1)Computer takes an input of a number N and X where X is a factor of the number N.N is 83p796161q and X is equal to 11 where o
1) 3
2) 2^44
are these ans correct..?
1) 3
2) 2^44
are these ans correct..?
Thnx..... but how to find euler number ?
any specific formula ??
the ruke is if N is the number then the euler no. will be
=N(1-1/p)(1-1/q)...
where p,q are the prime factors of N
@ ksagarwa: solutions please.
Hi guys,
I have a question though i don't think it was in the arun sharma book.
HCF of (77777.........2005 times, 3333333.......2010 times)?
Really need someone to solve this.
thanks
Bgoel
I have another one.
Suppose that a and b are digits, not both nine and not both zero and the repeating decimal 0.ab is expressed as fraction in lowest terms. How many different denominators are possible.? (PS: the the bar under ab is actually on top of it ie the the representation of repeating numerals).
options:
a) 3 b)4 c)5 d)8 e)9
Hi guys,
I have a question though i don't think it was in the arun sharma book.
HCF of (77777.........2005 times, 3333333.......2010 times)?
Really need someone to solve this.
thanks
Bgoel
Is the answer 7 ?
Hi guys,
I have a question though i don't think it was in the arun sharma book.
HCF of (77777.........2005 times, 3333333.......2010 times)?
Really need someone to solve this.
thanks
Bgoel
77777.........2005 times = 7(11111..... 2005 times)
2005 = 5*401
So this expression is divisible by 11111 and 1111.... 401 times.
3333333.......2010 times = 3(11111.... 2010 times)
2010 = 5 * 2* 3 *67
So this expression is divisible by 11,111,11111, and 1111...67 times.
EDIT:
Also, 3333333.......2010 times will be divisible by 7
Any digit repeated 6 times, is div by 7, and 2010 is a multiple of 6.
So i think Cumulative HCF = 77777
I have another one.
Suppose that a and b are digits, not both nine and not both zero and the repeating decimal 0.ab is expressed as fraction in lowest terms. How many different denominators are possible.? (PS: the the bar under ab is actually on top of it ie the the representation of repeating numerals).
options:
a) 3 b)4 c)5 d)8 e)9
x = .ababab.....
100x = ab + x
x = ab/99
The denominatore for this x, can be 99 or any factor of 99.
99 = 3^2 * 11
factors of 99 = 6
But we cannot have the denominator as 1, since ab cannot be 99 (as given in question)
So total different denomnators = 6-1 = 5
Hi Angadbir ,
thanks for the quick reply . A couple of points i didn't get though.
a)factors of 99 = 6 ?
b)2005 = 5*401
So this expression is divisible by 11111 and 1111.... 401 times. ?
c)2010 = 5 * 2* 3 *67
So this expression is divisible by 11,111,11111, and 1111...67 times. ?
can you please clarify these statements a bit further.
thanks.
Hi Angadbir ,
thanks for the quick reply . A couple of points i didn't get though.
a)factors of 99 = 6 ?
b)2005 = 5*401
So this expression is divisible by 11111 and 1111.... 401 times. ?
c)2010 = 5 * 2* 3 *67
So this expression is divisible by 11,111,11111, and 1111...67 times. ?
can you please clarify these statements a bit further.
thanks.
a)factors of 99 = 6 ?
if N = a^x * b^y * c^z ...
where a,b,c are prime no.s, then
Factors of N = (x+1)(y+1)(z+1)...
b)2005 = 5*401
So this expression is divisible by 11111 and 1111.... 401 times. ?
Take a smaller no. 111111
This no has six 1's
6 = 2*3
So 11 (2 1's) repeated 3 times is 11111.
AND 111 (3 1's) repeated 2 times is 11111.
So we can say that, 111111 is div by both 11 and 111
To understand better, take another no. 111111... 15 times. And observe that this will be divisible by 11111 (5 ones) and 111 (3 ones). Manually divide and observe if necessary.
Hi Angadbir ,
thanks for the quick reply . A couple of points i didn't get though.
a)factors of 99 = 6 ?
b)2005 = 5*401
So this expression is divisible by 11111 and 1111.... 401 times. ?
c)2010 = 5 * 2* 3 *67
So this expression is divisible by 11,111,11111, and 1111...67 times. ?
can you please clarify these statements a bit further.
thanks.
99 = 3^2 * 11 ^ 1
So, number of factors = (2+1)(1+1) = 6, i.e, 1,3,9,11,33,99
2005 = 5*401
That means the expression is divisible by 11... 5 times, i.e., 11111
and 11111111 ... 410 times, i.e., 401 times 1.
Similarly, for the other one..
2010 = 2* 3 * 5 * 67
So, divisible by 11, 111, 11111, 1111.... 67 times
Any clues for this one?
N=(41^4+31^4)-(21^4+11^4) then N is always divisible by ?
a.)42
b.)65
c.)30
d.)20
e.)65 & 20
Any clues for this one?
N=(41^4+31^4)-(21^4+11^4) then N is always divisible by ?
a.)42
b.)65
c.)30
d.)20
e.)65 & 20
e) 65 and 20
Any clues for this one?
