n =10,11,20,21,30,31,40,41
so 8 values
correct me if I am wrong...
Can u explain ur approach???
Can u explain ur approach???
If N=7k+3 and k is a natural number then for how many values of k from 10 to 25( both inclusive) is n composite?
kdn1512 SaysCan u explain ur approach???
my take :
unit digit will remain same only in case of an addition of a 0(zero). within the range of 50, we get 0, 10, 20, 30 & 40 ending with zero.
now if n=10, (n-1)=9
so getting the sum upto 10^10! gives the same unit digit as upto 9^9!
also if n=11, (n-1)=10
again getting the sum upto 11^11! gives the same unit digit as upto 10^10!
the same logic applies with the other numbers also...
If n is a natural number such that 1^1! + 2^2! + 3^3! + ...+ n^n! ends with the same unit digit as 1^1! + 2^2! + 3^3! + ...+(n-1)^(n-1)! and n
1) 3
2) 4
3) 6
4) 8
5) 9
For me Answer is 2)4
I am not sure about the answer..
Since last digit of n must be 0 for addition to be same
sets of n-1 an n would be (9,10),(19,20),(29,30),(39,40)...
Another possibility ..if both last digit add up to 10..
Let see cyclicity table
0-0000
1-1111
2-2468
3-3971
4-4646
5-5555
6-6666
7-7931
8-8426
9-9191
0-0000
I tried to see value diagnolly but didnt find any that add to 10:shocked::shocked:..
(hope i am correct :P)
Hence my bet on option 2..
Correct me if iam wrong..and please mention the process
For me Answer is 2)4
I am not sure about the answer..
Since last digit of n must be 0 for addition to be same
sets of n-1 an n would be (9,10),(19,20),(29,30),(39,40)...
Another possibility ..if both last digit add up to 10..
Let see cyclicity table
0-0000
1-1111
2-2468
3-3971
4-4646
5-5555
6-6666
7-7931
8-8426
9-9191
0-0000
I tried to see value diagnolly but didnt find any that add to 10:shocked::shocked:..
(hope i am correct :P)
Hence my bet on option 2..
Correct me if iam wrong..and please mention the process
sry diffrence should be 10
For me Answer is 2)4
Since last digit of n must be 0 for addition to be same
sets of n-1 an n would be (9,10),(19,20),(29,30),(39,40)...
for every distinct value of (n-1) we are getting a corresponding distinct value of n
here also you got 8 values of (n-1), so n should possess 8 values + 1 extra (I missed the number 1 in the initial solution)
kriti.v123 SaysIf N=7k+3 and k is a natural number then for how many values of k from 10 to 25( both inclusive) is n composite?
range chota hi to hai...k ka value put karke dkh lete hay
for k = 10, 14, 22 we get prime nos as 73,101,157
therefore other values will give composite nos
no of values of k as specified = 16 - 3 = 13
kriti.v123 SaysIf N=7k+3 and k is a natural number then for how many values of k from 10 to 25( both inclusive) is n composite?
N=7k+3
let k=10+a
==>N=73+7a
--N=72+7a+1==>(2^3*3^2)+(7a+1)
so if ais 0dd then 7a+1 is even..hence N becomes a multiple of 2..
so we can say for all odd values of a or k no. is composite..
For even numbr a=(0,2,4,6,8 etc..)..I guess we need to calculate them..:P
Thanks tamal for previous soln
N=7k+3
let k=10+a
==>N=73+7a
--N=72+7a+1==>(2^3*3^2)+(7a+1)
so if ais 0dd then 7a+1 is even..hence N becomes a multiple of 2..
so we can say for all odd values of a or k no. is composite..
For even numbr a=(0,2,4,6,8 etc..)..I guess we need to calculate them..:P
Thanks tamal for previous soln
wil it not b easier to just eliminate the prime numbers n the left ones wil b composite??
