How many four digit numbers are in base 7,if u are counting the numbers in the same number system
a) 2058 b) 5666 c) 6000 d) None of these
i got it as 2058 but it is apparently wrong....can someone please explain the answer...
thanks in advance 
How many four digit numbers are in base 7,if u are counting the numbers in the same number system
a) 2058 b) 5666 c) 6000 d) None of these
i got it as 2058 but it is apparently wrong....can someone please explain the answer...
thanks in advance :cheers:
1st no. is (1000)base 7 = 343
last no. is (6666)base 7 = 2400
Total nos. is 2058 but in the base 10.. so you have to convert it to base 7
So, Answer is (c) 6000.. Hope clear..
Know the CATs: Number Systems
puys..
this site has almost 100 good questions on number systems...
but unsolved..
can anyone give me the solutions?please
and do u like the site?
puys..
this site has almost 100 good questions on number systems...
but unsolved..
can anyone give me the solutions?please
and do u like the site?
a request to all mathematicians here. plz explain me the chinese remainder theorem. i saw a qsn on first page of this thread:
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
was totally bounced with the method they used to solve it
E(17) = 16, (38^16!)^1777 mod 17 = 38^16k mod 17 = 1
this was one more method they used.... i have no idea about this. do you?
PLEASE SOLVE THIS ONE:
Two different numbers when divided by the same divisor left remainders of 11 and 21 respectively. When the numbers' sum was divided by the same divisor, the remainder was 4. What was the divisor?
OPTIONS:
36
28
12
9
PLEASE SOLVE THIS ONE:
Two different numbers when divided by the same divisor left remainders of 11 and 21 respectively. When the numbers' sum was divided by the same divisor, the remainder was 4. What was the divisor?
OPTIONS:
36
28
12
9
Lets say the divisor be D, then
p = Dn + 11
q = Dk + 21
p + q = D(n + k) + 32 = Dr + 4
=> D has to be a factor of 28 (32 - 4) and also it should be greater than 21
=> D = 28
puys,
some simple no.system problems from TIME material:
please give me the solutions to these:
1.The product of a 7 digit,10 digit and a twelve digit no. is a digit no.
a.27..........b.28..........c.28 or 29..........d.27 or 28 or 29
2.The no. of digits n the square root of a thirteen digit no. is
a.6.......b.7...........c.8.................4.either option a or b
3.x^2-4x+1=0, find the value of x^4+1/x^4
a.42 b.68 c.84 d.194
4.x+1/x=4 , find the value of x^4-1/x^4 (x>1)
a.112 b.112root3 c.224 d 224root3
kindly give the explanations to ur answers too
Gurdeesh
puys,
some simple no.system problems from TIME material:
please give me the solutions to these:
1.The product of a 7 digit,10 digit and a twelve digit no. is a digit no.
a.27..........b.28..........c.28 or 29..........d.27 or 28 or 29
2.The no. of digits n the square root of a thirteen digit no. is
a.6.......b.7...........c.8.................4.either option a or b
3.x^2-4x+1=0, find the value of x^4+1/x^4
a.42 b.68 c.84 d.194
4.x+1/x=4 , find the value of x^4-1/x^4 (x>1)
a.112 b.112root3 c.224 d 224root3
kindly give the explanations to ur answers too
Gurdeesh
1.c
2.b
3.d
4.c
1.c
2.b
3.d
4.c
1.d
and
4.b
2 nd 3 r correct
but i need the solutions and explanations..
pls explain these fundas to me..
Gurdeesh
puys,
some simple no.system problems from TIME material:
please give me the solutions to these:
1.The product of a 7 digit,10 digit and a twelve digit no. is a digit no.
a.27..........b.28..........c.28 or 29..........d.27 or 28 or 29
i did it this way...let us take the product of small nos. say 11*111=1111(no. of digits is 4)
99*999=98901(no. of digits is 5)
so when a 2 digit no. is multiplied with a three digit no..max no. of digit is 5 and min no. of digits is 4
so when when 7 digit is multiplied with 10 digits no. the max no. of digit will be 17 and min no. of digits will be 16.
when 17 digit no. is multiplied with 12 digit no. --max digit=29 and min=28
when 16 digit no. is multiplied with 12 digit no. --max digit=28 and min=27
hence option d.
