Ways to Find Remainders | Quantitative for GMAT PDF Download

We've explored various scenarios by framing two distinct questions and delved into their solutions using diverse methods to highlight key insights.

Case 1

The Context

When determining the remainder of a general expression divided by a specific number, most people typically use traditional methods involving properties of remainders to arrive at the solution. While this approach is correct, there are instances where a quicker and simpler method can be employed to obtain the answer.
Let's consider an initial example to illustrate this concept:

Example: P is a two-digit number with exactly two factors. What is the remainder when (P² – 1) is divided by 6?
(a) 0
(b) 1
(c) 2
(d) 4
(e) 5
Ans: (a)

Understanding the Question

Let us read the question stem first and draw all the pertinent inferences from the given information.
In this question, we are given that P is a two-digit number, with exactly two factors.

• Now, if P has exactly two factors, can we specify anything about P?
• Yes, we can. We know that a number with exactly two factors is nothing but a prime number.
  • Hence, we can say that P is a two-digit prime number.
• Can we make any other inferences?
  • Yes, we can also say that P is an odd number, as any prime number greater than 2 is always an odd number.

Now, we need to find out the remainder, when P² – 1 is divided by 6.
We can solve this question in more than one way – let’s discuss them one-by-one.

Method 1

In this approach, we will use the general property of any prime number greater than 3.
We know that if we have a prime number greater than 3, we can express that number in 6k + 1 or 6k – 1 form, where k is a positive integer.
As P is a two-digit prime number, we can also express P in the form 6k + 1 or 6k – 1.
Case-1: P = 6k +1,
P² = (6k + 1)² = 36k² + 12k + 1
Or, P² – 1 = 36k² + 12k
Or, P² – 1 = 6 (6k² + 2k)
We can see that P² – 1 is a multiple of 6. Hence, if we divide P² – 1 by 6, the remainder will be 0.
Case-2: P = 6k – 1
P² = (6k – 1)² = 36k² – 12k + 1
Or, P² – 1 = 36k² – 12k
Or, P² – 1 = 6 (6k² – 2k)
We can see that P² – 1 is a multiple of 6. Hence, if we divide P² – 1 by 6, the remainder will be 0.
Therefore, we can see that whether P is 6k + 1 or 6k – 1, the remainder when P² – 1 is divided by 6 is always 0.
Hence, the correct answer is option A.
Now, let’s solve the same question using a different approach.

Method 2: (Intuitive way to solve using GMAT remainders concept)

This method is much simpler and more intuitive. Here we don’t need to know any specific property of prime numbers to solve the question.
If we look at the option choices, we are given five definite numbers and no option says the answer cannot be determined.
Hence, we can definitely say that for any value of P, our answer will be always the same, right?
So, let’s try to find the remainder, using some values of P.

We know that P is a two-digit prime number. Let’s assume P = 11.
So, P² – 1 = 11² – 1 = 121 – 1 = 120
Now, 120 is divisible by 6, hence, if P² – 1 is divided by 6, the remainder is 0.
Let’s try the same with a different value of P.
Assume that P = 19.
So, P² – 1 = 19² – 1 = 361 – 1 = 360
Again, 360 is divisible by 6, hence, if P² – 1 is divided by 6, the remainder is 0.
As we can see, for any value of P, the remainder when P² – 1 is divided by 6 is always 0. Hence, the correct answer choice is option A.

Case 2

The context

There are some specific cases of remainders, where we can apply our conceptual learning from other topics and find the solution in a faster way. Let’s see that in the following example.

Example: What is the remainder when 2²⁰²⁰ is divided by 5?
(a) 0
(b) 1
(c) 2
(d) 3
(e) 4
Ans: (b)

Understanding the Question

This remainder question belongs to one of the most conventional question types which are tested very frequently in the GMAT.
In this question, we are asked to find out the remainder, when the number 2²⁰²⁰ is divided by 5. We will solve it in two ways – let’s start discussing the first method.

Method 1

In our first method, we will use the conventional process of finding out the remainder.
As we are dividing 2²⁰²⁰ by 5, we will first try to find out the value of one such power of 2, which can be expressed in the form 5k + 1 or 5k – 1.

• From observation, we can say that 2⁴ is 16, which is 5 x 3 + 1.
  • Hence, we can say that if we divide 24 by 5, we get a remainder 1.

Now, we can express the given numerator in the form of 24.

• 2²⁰²⁰ can be written as (2⁴)⁵⁰⁵.
• So, the remainder when 2²⁰²⁰ is divided by 5 will be same as the remainder when (2⁴)⁵⁰⁵ is divided by 5.
  • Or, in other words, the remainder will be the same when (1)505 is divided by 5.
• As (1)505 is equal to 1, we can say the final remainder is 1.

Hence, the correct answer choice is option B.

Important Observation

Some of us may have questioned that is it really necessary to express 2²⁰²⁰ in terms of 2⁴ only? What happens if we express 2²⁰²⁰ in terms of any other powers of 2?
Well, we can express the numerator in terms of other powers of 2 also. In such cases, the calculation may differ and becomes lengthier, but the end result will always be the same. Let’ see how we can get that.
Assume that we have expressed 2²⁰²⁰ in terms of 2³. Hence, we can write 2²⁰²⁰ as (2³)⁶⁷³ x 2.
Now, we know, when 2³ is divided by 5, the remainder is 3.
Hence, we can rewrite the numerator as (3673) x 2.
We also know that 3², when divided by 5, gives a remainder 1.
So, we can write (3⁶⁷³) x 2 as (3²)³³⁶ x 3 x 2 or (1)336 x 3 x 2, which is equal to 6.
Now, if we divide 6 by 5, we get a remainder 1.
Hence, the correct answer choice is option B.
So, we can see that we can get the correct answer even if we express the numerator in other forms, apart from the conventional form of (denominator x k ± 1) form. However, the number of steps and the calculation involved may increase in such cases.

