Arithmetic

You have learned some basics of arithmetic at study plan for beginners, now you can move on to in-depth analysis of numbers and their properties. Remember that arithmetic is the key to success, as I said in study plan for beginners as well. So never skip this topic in any way even if you don’t like math at all.

Number Properties:

Odds and Evens:
Refer to the following table:

From the table above, it’s clear that in addition or subtraction, when both are same whether even or odd, the result is even; Otherwise it’s odd. In multiplication, when at least one even exists, the result would always be even, otherwise it’s odd. And in division, as the result may be an integer or may be a fraction, so we can’t determine what’s the result; in short Not Applicable (N/A).

Integers and Fractions:
Refer to the following table:

From the table above, it’s clear that in addition or subtraction, when both are different, the result must be a fraction; if both are integers, the result must be an integer, and if both are fractions, the result may be an integer or a fraction {For instance, when 2.6 and 0.4 are added, the result is an integer; and when 2.2 and 0.4 are added, the result would be a fraction}.

In multiplication, only when both are integers the result must be an integers, otherwise the result may be an integer or a fraction {For instance, when 2 is multiplied by 2.5, the result would be 5, an integer; but when 2 is multiplied by 2.2, the result would be 4.4, a fraction; Similarly, when both are fractions i.e when (⁵⁄₂ =) 2.5 is multiplied by (²⁄₅ =)0.4, the result would be 1, and integer, but when 2.5 is multiplied by itself, the result would be 6.25, a fraction}.

And in division, only when a fraction is divided by an integer, the result must be a fraction, otherwise the result may be an integer or a fraction.

Consecutive Integers:

As you have learned little about consecutive integers at beginners level study. Let’s understand how questions come from this topic and how to solve it.

If the average of 125 consecutive integers is 15. What is the sum of first and the last of these integers?

Solution:
As we have learned in states that in every evenly spaced set, average may also be quickly find as just the average of first and the last number of that set.

So, Average = 15

⇒ (^{first integer + last integer}⁄₂) = 15

⇒ Sum of first and last integers = 30 Answer

Similarly, If the average of 125 terms of some consecutive integers is 540, what is the sum of first four integers?

Solution:
Here we are not asked about the sum of first and last integers, rather here we are asked to find sum of first four integers. If you see questions like this, always do the following:

Suppose the consecutive integers are:
n, (n + 1), (n + 2), (n + 3), …………, (125^th term)

Now first we need to find the last integer of this series in term of n. for that purpose, you need to use the in following way after careful looking the sequence:

First term = n + 0 = n

Second term = n + 1

Third term = n + 2

So,

(125^th) term = n + 124

Now, as the set is evenly spaced, so

⇒ Average = (^{first integer + last integer}⁄₂)

⇒ 540 = {^{n + (n + 124)}⁄₂}

⇒ n + (n + 124) = 1080

⇒ 2n = 956

⇒ n = 478 ———– (eq. 1)

Now we need to find sum of first four integers i.e, (n + n+1 + n+2 + n+3 = 4n + 6)

Just put value of equation 1 here, we’ll get the answer as follows:

Sum of first four integers = 4n + 6 = 4(478) + 6 = 1912 + 6

⇒ Sum of first four integers = 1918 Answer

Let’s take another scenario of consecutive integers type.

How many integers exists between 400 and 950 inclusive?
Remember that the word inclusive has its unique importance. This only word may change the answer of the question. For instance, in this case when inclusive said in question you must always follow the method bellow:

Highest value – Lowest value + 1 = number of integers inclusive

If no word of inclusive is there then instead of adding 1, you need to subtract 1 as follows:

Highest value – Lowest value – 1 = number of integers between highest value and lowest value.

Consecutive Even Integers:
In case of consecutive integers, always use the following series as a starting point in solving:

2n, 2n+2, 2n+4, 2n+6, 2n+8, …….

Where n is any integer.

Let’s take another scenario of consecutive even integers type.

How many even integers exists between 400 and 950 inclusive?
Again you see word inclusive. But in case of finding even integers or odd integers, you must do the following:

Firstly, you must make the series such that it must start from even and end with an odd, or start with an odd and end with and even. In short in must never be even ….. even, or odd …. odd. For instance in question above, it’s a case of even(i.e 400) …… even(i.e 950). So let’s temporarily remove 400 from this to make series like odd … even, as follows:

401 ……. 950.

Now we have 1 even(i.e 400) out of this series that we’ll add later on.

