Vriti Edustore
Home »
Ask & Discuss
»
Mathematics
« Back to Discussion
Algebra
Comments (10)
ok firstly see that f (100) = 100(1 - 1/2).(1 - 1/5)= 40.Since 7 and 100 are coprime Hence, by Euler’s Theorem, 740≡1mod 100......Now, f (40) =40( 1- 1/2).(1- 1/5)=16, hence 716≡1mod40.... Now since 1000= 16 ·62+8. This means that 71000≡(716)62.78 ≡162.78 ≡(74)2≡12≡1mod40 This means that 71000= 1+40t for some integer t. now assembling all this we get 7^(71000)≡71+40t ≡7.(740)t≡7 mod 100
this means last two digits are 07
well dis type of question are very rarely asked in now a days in IIT..dese are problems of cat..
first you have to learn some concepts
1st......
Let R be the remainder when N is divided by D1.
Now the remainder when N is divided by D2, where D2 is a factor of D1 is either
1. R ( if R <D2) or
2. R1, where R1 is the remainder obtained when R is divided by D2 ( if R>D2)
Remember the converse is also true.
2nd.....
Let's look at an example here. ( let me give the converse example)
Let 23 be the remainder obtained when a number N is divded by 30.
Now the remainder obtained when the number is divided 90 must be either 23 or 30×1+23 or 30×2+23 i.e the remainder will be of the form 30p+23 where p =0 or 1 or 2( I think this is clear to everyone)
now see dis example....
Let's apply this rule to the follwoing question.
1. Last two digits of N = 3^999.
Divided by 100 the remainder is the last two digits of the number
So let's divide 3999 with 100 and find the remainder. First find the remainder obtained when it is divided by 25 ( since 25 is factor of 100).
3999 % 2= (27333) % 25 = (2333)%25 = [(102433) x 23] % 25
Remainder obtained when 1024 is divided by 25 is 24 but we can take it as -1 also (always take the smallest remainder irrespective of sign)
So finally the problem becomes -1 x 23 %25= -8 =17 ( finally give the positive remainder)
So when our number is divided by 25 the remainder is 17
So when it is divided by 100 the remainder R must be of the form 25p+17
Similarly when N is divided by 4 the remainder is 3.I am taking 4 here because the other factor of 100 is 4.
Hence 25p+17 must leave a remainder 3 when divided by 4. i.e p+1 must leave a remainder 3 when divided by 4. Hence the value of p is 2.
Therefore R = 25p+17 = 67. Hence the number N ends with 67.
u can follow your problem like dis it works.. i did it here.. in our case first take 50 as D1 and 2 as D2....
dude this thing can help u....and improve ur concept
71000 = (50-`1)500=500C0 50500-.......................................-500*50+1
=1000(k)+1
ab dekh 7 ki powers ke unit digit periodic hain 4 ke . e.g . 7 , 49 , 243 ,2401,then again 16807
So u can say without any doubt that unit digit of 77^(1000) would be 7 .
7 1000k+1 =7 ( 71000k) = 7 ( 50 -1)500k =7.(500kC050500k-................-500*50k+1)
all the first 500k numbers would be having atleast 3 zero on the last three places . 07 would be last two digit
Suggestion ::
Yaar have a knowlegde of number theory . It helps a lot .
Thank u
Preparing for JEE?
Kickstart your preparation with new improved study material - Books & Online Test Series for JEE 2014/ 2015
@ INR 5,443/-
For Quick Info
Find Posts by Topics
Physics
TopicsMathematics
Chemistry
Biology
Institutes
Parents Corner
Board
Fun Zone
|
| 1. |
|
Bipin Dubey
|
| 2. |
|
Himanshu
|
| 3. |
|
Hari Shankar
|
| 4. |
|
edison
|
| 5. |
|
Sagar Saxena
|
| 6. |
|
Yagyadutt Mishr..
|
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
5804 Students Currently Online |
Connect with Us  |
 |
.