Method 2 (Using GMAT Remainders concept of cyclicity)

Now, this method is specific to the values given in the question and may not be generalized for every remainder problem.
In this question, we are asked to find the remainder, when 2²⁰²⁰ is divided by 5.
First, let us see what remainders we get when we divide different powers of 2 by 5.

• Remainder (2¹/5) = Remainder (2/5) = 2
• Remainder (2²/5) = Remainder (4/5) = 4
• Remainder (2³/5) = Remainder (8/5) = 3
• Remainder (2⁴/5) = Remainder (16/5) = 1
• Remainder (2⁵/5) = Remainder (32/5) = 2
• Remainder (2⁶/5) = Remainder (64/5) = 4
• Remainder (2⁷/5) = Remainder (128/5) = 3
• Remainder (2⁸/5) = Remainder (256/5) = 1
• Remainder (2⁹/5) = Remainder (512/5) = 2
• Remainder (2¹⁰/5) = Remainder (1024/5) = 4 and so on…

Now, we can observe two important things from here.

Observation 1

The remainder can be calculated just by dividing the units digit of the number by 5. [For example, Remainder (2⁸/5) = Remainder (256/5) = Remainder (6/5) = 1]
This is always true when we divide any number either by 5 or by 10. While calculating the remainder, we can only find the units digit of the numerator and divide the units digit by 5 or 10, to get the answer.
For example, 2²⁰²⁰ has a units digit 6. If 6 is divided by 5, the remainder is 1. Hence, the remainder when 2²⁰²⁰ is divided by 5 is 1.

Observation 2

The value of the remainder follows a particular cycle and gets repeated after every 4th power. [For example, the remainder is same for 2¹, 2⁵, 2⁹ and so on…]

• As the remainder gets repeated after every 4th power, we can express the numerator in a cycle of 4.
• 2²⁰²⁰ can be written as (2⁴)⁵⁰⁵. Now, we know that when 2⁴ is divided by 5, the remainder is 1.
  Hence, we can also say that when (2⁴)⁵⁰⁵ is divided by 5, the remainder is also 1.

So, we have seen that we can find the answer from different possible methods by using our concept of remainders. Let’s now consolidate our learning in the form of final takeaways.

Final takeaways

• A general expression applies to any valid number. For instance, prime numbers greater than 3 can be represented as 6k ± 1, where k is a positive integer.
• Substituting appropriate values into a general expression can often lead to a straightforward solution.
• When dividing a number by 5 or 10, checking the unit digit alone helps determine the remainder.
• Dividing the unit digit by 5 gives the remainder for the divisor 5, while the unit digit itself is the remainder for the divisor 10.
• While dividing a number of the form an by b, expressing the numerator in the form (denominator x k ± 1) typically provides the correct answer. However, alternative forms may work but might involve more extensive calculations.

Solved Examples

Example 1: For which of the following values of a and b will the integer (3)^(5a-3b) have a remainder of 2 when divided by 5?
(a) a = 6, b = 2
(d) a = 9, b = 3
(c) a = 12, b = 3
(d) a = 10, b = 2
(e) a = 2, b = 2
Ans: (c)
Explanation: Remember that powers of 3 end in 3, 9, 7 or 1. A number will end with a 3 if the power is 1, 5, 9, etc; in a 9 if the power is 2, 6, 10, etc; in a 7 for powers 3, 7, 11, etc and in a 1 for powers 4, 8, 12, etc.
For any digit to have a remainder of 2 when divided by 5, it must end in either a 2 or a 7. Since no integer power of 3 ends in a 2, we are looking for a power that satisfies 3 + 4K for any integer K.
In other words, 5a-3b-3 must be a multiple of 4. In option A, 5a-3b-3 yields 30-6-3=21, not a multiple of 4; Option B yields 33, also not a multiple of 4; Option C yields 48 and is therefore the correct answer.

Example 2: When the positive integer n is divided by 6, the remainder is 3. What is the remainder when n is divided by 8?
Statement 1: n is divisible by 11.
Statement 2: n < 100.
(a) Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
(b) Statement (2) ALONE is sufficient, but statement (1) is not sufficient.
(c) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(d) EACH statement ALONE is sufficient.
(d) Statements (1) and (2) TOGETHER are NOT sufficient.
Ans: (e)
Explanation: Get a feel for any question like this by first picking a couple of low values for n (disregarding the statements) to see whether a logical solution or pattern exists. For this one, try n=9 and n = 27. Both numbers have remainder 3 when divided by 6, but the first one has remainder 1 when divided by 8, while the latter has remainder 3. When you can’t see a pattern immediately, make a table with the first few possible values of n (especially useful when the question has a limiting parameter such as “n < 100”)
By now you see a pattern: the sequence “3 1 7 5” repeats. Moreover, you can see that 33 is the only value of n in the table divisible by 11. Because the values of n increase in increments of 6 and 11 is a prime number, the next value of n that divisible by 11 will be 33 + (6*11) = 99.
But the remainder of n = 99 could be the same as n = 33, in which case the answer would be C (or A, if it remained the same for all succeeding multiples). It is easier to find the remainder of 99/8 by doing the actual calculation, but if that were too time-consuming (e.g. if 7000<n<7100), you could notice that when n=33 the remainder is 1, and the pattern repeats itself every four values of n.
Therefore, the remainder of the 12th value of n after 33 will also have remainder 1, and the 11th (the one we want) will have remainder 3. Since both n = 33 and n = 99 fit both conditions and have different remainders, the answer is E.