You cannot have an integer in place of the ‘?’

Hence answer is 22.