Problems on Numbers is one of the most common topics in which questions are often asked in SAT exams. In competitive exams, we will have problems in this topic which would be based on natural numbers, real numbers, whole numbers, and integers.

Students who are preparing to improve their aptitude skills and those who are preparing for this type of competitive test must prepare this topic in order to have better score. Because, today there is no competitive exam without the questions from the topic problems on numbers. Whether a person is going to write placement exam to get placed or a students is going to write a competitive exam in order to get admission in university, they must be prepared to solve problems with numbers. This is the reason for why people must study this topic.

As we mentioned in the above paragraph, a person who wants to get placed in a company and a students who wants to get admission in university for higher studies must write competitive exams like placement test and entrance exam. To meet the above requirement, it is very important to score more marks in the above mentioned competitive exams. To score more marks, they have to prepare this topic. Preparing this topic would definitely improve their marks in the above exams. Preparing this topic is not difficult task. We are just going to remember the stuff that we have already learned in our lower classes

Students have to learn few basic operations in this topic problems on numbers and some additional tricks. Already we are much clear with the four basic operations which we often use in math. They are addition, subtraction, multiplication and division. Even though we are much clear with these four basic operations, we have to be knowing some more stuff to do the problems which are being asked from this topic in competitive exams. The stuff which I have mentioned above is nothing but the tricks and shortcuts which need to solve the problems in a very short time.

Short cut is nothing but the easiest way to solve the problems on numbers in a very short time. In competitive exams, we will have very limited time to solve each problem. Then only we will be able to attend all the questions. If we do problems in competitive exams in perfect manner with all the steps, it will definitely take much time and we may not able to attend the other questions. So we need some other way in which the problems can be solved in a very short time. The way we need to solve the problem quickly is called as shortcut.

Let us consider the problem, “Find the numbers of numbers which are divisible by 5 in the first 1000 natural numbers”. To answer this questions, if we check each number for divisibility of 5, it very difficult to check for the first 1000 natural numbers. Some people would suggest the short cut “ Arithmetic Progression” to find the answer like 5,10,15,20…………1000. But there is a another shortcut through which we can easily get the answer in seconds.

Short cut==> Number of numbers divisible by = 1000/5 = 200.

Hence the number of numbers which are divisible by 5 in first thousand natural numbers is 200.

Here, we are going to have some problems with numbers . You can check your answer online and see step by step solution.

1. If a number when divided by 296 gives a remainder 75, find the remainder when 37 divides the same number.

Let the number be ‘x’

Then x = 296k + 75, where ‘k’ is quotient when ‘x’ is divided by ‘296’

In the above sentence we have 296 is multiplied by the constant "k", 75 is added to that. In this form , we consider the number 75 as remainder when the number x is divided by 296.

We want to find the remainder when we divide the number "x" by 37. To do this, we need to have 37 at the place where we have 296 in the above equation.

So we can write 296 as 37 times 8 and 75 as 37 times 2 plus 1. It has shown below.

x = 37 × 8k + 37 × 2 + 1

x = 37(8k + 2) + 1

Hence, the remainder is ‘1’ when the number ‘x’ is divided by 37.

Then x = 296k + 75, where ‘k’ is quotient when ‘x’ is divided by ‘296’

In the above sentence we have 296 is multiplied by the constant "k", 75 is added to that. In this form , we consider the number 75 as remainder when the number x is divided by 296.

We want to find the remainder when we divide the number "x" by 37. To do this, we need to have 37 at the place where we have 296 in the above equation.

So we can write 296 as 37 times 8 and 75 as 37 times 2 plus 1. It has shown below.

x = 37 × 8k + 37 × 2 + 1

x = 37(8k + 2) + 1

Hence, the remainder is ‘1’ when the number ‘x’ is divided by 37.

2. Find the number of prime factors of

From 6^{10} × 7^{17} × 55^{27}, we have to write each base in terms of multiplication of its prime factors.

That is,=(2X3)^{10}×(7)^{17}×(5X11)^{27}

=2^{10}X3^{10}X7^{17}X5^{27}X11^{27}

The no. of prime factors = sum of the exponents

= 10+10+17+27+27

= 91

Hence, the number of prime factors is 91.

That is,=(2X3)

=2

The no. of prime factors = sum of the exponents

= 10+10+17+27+27

= 91

Hence, the number of prime factors is 91.

3. Find the unit digit of

Step-1: Take the last two digits in the power and unit digit in the base. They are 43 and 2

Step-2: Divide the last two digits of the power by 4 (in all the problems) and get the remainder. That is, when 43 is divided by4, the remainder is 3.

Step-3: Now this remainder 3 has to be taken as exponent of the unit digit 2

That is, 2^{3} = 8

Hence the unit digit of the given number is 8

Step-2: Divide the last two digits of the power by 4 (in all the problems) and get the remainder. That is, when 43 is divided by4, the remainder is 3.

Step-3: Now this remainder 3 has to be taken as exponent of the unit digit 2

That is, 2

Hence the unit digit of the given number is 8

4. Find the square root of 123454321

In the given number, we have the first five natural numbers in ascending order up to 5.

After 5, we have the first four natural numbers in descending order.

Whenever we have a number like this and we want to find square root, we have to replace each digit by 1, up to the digit where we have the first n natural natural numbers in ascending order.

So, in our number 123454321, we have to replace each digit by 1 up to 5. That is the square root of 123454321.

Hence the square root of 123454321 is 11111.

After 5, we have the first four natural numbers in descending order.

Whenever we have a number like this and we want to find square root, we have to replace each digit by 1, up to the digit where we have the first n natural natural numbers in ascending order.

So, in our number 123454321, we have to replace each digit by 1 up to 5. That is the square root of 123454321.

Hence the square root of 123454321 is 11111.

5. Find the least number which when divided by 35, leaves a remainder 25, when divided by 45, leaves a remainder 35 and when divided by 55, leaves a remainder 45.

For each divisor and corresponding remainder, we have to find the difference.

35-25=10

45-35=10

55-45=10

we get the difference 10 (for all divisors and corresponding remainders)

Now we have to find the L.C.M of (35,45,55) and subtract the difference from the L.C.M.

L.C.M of (35,45,55) = 3465

Hence the required least number = 3465-10 = 3455

35-25=10

45-35=10

55-45=10

we get the difference 10 (for all divisors and corresponding remainders)

Now we have to find the L.C.M of (35,45,55) and subtract the difference from the L.C.M.

L.C.M of (35,45,55) = 3465

Hence the required least number = 3465-10 = 3455

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