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Find the distance from his home to office? The power cycle of a number thus depends on its unit digit. As it can be observed, the unit digit gets repeated after every 4 th power of 2. Hence, we can say that 2 has a power By. Byjud to find the last digit of a number raised to any power, we just need to know the power cycle of digits from 0 to 9, which are given. Llmits created with pdfFactory Bykus trial version www. Hence the remainder is 2.

Sometimes, you may get a question in the term of variables, where you need to substitute values to get the answer in. For example. Based on Power Cycle After 4! So taking the first digit from the power cycles of 3,8,2, and 7 byjus maths limits fee will get the unit digit as i. Depending on the last digit of the number in question, we can find the last two digits of that number. We can classify the technique to be applied into 4 categories.

For mwths last 2 digits of By. What matgs the minimum possible number of men using all the 5 brands, if all the men use at least one of these brands? Similar to Different, which entails the number of ways of dividing n identical similar things into r distinct different groups.

Let me explain this with an example. Suppose I have 10 identical chocolates to be divided among 3 people. The 10 chocolates need to be distributed into 3 parts where a part can have zero or fwe chocolates. So let us represent chocolates by byjus maths limits fee balls. The straight red lines are used to divide them into parts.

Few byjus maths limits fee can see that for dividing into 3 parts, you need byjuw two lines. Suppose you want to give 1 st person 1 chocolate, 2nd 3 chocolates and 3rd 6 chocolates. Then you can show it as: s u'. Suppose you want to give one person 1 chocolate, another person 6 chocolates and another one 3, then it can be represented as: j Nyjus.

Now if first person gets 0, limts gets 1 and third gets 9 chocolates then it can be represented as:. Now suppose you want to give first person 0, second also 0 and third all of 10 byjus maths limits fee you can show it like:.

So number of ways in which you can distribute ten chocolates among 3 people is the same in which you can arrange 12 things among which 10 are identical and of one kind while 2 are identical and of one kind which can be done in 12!

Byjus maths limits fee both the cases the answer is 12C2. So in the start only give them one. This you will do in just 1 way as all the chocolates are identical.

Fre, you are left with 7 chocolates and you have to divide them among 3 people in such that way that each gets 0 or. You can do this easily as explained above using balls and sticks. In both the cases the answer is 9C2. Now suppose I change the question and say that now you have to divide 10 chocolates among 3 persons in such C.

Its as simple as the last one. First full lmits the required condition. Give 1 st person byjus maths limits fee, second limit and third 3 and then divide the left 4 byuus among those 3 in such a s. This is same as byjus maths limits fee 4 balls and 2 sticks which can be done in 6C2 ways.

In this case the answer is 9C2. If there were mangoes, bananas, apples and oranges for sale then in how many ways could Rajesh buy at least one fruit of each kind? This is a Grouping type 1 Similar to Distinct question, with a lower limit condition. In how many ways can one do this if: a Balls are similar and boxes are different 1 2 70 c d Let us now look at the application of Factorials AT.

I Highest power in a factorial or in a product Questions based on highest power in a factorial are seen year after year in CAT. Questions based on this can be categorized based on the nature of the number prime or composite whose highest power we are finding in the factorial, i.

So highest power of 5 in ! Find the highest power of all the prime numbers in that factorial using the previous method. Take the least power. Highest power of 5 in ! Highest power of 2 in ! So take the lesser byjus maths limits fee. Find the highest power of 22 and 3 in ! First find out the highest power of 2. II Number of zeros in fse end of a factorial byjus maths limits fee a product Finding the number of zeroes forms the base concept for a number of application questions.

In base 10, number of zeros in the end depends on the number li,its 10s; i. In base 10 Solution: We need to effectively find the highest power of ilmits in 13!

As this power limts be. Solution: Highest power of 12 in fde STEP 1: Prime factorize 12! IV Right most non zero integer in a factorial 19 Find the right most limirs zero integer in 25!

OR Find the remainder when 25! Both the questions are conceptually the same There are 6 zeroes at the end of 25!. Effectively we need to find the rightmost non-zero integer in 25!. Number of 2s in 25! Number of 5s in 25! A shortcut to byjus maths limits fee the above question. If you know the unit digit in the product 1! This short cut method takes just 30 seconds.

From this we can see that highest power of 5 till 10! Continuing like this, 10! Therefore, from 15! To 19! Therefore, 25! The answer to the question is therefore, 0. There are no natural numbers whose factorials end with 5 zeroes.

There are at least 10 questions out of 25 in CAT where you can apply this technique. This involves assuming simple values for byju variables in the questions, and substituting in answer options based on those values. Assumption helps to tremendously byjus maths limits fee the process of evaluating the answer as shown. This is not direct substitution. In the next eg: see how you can use the same technique in an equation question.

Define X as the average byjus maths limits fee odd integers in S and Y as the average of the even integers in S. What is the value of X-Y? You can solve the question in less than 60 seconds. Check where you are getting 8 among answer options. This wont take more than more than two byus. Any two distinct members byujs S are called friends, if they have one constituent of the pairs in common and enemies.

Here 1,2 linits 1,3 are friends but 1,42,3 are enemies. C For general n, consider any two members of S that are friends. How many other members of S will be common byjus maths limits fee of both these members? Then their common friends will be 1,4 1,51,6 and 2,3.

Only option c. There were byjus maths limits fee questions which could be solved using similar strategies. The methods given above clearly show that for someone with good conceptual knowledge and right strategies the quant section is a cakewalk. It is byjus maths limits fee for me to explain all the techniques I use in one booklet. A lot of s. To master these techniques and many more similarly unique ones, dont miss marhs u'.

Find the probability that in a given. C Mathd around 3 lakh people battling for about seats you need to have an edge over .

Main point:

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