So let’s find number of even integers between 401 and 950 inclusive as follows:

Highest value – Lowest value + 1 = number of integers between 401 and 950 inclusive ———————– (eq. 1)

Now divide it by 2, as there are half even and half odd here. At this stage you realized why we eliminated one even to make highest and lowest values different in terms of even and odd. This is because every series of integer that starts with even and ends with odd or a series that starts with odd and ends with even always has half even and half odd integers. For example if you see 1, 2, 3, 4, 5, 6. This series starts with odd and ends with even. And there are six consecutive integers, then there is always half evens and half odds. Also note that here even and odd are equal. On other hand if a series starts with even and end with even, then here odd and even are not equal, and therefore we cannot easily find number of even integers unless we eliminate one of the even to make even and odd equal.

Now from equation 1,

950 – 401 + 1 = 500
Now as there are half as many evens as odds, so divide this by to we’ll get number of evens between 400 and 950 inclusive as follows:

500/2 = 250

Now as we temporarily removed one even integer (i.e 400) so let’s add this to get the answer.

250 + 1 = 251 Answer

Consecutive Odd Integers:
In case of consecutive integers, always use the following series as a starting point in solving:

2n+1, 2n+3, 2n+5, 2n+7, 2n+9, …….

Now here again similar to as discussed in even integers, It may be asked about how many odd integers exists between 401 and 999 inclusive. Here again as I said first temporarily remove one odd. After that do the same thing as discussed earlier i.e (highest – lowest + 1)/2 will give number of odd integers. Finally add the previously temporarily removed odd to find total odd integers inclusive.

Sequences and Series:
Some equations you must know in order to solve any question related to this topic.

a_n = a₁ + (n – 1) d

Where a_n is the nth term of the sequence, a₁ is the first term and n is the number of terms and d is common difference between any two consecutive terms. Also note that if a sequence doesn’t has the common difference, then the equation cannot be used.

In order to find the sum of all the term of the sequence, you must use equation of average as follows:

Avg = Sum ÷ count   or   = Total ÷ Numbers

And also in evenly spaced set or having a common difference, avg. of the whole series is calculated by dividing (first term + last term) by 2. Here’s how you may find avg. and number of terms you would know. There’s how you may get Sum by using equation above for average.

Prime and Divisibility:
As you learn in beginners study plan about prime numbers. These are those positive integers that has only two different factors.

For instance,

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 ….

Remember that prime numbers are the core in arithmetic. In other words these are the backbone of arithmetic. Every integer may be broken down into its prime numbers which is called its prime factors that we also learned in basic plan. Prime factors are also helpful in determining the number of factors and also L.C.M and G.C.F. We’ll discuss how to find factors from prime factorization before learning about prime factorization.

Prime Factorization:
Every positive integer can be broken down into its prime factors. This process of breaking down is called prime factorization. For instance,

So 12 = 2² × 3

This is called prime factorization.

Factors and Multiples:
As in beginners, I said you that we’ll discuss factors and multiples in detail in advance. So let’s discuss.

As I mentioned before in beginners an equation that you must remember which is:

Multiple ÷ Factor = Integer

Now, tell me that 0 is multiple of every integer, is this must be true, could be true or never be true?

Answer can be find while using the equation above. If you put 0 in place of multiple, and put every integer except 0 in place of factor. You’ll always get the result as an integer (i.e 0). So it’s must be true.

Many people may raise question that if 0 divided by 0 is not an integer, then how 0 is a multiple of 0. The answer is simple we know 0 divided by 0 is undefined, but 0 multiplied by 0 is 0. There’s multiple resulted.

Now, tell me 1 is a factor of every integer, is this must be true, could be true or cannot be true?

Obviously your answer again would be must be true if you plug in 1 to place of factor and place any integer in place of multiple.

Now let’s learn how to find number of factors from Prime Factorization. Also number of even factors and odd factors.

As we know Prime factorization of 12 is:

So 12 = 2² × 3

Now, we need to find number of total factors of 12, even factors of 12 and odd factors of 12.

Remember that from prime factorization, we cannot answer what are factors of 12 or what even factors 12 have, or what odd factors 12 have. We can just answer how many factors does 12 have, or how many even or odd factors does 12 have.

As,

So 12 = 2² × 3

Now, what you need to do is to add ‘1’ to the exponents of each distinct prime factor of 12 i.e:

So 12 = 2^{(2 + 1)} × 3^{(1 + 1)}

Now just ignore the base and take the values of exponents into base of the distinct prime factors of 12 and then multiply the exponents i.e:

(2 + 1) × (1 + 1)

= 3 × 2 = 6

So there are total 6 factors of 12.

Now, how many even and how many odd?

