Can I get the solution for this?
can ny1 tell how to find d last non- zero digit of any factorial number
prachi144 Sayscan ny1 tell how to find d last non- zero digit of any factorial number
i guess this is the best material available for probs related to factorial...
http://totalgadha.com/mod/forum/discuss.php?d=5224
ATB
the remainder is 1.
20^41 +21^41 +..25^41 divided by 180 leaves remainder??
ans is 45..I got 135
Please tell me how to solve:
(3003^9000) / 9000
consider the numerator
3003^9000=(3000+3)^9000
By Binomail expansion all the terms would be divisible by 9000 except the last term which is 3^9000. So the problem reduces to finding 3^9000 mod 9000
now 3^9000 mod 9000 = 9*3^8998 mod 9000 = 3^8998 mod 1000 ( the 9 which we have cancelled out as the common term will have to be multiplied to the answer of 3^8998 mod 1000 )
1000 = 125*8
so using the Chinese Remainder Theorem ( http://www.pagalguy.com/discussions/conceptstotal-fundas-25023536 )
3^8998 mod 8 = 9^4499 mod 8 = 1 .... (1)
Euler number of 125 is 100
so 3^8998 mod 125= 3^98 mod 125
now 3^100 mod 125 = (3^2 * 3^98 ) mod 125
1 = ( 9 *3^98 ) mod 125
Now we find the smallest number which when multiplied by 9 will give remainder 1 when divided by 125. The smallest such number is 14.
Thus 3^98 mod 125 = 3^8998 mod 125 = 14 ... (2)
Now using Chinese remainder theorem and 1 & 2
125a+14=8b+1
The smallest such number is 889
thus 3^8998 mod 125 = 889
Now multiplying by 9 will give us our final answer which is 889*9 = 8001
plz help me with this
(12)base n+(34)base n+....+() base n, where N is the base of number system, N=2M+1 and M>2. Express X+M+2 in base M
(12) base n = n + 2
(34) base n = 3n +4
(56) base n = 5n + 6
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.
.
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= (2M-1)n + 2Mn
adding all theses numbers we get
{1+3+5+....+(2M-1)}*n + 2*(1+2+3+....+m) .... (1)
now
1+3+5+..+(2M-1) is an AP with d=2 and 'm' terms
so Sum of these terms = m^2
Similarly 1+2+3+..+m = m(m+1) / 2
So eqn 1 translates to
m^2 * n + 2*m(m+1)/2 = m^2*n + m*(m+1)
now n =2m+1
so we get 2m^3+2m^2+m = (2210) base m
(12) base n = n + 2
(34) base n = 3n +4
(56) base n = 5n + 6
.
.
.
.
= (2M-1)n + 2Mn
adding all theses numbers we get
{1+3+5+....+(2M-1)}*n + 2*(1+2+3+....+m) .... (1)
now
1+3+5+..+(2M-1) is an AP with d=2 and 'm' terms
so Sum of these terms = m^2
Similarly 1+2+3+..+m = m(m+1) / 2
So eqn 1 translates to
m^2 * n + 2*m(m+1)/2 = m^2*n + m*(m+1)
now n =2m+1
so we get 2m^3+2m^2+m = (2210) base m
sry bro the answer to this question is (2222) base m
the procedure you adopted is right
(12) base n = n + 2
(34) base n = 3n +4
(56) base n = 5n + 6
.
.
.
.
= (2M-1)n + 2M
therefore x=nM^2+M(M+1)
substituting n=2M+1 we get X=(2M+1)^2+M^2+M=2M^3+2M^2+M
hence X+M+2=2M^3+2M^2+M+M+2=2M^3+2M^2+2M+2=(2222) base M
20^41 +21^41 +..25^41 divided by 180 leaves remainder??
ans is 45..I got 135
20^41 mod 180 = 140 or -40
21^41 mod 180 = 81
22^41 mod 180= 52
23^41 mod 180= 83
24^41 mod 180 = 144
25^41 mod 180 = 85
140+81+52+83+144+85 = 585
585 mod 180 = 45
....Double post
Just an easier method
consider the numerator
3003^9000=(3000+3)^9000............
By Binomail expansion all the terms would be divisible by 9000 except the last term which is 3^9000. So the problem reduces to finding 3^9000 mod 9000
now 3^9000 mod 9000 = 9*3^8998 mod 9000 = 3^8998 mod 1000 ( the 9 which we have cancelled out as the common term will have to be multiplied to the answer of 3^8998 mod 1000 )
from here we can reduce the no of steps..
we know 3^9000 mod 1000 = 1
3^8998 * 3^2 mod 1000 =1
x*9 = 1000k+1
so for k=8
we have x= 889
hence rem is 889*9 = 8001...................
The sum of 40 natural numbers is 801. What is the minimum possible value of their LCM?
puys,
i need some..if possible many solved and unsolved questions on LCM and HCF...
what good book should i refer for total thorough understanding of Number systems?
iv been told to start doing 6th class maths book...
is that ok?
Gurdeesh
somyasinha SaysThe sum of 40 natural numbers is 801. What is the minimum possible value of their LCM?
the options plz
somyasinha SaysThe sum of 40 natural numbers is 801. What is the minimum possible value of their LCM?
The idea is to maximize the number that has max factors possible.
24*33 + 1*6 +3*1= 801
Hence min LCM = 24
The idea is to maximize the number that has max factors possible.
24*33 + 1*6 +3*1= 801
Hence min LCM = 24
Sir, is there any approach to this type of questions or hit n trial.
I mean, how do you got to 33 24's n rest...??
Sir, is there any approach to this type of questions or hit n trial.
I mean, how do you got to 33 24's n rest...??
Dude this is not exactly by hit and trial.
See 801/40 is approxx 20. Now in order to minimze the LCM we must have one number, lets say X, needs to be repeated the max number of times that it can be. Now if the X stays close enough to 20 on either side we are looking at maximizing the number of times X can appear. Also, we cannot take X too less because other wise the occurence will itself surpass the limit of 40 numbers that we had.
Now there has to be at least one odd number included in the sum. Now you need to check among 18,20,22 and 24 individually; however here also u need to keep that number which has relatively more smaller factors such as 2, 3, 4. Now when you compare these 24 wins
Please help me out with this question:
Find the number of numbers between 200 to 300 which leave a remainder of 5 when divided by 7.
a.)16
b.)15
c.)14
d.)None of these
Ans: b.)15
Please help me out with this question:
Find the number of numbers between 200 to 300 which leave a remainder of 5 when divided by 7.
a.)16
b.)15
c.)14
d.)None of these
Ans: b.)15
we see numbers 203,210,....294, 301 divides 7 completely
to get a remainder 5 and no. b/w 200 and 300
no.s are 201,208,........299
this is an AP with a1= 201, l=200, c.d.=7
hence no. of terms can be calculated easily...
299=201+ (n-1)7
n=15
first num is 201 last is 299
total is 15 numbers.