@TootaHuaDil said:Last two digits of 78^2009?
08??
78^2009=(2*39)^2009=2^2009*39^2009
2^9*(2^10)^200*39*(39^2)1004
12*(24)^200*39*(21)^1004
12*76*39*81
12*39*81=08
@The_Loser said:divide your power wid 20.u will find 78^9 --> 08 last two digits.ya Tootahua dil.. acha ID name hai
Thanks 
by the way why divide by 20?? Explain me this approach...seems faster :)
by the way why divide by 20?? Explain me this approach...seems faster :)@sos2god said:41 ka kya 42 ans hain.... 16*2 + 3*2 +(210,420,630,840) ????
42 it is, are u considering 111 twice or once becoz it is both ap and gp
16*2 how?
an easy one.
a + b + c + d = 90
a,b,c,d =how many solutions?
@TootaHuaDil said:yes....08 now detail
(78)^2009
Lets write it as (13*3*2)^2009
For 2^2009 can be written as (2^10)^200 * 2^9..
For this we shall use the logic that 2^10 = 1024 ..and ...24^even then last 2 digits shall be 76 and if...24^odd then last 2 digits shall be 24...
So 2^10 last 2 digits shall be 76...and for 2^9 last 2 digits shall be 12..so 2^2009 last 2 digits shall be ...12
For 3^2009 , we shall use the logic that a number (a)^b if the units digit in a is 1 then the last 2 digits shall be tens digit in a * units digit in b..means for (...21)^32 = ..41
So 3^2009 can be written as (3^4) ^502 * 3 = (81)^502 * 3 = (..61)*3 = (..83)..
For 13^2009 = (13^4) ^ 502 * 13 = ( 169 * 169 ) ^ 502 * 13 = ( ....61) ^ 502 * 13 = (..73)
Now you know the last 2 digits of each ..and you shall get the answer to be 08..
@TootaHuaDil said:Last two digits of 78^2009?
lst 2 digits=|78^2009/100|
e[100]=40
|78^2009/100|=|78^9/100|=08...
@TootaHuaDil said:Thanks by the way why divide by 20?? Explain me this approach...seems faster
trick (a shorcut) - in case on even number. divide your no's power by 20.
for odd - a very short cut. but can only explain you wid an example.
if u have any ques dan post it.
for odd - a very short cut. but can only explain you wid an example.
if u have any ques dan post it.@The_Loser said:trick (a shorcut) - in case on even number. divide your no's power by 20.for odd - a very short cut. but can only explain you wid an example. if u have any ques dan post it.
Find the last two digits of 33^288??
Find the last two digits of 78^379??
Any one will serve as an example???
Find the last two digits of 78^379??
Any one will serve as an example???
@TootaHuaDil said:Find the last two digits of 33^288??Find the last two digits of 78^379??Any one will serve as an example???
41 ,28
@vbhvgupta
@vbhvgupta said:42 it is, are u considering 111 twice or once becoz it is both ap and gp 16*2 how?
there are 32 AP's and 6 GP's and 4 AP's with 0 in the end...so 42...i havet considered 111..i dont think its a GP /AP ...
@happy3475 said:lst 2 digits=|78^2009/100|e[100]=40|78^2009/100|=|78^9/100|=08...
Euler 😃 got this one!
@TootaHuaDil said:Last two digits of 78^2009?
78^2009 = 2^2009 . 39^2009 = (2^10)200. 2^9. (39^2)^1004.(39)
=24^200.(512)(..21)^1004.(39) (becoz 2^10 ends with 24 and 24^even is 76 And 39^2 ends with 21)
= 76.12.81.39 (Taking the last 2 digits only)
= 08
I referred the method i learnt here http://quantexpert.co.in/qenotes/quantnotes/numbers/44-secondlast-digit-of-any-power.html
=24^200.(512)(..21)^1004.(39) (becoz 2^10 ends with 24 and 24^even is 76 And 39^2 ends with 21)
= 76.12.81.39 (Taking the last 2 digits only)
= 08
I referred the method i learnt here http://quantexpert.co.in/qenotes/quantnotes/numbers/44-secondlast-digit-of-any-power.html
yes as 78 is even so dividing the power wid 20 wud serve.
for odd like 33, here is the trick.
33^288 = (33*33)^288
= (89)^144 = (89*89) = (21)^144
what i did is nth just extracted the last two digits from each bracket.
once i deduced a number wid it's unit digit one(here 21). My question is solved.
now unit digit is one. for ten's digit just multiply 2 (ten's digit f final number) with 4 (unit digit of power).
Might seeems a bit daunting bt try to solve any other random number like 37^ 76584 or wteva wid this method.
within seconds solution can be deduce.
@TootaHuaDil said:Euler got this one!
are u sure we can use euler theorem here...as 78 and 100 share a common factor 2....
@sos2god said:@vbhvgupta there are 32 AP's and 6 GP's and 4 AP's with 0 in the end...so 42...i havet considered 111..i dont think its a GP /AP ...
111,222,333.....nnn is both ap and gp IMO
@vbhvgupta said:111,222,333.....nnn is both ap and gp IMO
cant be ap for sure...as then d would be 0...
@vbhvgupta said:@ravi.theja AM I right?
no buddy..u are wrong..dey are only in A.P not in G.P
check 2b=a+c fro A.p and b^2 = a*c for g.p..the terms above jus satisfy condition for a.p but not g.p
check 2b=a+c fro A.p and b^2 = a*c for g.p..the terms above jus satisfy condition for a.p but not g.p
@fireatwill said:what will be the day of the week on 10th december of 1989?
400 years have 0 odd day
300 years have 1 odd day
400*4+300+88
0+1+88+22(leap years)=6 when divided by 7
31st dec 1988 is sat
Jan-3
Feb-0
March-3
April-2
May-3
June 2
July-3
Aug-3
Sep-2
Oct-3
Nov-2
Dec-10
Total =29..Remainder is 1 when divided by 7
6+1=7(sunday)