Always start from finding odd factors first, as shown bellow:

As in prime factorization, only odd is 3, so we’ll take it’s exponent and add 1 to it and then write down to base to know how many odd factors, i.e

Odd prime factor of 12 = 3^{(1 + 1)}

So odd factors of 12 = 1 + 1 = 2

Now to find even factor, don’t take even factor and do same as we did above. You’ll get wrong answer in that case. Just subtract the odd factors from total number of factors as follows:

Even factors of 12 = 6 – 2 = 4

Now, let’s learn how to answer what are factors of 12?

Just make T shaped figure and then start from 1 on the left side and multiply it with an integer that would give 12 in an answer as follows:

So factors of 12 are, 1, 2, 3, 4, 6, and 12 it self.

We see that 12 is the factor of itself. Similarly 12 is a multiple of itself. Every integer is a factor of itself and also every integer is a multiple of itself. It’s because 1 is a factor of every integer. i.e If we multiply 12 with 1, it gives 12. So the resulted 12 is multiple of both 12 and 1. Also 12 and 1 are factor of the resulted 12. This thing you have already learned in beginners. So let’s move on…

Important Note: It is important to note that if you need to get prime factorization of big integer (let’s say 16,000), you must first break down this into two other easy integers such that those also easily break down into further ones. If you are a good player of Snooker (my favorite game) you would be an expert in the breaking down big integers, because in snooker we always select that ball to put into the pocket such that after play shot, we could easily place other predefined ball whose number come next in order to get highest points. So for instance, 16,000 must break down into 16 and 1000 that further be easily broken down easily.

Divisibility:
Have you ever study math before your graduation or intermediate level. If yes you must know a forgotten equation with diagram as bellow:

And from here, you must remember the equation:

Dividend = Divisor × Quotient + Remainder

This equation is used to solve questions of the following type:

Where s and t are positive integers, what will be the remainder if s is divided by t?

Solution:

First you must split 2.75 as follows:

s/t = 2 + 0.75

⇒ s = 2t + 0.75t

Here, compare this with: Dividend = Divisor × Quotient + Remainder

So you’ll have 0.75t = Remainder

Now, you’ll be given some answer choices showing some integers, put those integers one by one. And see where t results to an integer. If putting an answer choice (let’s say choice D) in place of remainder in above equation of 0.75t = Remainder, results to an integer value of t, that choice would be right anwer. Otherwise you must plugin other choices one by one to see where value of t become an integer.

Now, let’s forward to other type.

How many positive integers less than 1000 inclusive that are divisible by 3, 5 and 7?

This question is most important one.

First tell me how many integers less than 10 inclusive that are divisible by 4. You quickly reply 2. Well how you get is long way. Just divide 10 by 4, you’ll get the required answer perhaps in fractional form as in this case 2.5. Just consider value before decimal point. i.e 2 in this case.

Now, 1000 divided by 3 gives integer value of 333. So there must be 333 integers that are divisible by 3. But question is asked integers that are divisible by 3, 5 as well as 7. For that purpose we need to take L.C.M of 3, 5 and 7. Taking L.C.M in short way will be explained later on. Here we just need to solve the question. So if you take L.C.M of 3, 5 and 7 by short cut explained later in this lecture, you’ll get 105. So just divide 1000 by 105, you will get the answer as bellow:

1000/105 ≈ 9     (As I said just take value before decimal point)

So there are 9 positive integers less than 1000 inclusive that are divisible by 3, 5 and 7.

Now how many integers

Now let’s move on to further scenarios of arithmetic in divisibility topic of arithmetic.

If m is divided by 5, the remainder is 2. If n is divided by 5 the remainder is 3. What would be the remainder if m + n is divided by 5?

As dividend is simply add up, so as the remainders .i.e 2 + 3 = 5. But according to the rules given bellow remainder would be 0. Now, check the following rules:

Remember there are some rules of divisibility as follows:

1. Remainder is always less than divisor.

2. Remainder cannot be negative, if it’s negative then add the divisor (or a multiple of divisor nearest to that negative integer) to that negative remainder unless it becomes positive, but remain less the divisor.

3. If dividends add up and divide with same divisor, then the individual remainders of those divisors would also add up, and then follow the above two rules to get remainder. (As shown in above examples of two dividends of m and n that divides by 5.)

