For last 1 digit, numbers show a cyclicity of 4 or less. For last 2 digits, the cyclicity is 20 or less. In fact more last-2-digit combos show cyclicities of 10 or 20 steps; very few show a 2 or 4 step repetition! But in the exam, they are more likely to ask such a number simply because the 20-step ones could be a bit too complicated…

Unlikely to come in CAT; if it does, it will be something in which the concepts above can be easily modified (as in, something which repeats soon or in which binomial expansion can be of use)

I am not able to understand why dint you do 68*68, instead you did 68*18. I get that we get the same result for last two digits for both, but is there any pattern like, for all numbers greater than 50 we can keep multiplying, Number*(Number-50) to find last two digits ?

PS – Thanks for the blog even though its an year old, It has been of great use to my preparation for CAT-14 ðŸ™‚

It will work for even numbers (as even * 50 will end in 00). I don’t remember why I didn’t mention it. Or possibly I took it as a given (it is a shortcut I use sometimes, it has become obvious to me but it isn’t that obvious really…sorry)

Hi J It will be helpful if you can explain the idea of taking 18 instead of 68 in a bit detail.

68= (50+18). Now for any even number E, E*50 will give 00, and so the last two digits of E*68 i.e. E*(50+18) will be the same as those of E*18

regards
J

Can binomial expansion be used to find the last two digits of 12^13?
(10+2)^13 = 13C1*10*2^12 + 2^13
Last digit of 2^12 = 6.. so 13*10*6= 80 and last digit of 2^13 is 2
so 80 +2 =82. But the answer is 72 by another method.
Can you tell me where exactly did I make a blunder?

While most of the numbers shows a behavioural pattern having a cyclicity of 4, why 12 shows a cyclicity of 20?

For last 1 digit, numbers show a cyclicity of 4 or less. For last 2 digits, the cyclicity is 20 or less. In fact more last-2-digit combos show cyclicities of 10 or 20 steps; very few show a 2 or 4 step repetition! But in the exam, they are more likely to ask such a number simply because the 20-step ones could be a bit too complicated…

regards

J

Any post on cylicity of last 3 digits?

Unlikely to come in CAT; if it does, it will be something in which the concepts above can be easily modified (as in, something which repeats soon or in which binomial expansion can be of use)

regards

J

I am not able to understand why dint you do 68*68, instead you did 68*18. I get that we get the same result for last two digits for both, but is there any pattern like, for all numbers greater than 50 we can keep multiplying, Number*(Number-50) to find last two digits ?

PS – Thanks for the blog even though its an year old, It has been of great use to my preparation for CAT-14 ðŸ™‚

It will work for even numbers (as

even * 50will end in 00). I don’t remember why I didn’t mention it. Or possibly I took it as a given (it is a shortcut I use sometimes, it has become obvious to me but it isn’t that obvious really…sorry)regards

J

Hi J It will be helpful if you can explain the idea of taking 18 instead of 68 in a bit detail.

68= (50+18). Now for any even number E, E*50 will give 00, and so the last two digits of E*68 i.e. E*(50+18) will be the same as those of E*18

regards

J

Can binomial expansion be used to find the last two digits of 12^13?

(10+2)^13 = 13C1*10*2^12 + 2^13

Last digit of 2^12 = 6.. so 13*10*6= 80 and last digit of 2^13 is 2

so 80 +2 =82. But the answer is 72 by another method.

Can you tell me where exactly did I make a blunder?

You missed the second last digit of 2^13….that will also affect the answer…

regards

J

Ok Got it Thanks a lot..

Right! Thank you! ðŸ™‚

Hello J sir

Could you please guide me whether there is a post for Indices& Surds and Functions anywhere in the blog.

Thank You

Not yet, but there are some posts on Sequences specifically.

regards

J