try writing down 3^1,3^2...there's a pattern of last digits 3,9,7,1,3,9... and for 2^1,2^2...it is 2,4,8,6... So in both cases the pattern repeats after 4th power. when power is 2001=2000+1 the last digits will be 3 and 2 for the reqd numbers.Hence you get the answer
hey abhishek... not a tough q 2 expln... although anchal has done the job... but i do remember ( kinda) wat krishnan once told me.... the sequence thing has no fault in at all... but as an addition and faster reasoning... one can take the followin method-- instance its 3^2001 obvious it is 3^2000 * 3.... and 3^2000 is (3^4)^500... which is 81 ^ 500... which obviosly ends with 1... so 3^2001 ends with 1*3=3 similar case with 2^2001= (2^2000)*2 = (2^4)^500 * 2 that is 6 * 2 = 12... so the digit is 2.. so the answer 3-2 =1... well its d same thing... but wit little diff... n its defn faster than taking out a sequence cause it can even work with other numbers as well and one more thing... ( ie to whosoever reads this comment) im abhishek jain... diff from abhishek nanda( whos the member and whos posted the q)
i wanted to see the method b'coz i wanted to compare it to mine.anyway i think dat the catch is to express such a n.o to a certain power that whatever be the power the last digit remains same.it could be n.os ending with 1,5,6.isn't it?
The answer to the q concerning the number of times the hands wud cross each other. i suppose the needles would meet 12 times (2 times during 12 noon and 12 midnight included). u could find the exact time of their meeting as well.
i)At 12 noon they are together.
ii)between 1pm and 2pm: at 1 PM, the min hand is 5 min behind the hour hand[min hand is at 12 and hour hand at 1]. So it has to move 5 min relative to the hour hand to meet it. In 60 mins, the min hand moves 55 min relative to the hour hand. In 60/55 min it moves 1min relative to the hour hand => in 60/11min it moves 5min relative to the hour hand. So at 60/11 min past 1 the hands are together again. So we can find 12 such times b/w 12 hours. Hope my language is legible coz im not much used to writing proofs over the net [it is usualy confined to the answer sheets] And i was wandering whether i cud get an invite...abhijit89@gmail.com if you agree to it.
And hey if you try this question with 'Logaritm' it might not work properly. It'll also be too long involving use of tables for taking out log and antilog..both which are reqd. twice. Also..log tables are not provided anywhere except CBSE noard exam where this qs is least likely to come.
the last digit will be 0. let S = 2001^3 - 2001^2 = 2001^2(2001 - 1) = 2001^2(2000) = Q*1000 anythin multiplied by 10^n (n is a natural no.) ends with 'n' zeros.
17 Comments:
its the subtraction sign in between
The last digit of 3^2001 is 3 and of 2^2001 is 2.
So ur answer is 1.
could u explian how u got dat in short
try writing down 3^1,3^2...there's a pattern of last digits 3,9,7,1,3,9... and for 2^1,2^2...it is 2,4,8,6...
So in both cases the pattern repeats after 4th power.
when power is 2001=2000+1 the last digits will be 3 and 2 for the reqd numbers.Hence you get the answer
hey abhishek...
not a tough q 2 expln...
although anchal has done the job... but i do remember ( kinda) wat krishnan once told me....
the sequence thing has no fault in at all...
but as an addition and faster reasoning... one can take the followin method--
instance its 3^2001 obvious it is 3^2000 * 3.... and 3^2000 is (3^4)^500... which is 81 ^ 500... which obviosly ends with 1... so 3^2001 ends with 1*3=3
similar case with 2^2001= (2^2000)*2
= (2^4)^500 * 2
that is 6 * 2 = 12... so the digit is 2..
so the answer 3-2 =1...
well its d same thing... but wit little diff... n its defn faster than taking out a sequence cause it can even work with other numbers as well
and one more thing... ( ie to whosoever reads this comment) im abhishek jain... diff from abhishek nanda( whos the member and whos posted the q)
easiest solution is take log find the ans and enjoy life
i wanted to see the method b'coz i wanted to compare it to mine.anyway i think dat
the catch is to express such a n.o to a certain power that whatever be the power the last digit remains same.it could be n.os ending with 1,5,6.isn't it?
anyway anymore answers to my previous post?
This comment has been removed by a blog administrator.
The answer to the q concerning the number of times the hands wud cross each other.
i suppose the needles would meet 12 times (2 times during 12 noon and 12 midnight included).
u could find the exact time of their meeting as well.
i)At 12 noon they are together.
ii)between 1pm and 2pm:
at 1 PM, the min hand is 5 min behind the hour hand[min hand is at 12 and hour hand at 1].
So it has to move 5 min relative to the hour hand to meet it.
In 60 mins, the min hand moves 55 min relative to the hour hand. In 60/55 min it moves 1min relative to the hour hand => in 60/11min it moves 5min relative to the hour hand. So at 60/11 min past 1 the hands are together again.
So we can find 12 such times b/w 12 hours.
Hope my language is legible coz im not much used to writing proofs over the net [it is usualy confined to the answer sheets]
And i was wandering whether i cud get an invite...abhijit89@gmail.com if you agree to it.
And hey if you try this question with 'Logaritm' it might not work properly. It'll also be too long involving use of tables for taking out log and antilog..both which are reqd. twice.
Also..log tables are not provided anywhere except CBSE noard exam where this qs is least likely to come.
I gave the the method of taking log for the kind of people who are interested in finding the correct ans using calculators etc.
hey culd i possibly get an invite... i have an intresting q to share with all of you..
my email id- abhishekdipsite@gmail.com
the grid one?shall i post it?
the last digit will be 0.
let S = 2001^3 - 2001^2
= 2001^2(2001 - 1)
= 2001^2(2000)
= Q*1000
anythin multiplied by 10^n (n is a natural no.) ends with 'n' zeros.
Arjun, you got the question wrong.
Its 3 to power 2001 and NOT 2001 to power 3.
OOPS!!!!
sorry people!!
stupid mistake
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