4. If dividends subtracted and divided with same divisor then the individual remainders would also be subtracted. And then follow the first two rules. For instance, if (m – n) is divided by 5 in addition to above question, then remainder would be 2 – 3 = -1. As remainder cannot be negative so in that case according to the rule, we must add the divisor to this remainder till it become positive or 0. So -1 + 5 = 4, so 4 is the remainder when m – n is divided by 5. Remember that if remainder is let’s say -233 at some case while dividing by 5. You just add a multiple of divisor (i.e 5) that is nearest but greater than -233 to make remainder positive, rather than adding 5 to -233 again and again to get positive remainder. So remainder would be -233 + 235 = 2.

Now, let’s solve an actual GRE question, which might also be tested in other exams.

If m is divided by 5, the remainder is 1, while if it is divided by 6 the remainder is 0. What will be the remainder when m is divided by 30?

First try this by your own, then go ahead for answer with explanation.

It’s a very good question that will further clarify your concepts. Let’s solve it.

From the information, you can quickly decide that m must be a multiple of 6. Right? Great!

Now, another information that we can extract from the question is that m must be from the series of the numbers below:

6, 11, 16, 21, 26, 31, 36, 41, 46, …..

Because the above series when divided by 5 gives remainder 1.

Now, if we combine both of the extracted information, we will conclude that m must be any number from the series below:

6, 36, 66, …

Now, if you divide any of the number by 30, you’ll get remainder 6. 🙂

Therefore, Remainder = 6 Answer

This question might be changed with addition of (let’s say) if m > 100, and remaining similar question. The answer will remain 6, because you know why. If not ask me on WhatsApp.

LCM and GCF:
Remember that never use long method of taking L.C.M. Always use the following method of taking L.C.M/G.C.F.

For instance we need to take L.C.M of 8 and 12.

First take common from 8, and 12 as maximum integer value as you can. We can take 4 maximum integer common from 8 and 12 as follows:

4 (8/4, 12/4)

Now we’ll left

4 (2, 3)

Remember that this highest values that we take common is G.C.F / H.C.F named as Greatest Common Factor / Highest Common Factor. So here 4 is G.C.F.

Now, we cannot take common anymore from 2 and 3. At this point just multiply all the three integers as follows:

4 × 2 × 3 = 24, which is the L.C.M of 8 and 12.

Now, in case there are couple of integers for whom we need to take L.C.M of all, just take L.C.M of any two integers, then use that L.C.M with another integer to take L.C.M and so on.

Arithmetic Expressions:
If it’s expressed in question that a, b and c are 4 distinct digits you have to take it from the 10 digits (i.e 0-9 inclusive) such that no digit repeats (as the word distinct suggests). But if there’s no word like distinct, unique or different is there, you may repeat digits. Each words of question is very important to understand and solve the question accurately. Failure to do so results in silly mistakes.

Now, let’s learn a more scenarios:
This question came in GRE exam.

ab is a two digit integer. The sum of it’s digits is 12. If its digits are reversed, the resulted integer is 18 less than the original integer. What is the original integer?

A) 93
B) 84
C) 75
D) 66

Well you don’t need to think lot about which integers must I assume? Just go through answer choices and get the answer. i.e

Choice A: Is 93 – 39 = 18? No, so move on

Choice B: Is 84 – 48 = 18? No, so move on

Choice C) Is 75 – 57 = 18? Yes, that’s right answer.

Similarly, this question has once came in GAT,

What is the greatest difference between four distinct digit positive integer abcd and another four distinct digit positive integer efgh?

Well, first the greatest distinct four distinct digit positive integer is 9876. Now the least four distinct digit positive integer must be subtracted from this. The tough point came here now. Most of the students say the least four distinct digits positive integer is 1234, which is wrong. The least four distinct digit positive integer is 1023. That’s why you must be very careful while making plugin values. 0 is also a digit, that is not considered a four digit integer if it’s placed at first position from left, which we say thousands digit. But 0 may be placed at hundreds digit, tens digit or unit digit. So if it’s placed at hundreds digit here it’ll give the least result.

So 9876 – 1023 = 8853 is the greatest difference between a four distinct digit positive integer abcd (i.e 9876) and another four distinct digit positive integer efgh (i.e 1023).

Similarly, it may be ask if efgh is the least four distinct digits positive integer, what is the sum of its digits?

You know the answer is 6.

Unit Digit:

Let’s learn how to find out the unit digit of an exponential expression, e.g 2²⁴, 13³² and 19²⁹ etc.
As we know that the unit digit of any number is right most digit before fractional point. For instance unit digit of 23089 is 9, and 13082.95 is 2.

Let’s find out the unit digit of 5¹⁵⁰.
If we take 5 raised to power 1, it gives unit digit = 5
If we take 5 raised to power 2, it gives unit digit = 5 again               (As it results 25)
If we take 5 raised to power 3, it gives unit digit = 5 again               (As it results 125)

You see that 5 raised to power any integer gives a unit digit 5. Therefore 5¹⁵⁰ has unit digit 5.

