In this article certain questions have been randomly taken from different topics of the quantitative section of CAT. Each question has been selected keeping in mind certain challenges you need to manage as a test taker. These challenges may range from concept basics to application orientation and test taking strategies.

7^{1}---- last 2 digits are 01

7^{2}---- last 2 digits are 49

7^{3}---- last 2 digits are 43

7^{4}---- last 2 digits are 01

Now this process has a cyclicity of 4 with the pattern 01, 49, 43 and 01.

Divide the index (2008) by 4 and the remainder decides the last two digits. Since 2008 is perfectly divisible, we are the 4th set which corresponds to the digits 01. Hence the answer is 01.

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Now remove any two numbers, say 3 and 5 and replace these with 3+5-1=7. Thus the set is 1,2,4,7

Now remove any two numbers, say 1 and 7. The new set is 2,4,7.

Now remove any two numbers, say 4 and 7. The new set is 2,10.

Now remove both and we are left with 11 as the last number.

If you note vigilantly, you can appreciate that the number left (11) is the sum of the first 4 numbers of the set plus 1. Validate the pattern through another set of numbers.

Now if we repeat this with the given set, the number left will be:

(1+2+3+….39)+1 = (39*40)/2 + 1= 781

Thus the general term of the common sequence will be

21+ (n-1) 20 = 20n+1 and this will be less than equal to the lesser of the last term of the two sequences i.e. 417.

Thus, n is less than equal to 416/20.

So, n= 20

The second such number will be 18 as seed(18)= seed(1+8)=seed(9)=9

Thereafter all subsequent numbers will be multiples of 9 only as divisibility of a number by 9 is determined by the sum of the digits of the number.

Hence the required numbers will be:

9,18,27……..495

This is an A.P whose first term is 9 and last is 495 with common difference as 9.

Thus 495= 9+ (n-1)9

Or n= 55

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Such numbers will have a format ABCD

A can be occupied by 1,2 and 3 (leaving 4000 aside). Thus there are 3 ways of occupying A

B, C and D can be occupied by all 0,1,2,3,4. Thus there are 5 ways each of occupying B,C and D

So the total number of integers will be 3*5*5*5= 375

However 4000 is also a possibility. So the total number of integers will be 375+1 =376