2.The no. of digits n the square root of a thirteen digit no. is
a.6.......b.7...........c.8.................4.either option a or b
i did this sums by using the options...
a 6 digit no. square will give me max. (6+6) digits=12
a 7 digit no. square will give me min 13 and max.14 digits
a 8 digit no. square will give me a no. with max 16 digits and min 15 digits.
so 7 will be the answer....
3.x^2-4x+1=0, find the value of x^4+1/x^4
a.42 b.68 c.84 d.194
we can write is as x+1/x=4
squaring both sides you'll get x^2+1/x^2=14
again squaring both sides you'll get x^4+1/x^4=194
4.x+1/x=4 , find the value of x^4-1/x^4 (x>1)
a.112 b.112root3 c.224 d 224root3
x+1/x=4
we can write is as x^2-1=4x
now x^4-1/x^4=(x^2+1/x^2)(x^2-1/x^2)=14*(x+1/x)(x-1/x)
=14*4*(x^2-1)/x=56*4x/x=224
Hi puys,
these r some problems on number systems from Arun Sharma:
im giving the answers too but i need the solution steps:
please give the explanations and solutions to these:
1.What is the remainder when (1!)^3 + (2!)^3.......(1152!)^3 is divided by 1152
ans is 225
2.find the 238383rd term in the series 123456789101112
ans is 9
3.what is the remainder when 9+9^2+9^3............9^(2n+1) is divided by 6.....ans is 3
4.Find the last 2 digits in the following nos.:
101X102X103X197X198X199
ans is 54
5.last 2 digits of following nos.:
65X29X37X63X71X87
ans is 95
6.what is the right most digit preceeding zeros in the value of 20^(53)
ans is 2
7.find the remainder when 123456789101112.....50 is divided by 16
ans is 6
8.the last 2 digits in the multiplication 122X123X125X127X129 will be...
ans is 50
9.last digit of the no. 1^2+2^2+3^2.........99^2
ans is 0
10.the last digit of the no. 1^3+2^3+3^3.......99^3 is...
ans is 0
11. when (2222)^5555+(5555)^2222 is divided by 7,the remainder is...
ans is 0
12.what is the remaindr of 2^(100) when divided by 3
ans is 1
Gurdeesh
Hi puys,
these r some problems on number systems from Arun Sharma:
im giving the answers too but i need the solution steps:
please give the explanations and solutions to these:
9.last digit of the no. 1^2+2^2+3^2.........99^2
ans is 0
Sol: unit digit squareof 1 =1
2=4
3=9
4=6
5=5
6=6
7=9
8=4
9=1
This end up 45 and same cycle repeats from 11.
SO 45*10=450
which gives 0 as unit digit
10.the last digit of the no. 1^3+2^3+3^3.......99^3 is...
ans is 0
Sol: same as above just take cube and see the cycle
11. when (2222)^5555+(5555)^2222 is divided by 7,the remainder is...
ans is 0
Sol: Euler of 7 is 6
so 5555/6 gives 5 so we have 3^5
and 2222/6 gives 2
so it gives (3^5+4^2)/7 which give us 0 remainder
12.what is the remaindr of 2^(100) when divided by 3
ans is 1
Euler of 3 is 2 so 2^100/3 will give 1 remainder
5.last 2 digits of following nos.:
65X29X37X63X71X87
ans is 95
Same sol: divide by 100
so it gives 3 remainder when divided by 4 and 20 when divided by 25
so 95 satisfy this
6.what is the right most digit preceeding zeros in the value of 20^(53)
ans is 2
Sol: ( 2^53 * 10^53 )
Power of 2 ends with 2,4,8,6 and it repeats so 2^53 ends with 2 and it is the ans
7.find the remainder when 123456789101112.....50 is divided by 16
ans is 6
Sol: check last 3 digits
950/16 gives 6 remainder
8.the last 2 digits in the multiplication 122X123X125X127X129 will be...
ans is 50
Sol: divide by 100......break into 25 and 4
with 4 we get 2 remainder and it is perfectly divisible by 25
so applying chinese remainder theorem least no. we got 50 which satisfy both conditions
puys,
some more questions:
please give detailed solutions to these:
1.What is the remainder when 128^(1000) is divided by 153.
