********************************************************** There are many doubts related to this question's plus 3 more question : 1)21^875/17 : Can't it be done as 1^875 is one so dividing by 17 will give me one is that can't be found out 2)In the above you solved by diving the digit by 17 two times 21/17 first case and 16/17 = -1 as second case as why it was done. Sorry for asking again and again but the problem is i am having lot of issues related to these question and getting confused. Can you please use the same method above and solve these questions: 1)Find the remainder when 54^124 is divisible by 17 a)4 b)5 c)13 d)15 can it be done as 4 to the power 4k(124) is 6 so diving 6 with 17 will give 6 as remainder but there is no such option 2)Find the remainder when 83 ^261 is divided by 17 a)13 b)9 c)8 d)2 Can't it be done by this method: 3 ^261 is equal to 3 ^4k+1 which is 3 so the remaider should be 3 but no such option. I request who so ever gives the answer please do 2 things: Solve by the method that is explained earlier like: ***************************************************** 16/17 = -1
-1^437 *4 =-4
and 17-4= 13 ans
Can you please explain me the above steps as why it was done like this only.
*************************************************** And let me know the method that i mentioned why it was not correct Thanks
i dint get ur post wat u wanted to say but im solving the questions u asked:********************************************************** There are many doubts related to this question's plus 3 more question : 1)21^875/17 : Can't it be done as 1^875 is one so dividing by 17 will give me one is that can't be found out 2)In the above you solved by diving the digit by 17 two times 21/17 first case and 16/17 = -1 as second case as why it was done. Sorry for asking again and again but the problem is i am having lot of issues related to these question and getting confused. Can you please use the same method above and solve these questions: 1)Find the remainder when 54^124 is divisible by 17 a)4 b)5 c)13 d)15 can it be done as 4 to the power 4k(124) is 6 so diving 6 with 17 will give 6 as remainder but there is no such option 2)Find the remainder when 83 ^261 is divided by 17 a)13 b)9 c)8 d)2 Can't it be done by this method: 3 ^261 is equal to 3 ^4k+1 which is 3 so the remaider should be 3 but no such option. I request who so ever gives the answer please do 2 things: Solve by the method that is explained earlier like: ***************************************************** 16/17 = -1
-1^437 *4 =-4
and 17-4= 13 ans
Can you please explain me the above steps as why it was done like this only.
*************************************************** And let me know the method that i mentioned why it was not correct Thanks
83^261/17
first thing that comes to my mind is that 17*5=85 which is nearest to 83
so rem 83/17 =-2 (rem is 15 but 15/17 =-2)
so it now becomes -2^261/17
now i know that 2^4 =16 is nearest to 17
4^130 * (-2)/17
or
16^65 * (-2)/17
or
-1^65 * (-2)/17
so the rem is (-1)*(-2)=2
for the other ques...54^124/17.......the ans is 4
tell me if the ans r not correct........i do them roughly...
for y they r done lik this...the very ans lies in arun sharma's book....go thru its concepts ...these are the short tricks
********************************************************** There are many doubts related to this question's plus 3 more question : 1)21^875/17 : Can't it be done as 1^875 is one so dividing by 17 will give me one is that can't be found out 2)In the above you solved by diving the digit by 17 two times 21/17 first case and 16/17 = -1 as second case as why it was done. Sorry for asking again and again but the problem is i am having lot of issues related to these question and getting confused. Can you please use the same method above and solve these questions: 1)Find the remainder when 54^124 is divisible by 17 a)4 b)5 c)13 d)15 can it be done as 4 to the power 4k(124) is 6 so diving 6 with 17 will give 6 as remainder but there is no such option 2)Find the remainder when 83 ^261 is divided by 17 a)13 b)9 c)8 d)2 Can't it be done by this method: 3 ^261 is equal to 3 ^4k+1 which is 3 so the remaider should be 3 but no such option. I request who so ever gives the answer please do 2 things: Solve by the method that is explained earlier like: ***************************************************** 16/17 = -1
-1^437 *4 =-4
and 17-4= 13 ans
Can you please explain me the above steps as why it was done like this only.
