Find the number of divisiors of 1080 excluding the throughout divisors which are perfect squares
a)28
b)29
c)30
d)31
Find the sum of divisiors of 544 which are perfect squares
32
64
42
21
Kindly reply
Find the number of divisiors of 1080 excluding the throughout divisors which are perfect squares
a)28
b)29
c)30
d)31
Find the sum of divisiors of 544 which are perfect squares
32
64
42
21
Kindly reply
The first question:
1080=(2^3)*(3^3)*(5)
thus number of factors=4*4*2=32
out of these we have (2^2),(3^2) and (2^2*3) as perfect squares so 29 divisors.
The second question:
544=((2^5)*17)
perfect squares can only be 2^2=4 and 2^4=16; thus the sum should be 20 which is not in the options!
but the answer to the first question is 28
and the answer to the second question is 21
So is the answer given in the book is wrong or there is any better approach for this.
Kindly Help
Find the number of divisiors of 1080 excluding the throughout divisors which are perfect squares
a)28
b)29
c)30
d)31
Find the sum of divisiors of 544 which are perfect squares
32
64
42
21
Kindly reply
(1) 1080 = 2^3 * 3^3 * 5
=> Total number of factors = 4*4*2 = 32
=> Perfect sqaure part is 2^2 * 3^2
=> Sqaure root = 2*3 => Number of factors = 2*2 = 4
So, factors which are non-perfect sqaures = 32-4 = 28
Note: Perfect sqare factors are 1, 2^2, 3^2 and 2^2 * 3^2
(2) 544 = 2^5 * 17
=> Perfect sqare part = 2^4
=> Perfect sqaure factors => 1, 2^2 and 2^4
=> Sum = 21
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Find the maximum value of nsuch that 77x42x37x57x30x90x70x2400x2402x243x343 is perfectly divisible by 21^n.
a) 9 b) 11
c) 10 d) 7
Ans: is 7 (d).
But I am getting 6 as answer.
Can anyone please explain how it could be 7? And where I could be wrong in calculating!
Thanks,
:)
77,42,70 contain 1's 7.
343 =7*7*7
total 6, answer should be 6 only.
Find the number of divisiors of 1080 excluding the throughout divisors which are perfect squares
a)28
b)29
c)30
d)31
Find the sum of divisiors of 544 which are perfect squares
32
64
42
21
Kindly reply
1. 1080 = 2*5*3*3*3*2*2
total divisor 4*4*2 =32
now perfect square 1,9,4,36 so total 28
2. 544 = 2*2*2*2*2*17
4+16+1 =21
Find the maximum number of n such that 157! is perfectly divisible by 12^n
a)77
b)76
c)75
d)78
The factors of the number 12 are 2*2*3
if we apply 157/3+ 157/9+ 157/27+157/81 we get 77
and what I think is that I need to take 2^2 into account too
so by it
157/2+157/4+157/8+157/16+157/32+157/64+157/128
the total will give 132
since we have 2*2 as factor so we need 2 2's in the factor
soo it will give 33
but the answer is 75
What am I doing wrong?
Kindly Help
Find the maximum number of n such that 157! is perfectly divisible by 12^n
a)77
b)76
c)75
d)78
The factors of the number 12 are 2*2*3
if we apply 157/3+ 157/9+ 157/27+157/81 we get 77
and what I think is that I need to take 2^2 into account too
so by it
157/2+157/4+157/8+157/16+157/32+157/64+157/128
the total will give 132
since we have 2*2 as factor so we need 2 2's in the factor
soo it will give 33
but the answer is 75
What am I doing wrong?
Kindly Help
dude answer in book is correct from my point of view....
you need to check power of 3...which will be there..which will come to 75....
@thinkace bhai correct me if i am wrong......
Thanks for taking time to reply
but why wont we take 2 into consideration?
its also a prime number and is in power as in there are 2 multiples of 2
Thanks for taking time to reply
but why wont we take 2 into consideration?
its also a prime number and is in power as in there are 2 multiples of 2
dude there is nothing like thanks chill out...
and we always take only that number into consideration whose power will be less in the number....jaise
if we wanna check how many 30 will be there in any given factorial then we will only check for 5...
edit-please check calculation i think you have done some mistakes
yes by 3 I get 75. Thats a mistake on my part.
But in one solved example
Find the highest power of 175 which divides 344!
here in the solution hints its given that
175 = 5*5*7
So we will need to check the number of 5^2 and the number of 7's
Thats why I am confused that why are we taking 5^2 into consideration
Find the maximum number of n such that 157! is perfectly divisible by 12^n
a)77
b)76
c)75
d)78
The factors of the number 12 are 2*2*3
if we apply 157/3+ 157/9+ 157/27+157/81 we get 77
and what I think is that I need to take 2^2 into account too
so by it
157/2+157/4+157/8+157/16+157/32+157/64+157/128
the total will give 132
since we have 2*2 as factor so we need 2 2's in the factor
soo it will give 33
but the answer is 75
What am I doing wrong?
Kindly Help
You are using wrong formula.....
The factors of the number 12 are 2*2*3
To find the highest power of 3 which divides 157! completely....
apply + + +
where [] is the greatest integer less than or equal to....i.e. greatest integer function....as per formula given in CL material
we get 75....
To find the highest power of 2 which divides 157! completely....
apply + + ++++
we get 152....
Thus highest power of 2*2 which divides 157! completely is 76
So, answer is 75
Problem Based on Remainder theorem:
Find the Remainder when 51^203 is divided by 7.
a) 4 b) 2
c) 1 d) 6
Please give detailed explanation.
Problem Based on Remainder theorem:
Find the Remainder when 51^203 is divided by 7.
a) 4 b) 2
c) 1 d) 6
Please give detailed explanation.
It is a)4
49 is div. by 7 so rem 2
2^203 / 7
8^ 61 * 4 / 7
so remainder :4
Regards,
Never Back DownIt is a)4
49 is div. by 7 so rem 2
2^203 / 7
8^ 61 * 4 / 7
so remainder :4
Regards,Never Back Down
Thanks... I calculated the same way but not sure if the way was correct or not.
Find the maximum number of n such that
42*57*92*91*52*62*63*64*65*66*67
is perfectly divisible by 42^n
4
3
5
6
Kindly Help
Find the maximum number of n such that
42*57*92*91*52*62*63*64*65*66*67
is perfectly divisible by 42^n
4
3
5
6
Kindly Help
my take is n=3...what is the correct answer...
i will post my approach if my answer is correct
Hi.
The answer is 3
Can you tell me whats the approach?
Find the maximum number of n such that
42*57*92*91*52*62*63*64*65*66*67
is perfectly divisible by 42^n
4
3
5
6
Kindly Help
solution is simple we just have to look for the number of 42 we can form by these digits...
42=2*3*7
now by looking at number you can see that there are good amount of 2 and 3 are present....
so just for the number of 7 present in this multiplication....that is there are three 7 are present...
hence n=3
i hope this help..
Find the number of zeros in
1! *2! *3! *4!*5!*-------------------*50!
235
12
262
105