2.Find remainder when 32^33^34 is divided by 11.
3.Find remainder when 5^38 is divided by 11.
4.arrange the following in ascending order 100^300 ,
under root(300^300) , 300!
5.what digit will exist in the units place:
(347)^42-(763219)^2X(53213)^4
6.How many zeroes exist at the end of the product:
10^4X12^3X15^4X21^3X16^3X25^3
7.what digit exists at the units place of
39^42X27^23X36^12
8.find the remainder when 24^5 is divided by 5
thnx,
Gurdeesh
puys,
some more questions:
please give detailed solutions to these:
1.What is the remainder when 128^(1000) is divided by 153.
2.Find remainder when 32^33^34 is divided by 11.
3.Find remainder when 5^38 is divided by 11.
4.arrange the following in ascending order 100^300 ,
under root(300^300) , 300!
5.what digit will exist in the units place:
(347)^42-(763219)^2X(53213)^4
6.How many zeroes exist at the end of the product:
10^4X12^3X15^4X21^3X16^3X25^3
7.what digit exists at the units place of
39^42X27^23X36^12
8.find the remainder when 24^5 is divided by 5
thnx,
Gurdeesh
1)128^(1000) is divided by 153.
153=17x9
128^1000/17
=>9^1000/17
(81)^500/17
=>13^500
=>(169)^250
=>1
=>128^1000/9
=>2^1000/9
=>2^996/9=1
=>7/9
Now you have to find smallest number which gives 7 as remainder with 9 and 1 with 17. With options u can easily make out.
2) 32^33^34 is divided by 11
32^33^34/11
=>(-1)^33^34/11
=>33^34 is always odd
=>-1/11
=>10 is the remainder
3)5^38 is divided by 11.
=>apply format little theorem
5^10/11=1
=>5^8/11
=>4
4)arrange the following in ascending order 100^300 ,
under root(300^300) , 300!
300! has 300!/5+300!/25+300!/125
=>60+12+2=>74 Zeroes
=>under root(300^300), 300!, 100^300
5)(347)^42-(763219)^2X(53213)^4
unit digit of (347)^42=>9
unit digit of (763219)^2=>1
unit digit of (53213)^4=>1
=>9-1x1
=>8
6)How many zeroes exist at the end of the product:
10^4X12^3X15^4X21^3X16^3X25^3
no of 2`s is 4+6+12=>22
no of 5`s is 4+4+6=>14
no of zeroes is 14
7).what digit exists at the units place of
39^42X27^23X36^12
=>1x3x6
=>8
8 )find the remainder when 24^5 is divided by 5
24^5 /5
=>4^5/5
=>4
whats the OA
from morning_star
"Find the remainder if 30^40 is divided by 17."
REPLY - I cam 2 knw such questions won't b asked in CAT.
divisibility of 17 & 19 has no rule... v hav 2 check it & v can't check suh no. 30^40.
from morning_star
"Find the remainder if 30^40 is divided by 17."
REPLY - I cam 2 knw such questions won't b asked in CAT.
why do u say so, any sort of prob can be asked..Dont have any such assumptions
for this, 30^40/17
=>13^40/17
=>(13^2)^20/17
=>(-1)^20/17
=>1
why do u say so, any sort of prob can be asked..Dont have any such assumptions
for this, 30^40/17
=>13^40/17
=>(13^2)^20/17
=>(-1)^20/17
=>1
did not understand how tat 13^2 become -1 plz could u explain tat