Originally Posted by tamal View Post
For me Answer is 2)4
Since last digit of n must be 0 for addition to be same
sets of n-1 an n would be (9,10),(19,20),(29,30),(39,40)...
for every distinct value of (n-1) we are getting a corresponding distinct value of n
here also you got 8 values of (n-1), so n should possess 8 values + 1 extra (I missed the number 1 in the initial solution)
ans should be (2)4 as n=1 not possible bcoz (1^1!)=1 (0^0!)=(0^1)=0 so they are not equal ........moreover if u calculate for n=11,21,31,41 then its adding extra 1 to unit digit
kriti.v123 Sayswil it not b easier to just eliminate the prime numbers n the left ones wil b composite??
riginally Posted by tamal View Post For me Answer is 2)4
Since last digit of n must be 0 for addition to be same
sets of n-1 an n would be (9,10),(19,20),(29,30),(39,40)...
for every distinct value of (n-1) we are getting a corresponding distinct value of n
here also you got 8 values of (n-1), so n should possess 8 values + 1 extra (I missed the number 1 in the initial solution)
ans should be (2)4 as n=1 not possible bcoz (1^1!)=1 (0^0!)=(0^1)=0 so they are not equal ........moreover if u calculate for n=11,21,31,41 then its adding extra 1 to unit digit
suppose i start number a,b,c,d ....................z as 1 -26 and then on reverse from y ,x............a as 27to 51 and then again b,c,d.............z as 52-76 and then again from y to a as 77-101 ...........................the process continues ..............find the 777th character ??
Options:-
1.a
2.b
3.r
d.non of these
i am getting y which means d.none of these but ans given is a 😞 😞
mukulsriv10 Sayscan anybody pls help me..what will be the last digit of 27 raised to power 36. :O
27^36
Consider the last digit here it's 7...
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
___________
7^5 = 7
So actually the cyclicity of 7 is 4....
Now with the problem :
27^36
Ignore the ten's digit and proceed with the units digit ( Coz U r asked to find the unit's digit )
7^36
Can b written as
7^4 * 7^4......... ( 9times )
Value of 7^4 = 1 (We derived it )
So value of 7^4 * 7^4......... ( 9times ) is 1 * 1* 1 ( 9 times ) = 1....
mukulsriv10 Sayscan anybody pls help me..what will be the last digit of 27 raised to power 36. :O
It will be 1.
How to solve :
{1/(1+3)} + {1/(3+5)} + {1/(5+7)} + ....upto 50 terms.
ANS: (101 - 1 )/2
chilli bhai pls help me with these question
2) A = 1^1 x 2^2 x 3^3 x 4^4 x ...100^100 . How many zeroes will be there at the end of A?
a) 1300
b) 1320
c) 1325
d) 1050
e) None of these
Answer is 1300.
We just need to calculate the number of exponents of 5 for such type of questions.
1^1 x 2^2 x 3^3 x 4^4 x 5^5... X 25^25 X .... X 50^50 X...X 75^75 X... X100^100 .
Now, powers of 5 would be.. as 5^5 + 10 ^ 10 + 15^15+...100^100
So, exponents are in the form 5+10+15+20+25+...
So, 5(1+2+3+4+...20) = 5*20*21/2 = 1050.
Now, for 25,50,75 & 100, we will have extra 5 component.. coz 25 = 5*5, 50 = 5*5*2, 75 = 5*5*3, 100=5*5*4.
So, adding their exponents once again, 25+50+75+100 = 250.
So, total = 1050+250 = 1300
2) A = 1^1 x 2^2 x 3^3 x 4^4 x ...100^100 . How many zeroes will be there at the end of A?
a) 1300
b) 1320
c) 1325
d) 1050
e) None of these
ans is a:
is it??/
pkaman SaysIs the answer (d) .. ? Pls verify.
@pkman, it should be 1300, you forgot to consider extra 5 at 25, 50, 75, and 100
How to solve :
{1/(1+3)} + {1/(3+5)} + {1/(5+7)} + ....upto 50 terms.
ANS: (101 - 1 )/2
1/(1+3) is of the form
Multiplying the numerator and denominator by {(2n+1) - (2n-1)}, we will get,
(-1+3-3+5-5+7+....+101) / 2
= (101 - 1 )/2
Hope it helps ...
How to solve :
{1/(1+3)} + {1/(3+5)} + {1/(5+7)} + ....upto 50 terms.
ANS: (101 - 1 )/2
rationalize each factor the denominator will be 2 and in the numerator everything will get cancelled except root(101) and 1
hope you understand it
abhi0988 Says@pkman, it should be 1300, you forgot to consider extra 5 at 25, 50, 75, and 100
Oh, sorry. actually I was getting 1150. I missed out 50 & 100. Thanks for correcting. 😃