N=(41^4+31^4)-(21^4+11^4) then N is always divisible by ?
a.)42
b.)65
c.)30
d.)20
e.)65 & 20
N=(41^4+31^4)-(21^4+11^4)
=> (41^4 - 21^4) + (31^4 - 11^4)
=> (41-21)k + (31-11)k' --------> (a^4 - b^4) is always div by (a-b)
=>20k''
Also, we can re-write it as
(41^4 - 11^4) + (31^4 - 21^4)
=> (41+11)k + (31+21)k' ---------> (a^4 - b^4) is always div by (a+b)
=>52k''
N is divisible by 20 (4*5), and N is divisble by 52 (13*4).
We can safely say that, N is divisible by 5 and 13.
That means N is definitely divisible by 65 (13*5) as well.
So option e)
Brilliant!!
Q.) X needs to pay 2010 for her dinner. She has an unlimited supply of 2,5 and 10 rupees notes.In how many ways can she pay?
a.)20403
b.)20503
c.)20203
d.)20303
e.)none
Q.) Find the sum of the perimeters of all rectangles with sides as integral values and area 120 units?
a.)720
b.)480
c.)1440
d.)360
e.)62
a.)20403
b.)20503
c.)20203
d.)20303
e.)none
Q.) Find the sum of the perimeters of all rectangles with sides as integral values and area 120 units?
a.)720
b.)480
c.)1440
d.)360
e.)62
what is the remainder when
1. 128^1000 is divided by 153
2.50^56^52 is divided by 11
33^34^35 is divided by 7
what is the remainder when
1. 128^1000 is divided by 153
is this answer=1
katariags Saysis this answer=1
explain plz
Q.) X needs to pay 2010 for her dinner. She has an unlimited supply of 2,5 and 10 rupees notes.In how many ways can she pay?
a.)20403
b.)20503
c.)20203
d.)20303
e.)none
2a + 5b + 10c = 2010
a should be a multiple of 5, say, a = 5k, then
2k + b + 2c = 402
b should be even
=> b = 2n
=> k + n + c = 201
=> C(201 + 3 - 1, 3 - 1) = C(203, 2) = 20503 ways
Q.) Find the sum of the perimeters of all rectangles with sides as integral values and area 120 units?
a.)720
b.)480
c.)1440
d.)360
e.)62
Possible rectangles = (1, 120), (2, 60), (3, 40), (4, 30), (5, 24), (6, 20), (8, 15), (10, 12)
Sum of perimeters = 2*Sum of all factors = 2*(1 + 2 + 4 + 8 )(1 + 3)(1 + 5) = 2*15*4*6 = 720
what is the remainder when
1. 128^1000 is divided by 153
2.50^56^52 is divided by 11
33^34^35 is divided by 7
128^1000(mod 153)
153 = 9*17
9 = 2^3 + 1
=> 128^1000 = 2^7000 = 2*2^6999 = 7(mod 9)
17 = 2^4 + 1
=> 128^1000 = 2^7000 = 1 (mod 17)
=> N = 9k + 7 = 17n + 1
=> k = 2n - (n + 6)/9, k will be integer for n = 3
=> N = 153k + 52
50^56^52 is divided by 11
Eulers number of 11 is 10
56^52 = 6 (mod 10)
=> 50^56^52 = 50^6(mod 11) = 6^6 (mod 11) = 3^3 (mod 11) = 5 (mod 11)
33^34^35 is divided by 7
33^34^35 = 2^34^35 (mod 7)
We know that 2^3 = 1 (mod 7)
34^35 = 1 (mod 3)
=> 2^34^35 = 2^(3k + 1) (mod 7) = 2 (mod 7)
Q.) X needs to pay 2010 for her dinner. She has an unlimited supply of 2,5 and 10 rupees notes.In how many ways can she pay?
a.)20403
b.)20503
c.)20203
d.)20303
e.)none
X needs to pay total of 2010 using Rupee notes of 2,5 and 10
2x + 5y + 10z = 2010
where x,y,z are integers.
Rearranging we get,
5y+10z = 2010 - 2x
y + 2z = 402 - 2x/5
Since y,z are integers LHS is an integer value, so RHS also needs to be an integer. Thus, x is a multiple of 5. x=5x'
2x = 10x'
Again rearranging we get,
10x'+10z = 2010 - 5y
x' + z = 2010 - y/2
Since x' and z are integers, LHS is an integer value. So y/2 should be an integer. Thus, y is a multiple of 2. y = 2y'
5y = 10y'
The equaiton now becomes:
10x' + 10y' + 10z = 2010
x'+y'+z=201
Total non negative integral solns for this = C(201 + 3 -1, 3-1) = C(203,2) = 20503
Q.) Find the sum of the perimeters of all rectangles with sides as integral values and area 120 units?
a.)720
b.)480
c.)1440
d.)360
e.)62
Let sides be a,b
Both a,b are integers.
Perimeter = 2(a+b)
So we are basically lookin for all distinct values of (a+b)
ab = 120
We can simply write the pairs:
(a,b)......SUM
(1,120) ->121
(2,60) -> 62
(3,40) -> 43
(4,30) -> 34
(5,24) -> 29
(6,20) -> 26
(8,15) -> 23
(10,12) -> 22
Total = 360
Perimeter = 2*360 = 720