*************************************************** And let me know the method that i mentioned why it was not correct Thanks
The method you're using is totally wrong. You can't just take the unit's digit of a number and use it to find the remainder for the whole number. I'll just take an example to illustrate why you're wrong:
Eg: What's the remainder of 5^2/17?? It should be 25/17 = 8
Now, what's the remainder of 15^2/17?? If we go by your method, it should again be calculated as 5^2/17, which would give us 8. But the actual remainder is 4 (225/17)
Hope now you get the point that you can't just pull out the last digit to find the remainder for the whole number. Now although Nirala has solved both the questions, I'll still solve them again. Probably you would understand the concept now:
1. Remainder of 54^124/17
54 = 3 mod 17 (51 is divisible by 17)
54^4 = 3^4 mod 17 = 81 mod 17 = (-4) mod 17 (85 is divisible by 17)
54^8 = (-4)^2 mod 17 = 16 mod 17 = (-1) mod 17
Now, 54^120 = (54^8 )^15 = (-1)^15 mod 17 = (-1) mod 17
54^124 = 54^120 * 54^4 = (-1) * (-4) mod 17 = 4 mod 17 (Multiplying the remainders obtained in the case of 54^120 and 54^4 respectively)
Thus, the remainder is 4.
2. Remainder of 83^261/17
83 = -2 mod 17 (85 is divisible by 17)
83^4 = (-2)^4 mod 17 = 16 mod 17 = (-1) mod 17
Now, 83^260 = (83^4)^65 = (-1)^65 mod 17 = (-1) mod 17
83^261 = 83^260 * 83 = (-1) * (-2) mod 17 = 2 mod 17 (Again multiplying the remainders obtained in the case of 83^260 and 83 respectively)
Thus, answer is 2.
PS: Just space your post a bit whenever you post next time. It becomes really hard for us to read.
20 can be divided into 4 parts in the following way
2,4,6,8----sum it you get 20 and they are in AP
and also product of 1st and 4th part is 2*8=16
product of 2nd and 3rd part is 4*6=24
and their ratio is 2:3
so the largest part among the four parts which satisfy the condition is 8
Hi Puys
I am a CAT 2010 aspirant and have joined TIME classes.I am very confused about what and where to practice from...only TIME study materials are enough or for Quant I should do Arun SHarma also???
My maths is not great.What should I bank on..TIME study material or also practice AS??
Pls Reply.
Thnx Friends
all d best
Hi Puys
I am a CAT 2010 aspirant and have joined TIME classes.I am very confused about what and where to practice from...only TIME study materials are enough or for Quant I should do Arun SHarma also???
My maths is not great.What should I bank on..TIME study material or also practice AS??
Pls Reply.
Thnx Friends
all d best
I personally preferred Arun Sharma over coaching material to clear up my fundamentals. It's explained in an easy to understand format and you should be able to follow it easily too. Just that it has quite a few wrong answers to the questions mentioned, but of course, that will only come in the next stage.
I would suggest you to first go through the concepts in Arun sharma and try to solve LOD-1 questions and solve the Time material exercises(by this time you would be quite acquainted with the concepts).Do this for every chapter,and finally when u r done with all the exercises in Quant from TIME and LOD-1 exercises from Arun sharma,then try to solve LOD-2 and LOD-3.these would come handy as additional practice exercises.If you could solve LOD-2 questions with greater ease,I would say you could easily crack CAT quant section
hi! need some help...
the sum of the series is represented by..
1/(1x5) +1/(5x9) +1/(9x13)......1/(221 x225)= ???
another question
find the infinite sum of the series1/1 +1/3 +1/6+1/10.....
hi! need some help...
the sum of the series is represented by..
1/(1x5) +1/(5x9) +1/(9x13)......1/(221 x225)= ???
another question
find the infinite sum of the series1/1 +1/3 +1/6+1/10.....