Remember that there are four digits that is 0, 1, 5 and 6 whose raised to power any integer always give same unit digit.

Also remember that the unit digit of any exponential number is always equal to the unit digit of exponential unit digit. This point is hard to grasp I know. Let’s take an example to understand its meaning.

The above statement means that, for instance, the unit digit of 4927835⁴⁵ is same as the unit digit of 5⁴⁵. That means the unit digit of 4927835⁴⁵ is the same is the unit digit of exponential unit digit (i.e 5⁴⁵ that we got from the bold faced thing as here: 4927835⁴⁵).

I know few of you still unable to understand what’s going on 😀
Let’s take another easy example. The unit digit of 213¹⁹ is same as the unit digit of 3¹⁹, because 3 is the unit digit of 213.

So we can say that the unit digit of 2486³⁴, which is same as unit digit of 6³⁴, must give answer 6. As we know the unit digit of 6 raised to power any integer is always remains 6. Similarly unit digit of 3450²³ must have unit digit 0, as 0 raised to power any integer remains 0.

Now let’s discuss treatment with other digits except 0, 1, 5 and 6.

Let’s find out the unit digit of 213¹⁹, which must be the same as the unit digit of 3¹⁹.

3 raised to power 1 gives unit digit = 3
3 raised to power 2 gives unit digit = 9               (As it results 9)
3 raised to power 3 gives unit digit = 7               (As it results 27)
3 raised to power 4 gives unit digit = 1               (As by multiplying unit digit of 27 (i.e 7) with 3 gives unit digit 1)

After you get 1 unit digit, then sequence repeats again and again. i.e

3 raised to power 5 would give unit digit = 3         {As by multiplying previous unit digit (i.e 1) with 3, gives unit digit 3}.
3 raised to power 6 would give unit digit = 9

And so on…

so we got sequence as:

3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, …….

After every four terms, the sequence repeats. So let’s split the sequence

(3, 9, 7, 1), (3, 9, 7, 1), (3, 9, 7, 1), …….

At this stage always see the highest exponent where the sequence repeats. For instance in above case, the sequence repeats after every 4th exponent. So you must take multiple of 4 as reference.

Now you see from sequence that 3 raised to power any multiple of 4 (i.e 4, 8, 12, 16, 20 ….) gives unit digit 1. We need to find the unit digit of 3 raised to power 19, which is the previous term of 20th term in the sequence above. So we must conclude that the 3 raised to power 19 must give unit digit 7. Because 7 is in the previous of 1, as 19 is in previous of 20th term. There’s how you need to solve it.

Few of you still unable to understand this. Let’s take another example and scenario.

Let’s find out the unit digit of 29⁶⁷

9 raised to power 1 gives unit digit = 9
9 raised to power 2 gives unit digit = 1

As we have discussed, whenever 1 come, the sequence repeats. Because when 81 is multiplied with 9 again to get unit digit of 9 raised to power 3, it would again give unit digit 9. And then 9 raised to power 4 would again give unit digit 1, and so on…

So we got sequence here as bellow:

9, 1, 9, 1, 9, 1, ……..

Let’s split the sequence:

(9, 1), (9, 1), (9, 1), ……..

We have seen that after 2nd exponent, the sequence repeats, So we’ll take 2 as reference.

Therefore, 9 raised to power any multiple of 2 would give unit digit 1, other wise the unit digit is 9. But here we can also shortly say 9 raised to power even number gives unit digit 1, while 9 raised to power odd number gives unit digit 1. So we need to find unit digit of 9 raised to power 67, which is an odd. So the required unit digit is 9, as 67 is previous term of 68, as 9 is in previous term of 1 in the sequence. So answer is 9.

Approximation of numbers:

In approximation of numbers, always remember the rule bellow:

Suppose a, b, c and d are digits.

ab.cd = ab.(c+1), if d is 5 or greater.

But

ab.cd = ab.c, if d is 4 or less.

If x = 2.0453 and x* is the decimal obtained by rounding x to the nearest hundredth, what is the value of x* – x?

(A) –0.0053
(B) –0.0003
(C) 0.0007
(D) 0.0047
(E) 0.0153

Solution: x* = 2.05, as by rounding to hundredth digit we have to limit the number two digits after decimal point. (Remember, if it were asked to round to nearest tenth digit, then we need to limit the digits to one digit after decimal point).

Now x* – x = 2.0500 – 2.0453 = 0.0047 Answer.