Since N has 388 digits and has 2 and 5 as factors,the 2's and 5's would combine to give zeroes.So N would have 386 zeroes at the end.
Now,remaining digits= 388-386=2
Now,power of 2 which give only 2 digits are 4,5,6(2^4=16,2^5=32,2^6=64)
Hence M takes 3 values.
Find the reminder
3^32/50
dumb confused SaysI think u r forgetting one point. it takes a 5 and a 2 to make a 0. so to make 386 zeroes out of 386 5s, u need 386 nos of 2.
Find the ten's place digit
6^11^5 .
please do explain the method of finding ten's place digit.
Find the sum up to 25'th term of the series
3+6+11+18+...
3+6+11+18+....
1^2+2 + 2^2+2 + 3^2+2 +.......
2x25 + sum of first 25 squares ..
----------------------------------------------------------
6^11^2 remainder from 100 will serve the purpose .
6^1 = 06
6^2 = 36
6^3 = 216
6^4 = 1296
6^5 = xxx76
6^6 = xxxx56
6^7 = xxxx36
So after 6^6 last two digits starts repeating ..
Now 11^5 == 5k+1=6 or 11 or 17 ..
so answer is same as 6^6 =xxxx56
Others plz correct me or Give some other fundooo method ;)
Find the ten's place digit
6^11^5 .
please do explain the method of finding ten's place digit.
Find the sum up to 25'th term of the series
3+6+11+18+...
3+6+11+18+....
1^2+2 + 2^2+2 + 3^2+2 +.......
2x25 + sum of first 25 squares ..
----------------------------------------------------------
6^11^2 remainder from 100 will serve the purpose .
As 6 and 100 are nor coprime .reduce them to there lowest form
=> {6^(11^5 -2) } / 25
eulers theorm ....--> 25 has 20 coprimes less then it .
[ 6^(20k+11) - 2 ] . 3. 3 / 25
Answer will be 44 ..
Others plz correct me or Give some other fundooo method ;)
6^11^2 remainder from 100 will serve the purpose .
As 6 and 100 are nor coprime .reduce them to there lowest form
=> {6^(11^5 -2) } / 25
eulers theorm ....--> 25 has 20 coprimes less then it .
[ 6^(20k+11) - 2 ] . 3. 3 / 25
can some body explain Euler's theorem ??
Is the answer 41.....???
(3^5)^6*3^2= 243^6*9
((250-7)^6*9)/50= (-7)^6*9
(7^2)^3*9=49^3*9
(50-1)^3*9=(-1)^3*9
-9 is the remainder so 50-9 = 41 answer...
AIMING TO BELL THE CAT....
what is the reminder when (787878......78(31 times)) is divided by 35?
please explain the method
what is the reminder when (787878......78(31 times)) is divided by 35?
please explain the method
is the answer 43 ?
Find the Ten's place digit of 7^99??
please explain method of finding ten's place digit..
Find the Ten's place digit of 7^99??
please explain method of finding ten's place digit..
what is the tens digit of the number 7^2002 ???
a) 1
b) 2
c) 3
d) 4
what is the tens digit of the number 7^2002 ???
a) 1
b) 2
c) 3
d) 4
The answer is 4
the cyclicity for 7 repeats as...07,49,43,01...
7^1=07
7^2=49
7^3=343
7^4=2401
7^5=...07
so the answer for Ten's place digit of 7^99 is 4
2- sq rt.(6407522209/3600840049)
1.0.666039
2.0.666029
3.0.666009
4. none of these
this is a que from iift 2009 paper.
One concept.....suppose the sqaure of a number (1x) ends with ac......
Then the square of nx will have last digit as the last digit of x^2 i.e c....but second last digit will be (second last digit of 1x^2+(n-1)*x*2).......
Example.....12^2 ends with 44.......Last digit of squares of all numbers ending with 2 will be 4.....now to determine second last digit.....formula is (second last digit of 1x^2+(n-1)*x*2)....here x=2....
For 22^2 (484) it will be 4+1*2*2=8...
For 32^2 (1024) it will be 4+2*2*2=12....ignore everything other than units digit....so 2 remains.....
For 62^2 (3844) it will be 4+5*2*2=24....i.e 4....
Similarly 17^2=289.....x=7......
37^2 (1369) will have second last digit 8+2*7*2=36...i.e. 6....
Also the second last digit of all squares have a cyclicity of 5....other than those ending with 0 of course.....so 12^2 ends with 44.....(12+50)^2 will end with 44......(12+100)^22 will end with 44 and so on.....
Coming back to the question......
First square of five digits is (100)^2......
Now 32^2=1024......so 320^2=102400.....six digits....and 310^2=96100.....
Now all numbers ending with 0 will satisfy the condition......so multiples of 10 between 100 to 310=22.....
Now if a square ends with an odd number then its tens digit will be even.....so odd numbers are ruled out.....
Now consider even numbers......the squares will end with an even number....as said earlier, second last digit has a cyclicity of 5.....also second last digit of nx is even if and only if second last digit of 1x is even.....only 12 and 18 have their second last digit as odd amond even numbers between 10 and 20......
So the numbers satisfying the condition given in question will end with 2 or 8.....
Case 1: Numbers ending with 2.....12^2 ends will 44....so will 62^2....112^2 and so on till 312^2......here you have to check and you will find out that 312^2 has 5 digits......so number of such numbers=7....but we only have to take numbers which have five digits in squares i.e numbers greater that 100....so 12 and 62 are eliminated......so 5 numbers are there......
Case 3: Numbers ending with 18......18^2 ends with 24.....
Consider the number n8......second last digit will be 4 when last digit of (second last digit of 18^2+(n-1)*8*2) is 4......i.e 2nd last digit of (n-1)*8*2 is 2....n-1=2 satisfies this....so n=3......38^2 will have last two digits as 44........next will be 88^2......then 138^2 till 288^2.......338^2 will have six digits and hence is left out......again numbers below 100 i.e 38 and 88 are eliminated.....so total 4 numbers.....
So total number of numbers satisfying condition given in statement are (22+5+4)=31.....
Please verify the answer......