2. Is the answer to this question 2?? I think it should be. Here's my solution:
We can break 1 + 1/3 + 1/6 + 1/10..... like this:
2*
As we can see, apart from 1, which is the first number in the bracket, all the others get cancelled. So the final answer should be, 2*1 = 2
1. This question can also be solved similarly. We can break the equation like this:
1/4*
The middle terms get cancelled out, so the final answer would be 1/4*(1 - 1/225) = 56/225
Let me know whether the answers are correct. Been a long time since I've solved something of this sort. Took some while 😃
hi! thanks a lot. the answer is correct
but one question how do i get to thinking of arranging the terms in this fashion to arrive at the answer???
hi! thanks a lot. the answer is correct
but one question how do i get to thinking of arranging the terms in this fashion to arrive at the answer???
That you'll get by practicing such questions 😃 You basically need to arrange them in such a way that most of the terms cancel out and your calculation becomes simple.
ok so there are no shortcuts...:( got it buddy.
hi! need some help...
the sum of the series is represented by..
1/(1x5) +1/(5x9) +1/(9x13)......1/(221 x225)= ???
another question
find the infinite sum of the series1/1 +1/3 +1/6+1/10.....
the first one can be written as
(1/1 -4/5 )+ (4/5-7/9) + (7/9-10/13)+ (10/13 - 13/17)+....(166/221 - 169/225)
1-169/225= 56/225
if a man saves rs 1000 each year and invests at the end of the year at 5% compound interest, how much will the amount be at the end of 15 years??
one more question
the sum of the series:
1/(rt2+rt1) +1/(rt2+rt3) +......+ 1/(rt120+ rt121)
p2w Saysif a man saves rs 1000 each year and invests at the end of the year at 5% compound interest, how much will the amount be at the end of 15 years??
1000*1.05 +1000*1.05^2 +1000*1.05^3 +...1000*1.05^15
1000*/(1.05-1) = 22657.. used calci 😃
p2w Saysif a man saves rs 1000 each year and invests at the end of the year at 5% compound interest, how much will the amount be at the end of 15 years??
This can be solved in the form of a sum of a GP.
Rs. 1000 invested in the 1st year would become 1000*(1.05^14) at the end of the 15th year.
Rs. 1000 invested in the 2nd year would similarly, become 1000*(1.05^13) at the end of the 15th year, and so on.
So this will form a GP, arranging in the reverse order (i.e. from year 15 to year 1):
1000 + 1000*1.05 + 1000*(1.05^2) + .... (1000*1.05^14)
Here, a = 1000, n = 15, r = 1.05
Sum = a* (r^n-1)/(r-1) = 21579 (approx)
Again, I'm not sure of the answer, but this time, I'm sure of the approach 😃
how do u solve r^15, as that would take very long...
p2w Sayshow do u solve r^15, as that would take very long...
Well, actually I didn't. Used the computer's calculator for that. At this point, I can't remember how to get this done manually. Probably could have helped had I still been in practice. Possibly you can get this solved by someone else.
thanks a lot, u've actually helped a lot with the previous problems.
Please help me to do these questions from number system thanks:
Amitesh buys a pen,pencil and an eraser for RS 41. if the least price is of pen that is 12 and eraser cost more than the pencil and pen costthe least.
Answer the following questions:
If it is known that the eraser cost is not dividible by 4 then the cost of the pencil is :
a)12 b)13 c)14 d)15 The answer is a)
A man sold 38 pieces of clothing.Ih he sold at least 11 pieces each item and he sold more shirts than trousers and more trousers than ties then the number of ties that he sold :
a) Exactly 11 b) Atleast 11 c)At least 12 d) Cannot be determined The answer is a) but i am getting b)
Find the least number which must be subtarcted from 7147 to make a perfect square
a)86 b)89 c)91 d) 93 The answer is c)
Find the least sqaure number divisible by 6,8 and 15
The answer is 3600
Find the least number by which 30492 must be multiplied or didvided so as to make a perfect square:
a)11 b)7 c)3 d) 2
the answer is b
The power of 45 that will divide 123! is
a)28 b)30 c)31 d)29
The answer is 31 but i am getting 28 as 123/5+ 123/5^2
Find the smallest natural number n such that n! is divisible by 990
a)3 b) 5 c)11 d) None of these
The answer is d)