Gandhi88 SaysHow many factors does 16! have??
16!=2^15.3^6.5^3.7^2.11^1.13^1
5376---ans.?
16!=2^15.3^6.5^3.7^2.11^1.13^1
5376---ans.?
Gandhi88 SaysDont know whats the shortest sol for it. But I managed the answer after some time by trying combinations and adding em up..
You could it this way..
1,3,5,7,9
so you can have 120 possible 5 digit numbers
which can be divided into 24 sets of 5 numbers each,you'll see that you can have one set as below
13579
35791
57913
79135
91357
which adds up to 277775
you'll have 23 other sets which add up to the same sum(since no digit can be repeated) i.e each column and each row has the digits 1,3,5,7,9
so the final sum is gonna be 277775 * 24 = 6666600
can anyone sove this for me..
If we divide a two-digigt number by the sum of its digits,we get 4 as quotient and 3 as a remainder.Now if we divide that two-digit number by the product of its digits,we get 3 as a quotient and 5 as remainder.Find the two dogit number
In an organization, the daily average wages of 20 illiterate employees is decreased from Rs 25 to Rs 10, thus the average salary of all the literate and illiterate employee is decreased by Rs 10 per day. The number of literate employees working in the organisation are:
a) 15 b) 20 c)10 d)25
please show the working.. thanx in advance 😃 if possible please also answer(with working) the question number 19 n 20 in Pg-92 from the Quant book by arun sharma..
can anyone sove this for me..
If we divide a two-digigt number by the sum of its digits,we get 4 as quotient and 3 as a remainder.Now if we divide that two-digit number by the product of its digits,we get 3 as a quotient and 5 as remainder.Find the two dogit number
let the 2 digit number be 10x+y.. so according to question 10x+y=4(x+y)+3.. where 4 is the quotient, 3 the remainder and x+y is the sum of digits.. also according to question 10x+y=3xy+5.. so by eqating thz 2 eqnz.. we get 4x+4y-3xy=2.. by a few hit and trials, we can get the number as 23.. :)
this is kinda more eqation based, there must b a bttr way to answer this.. lemme work on it.. 😃
In an organization, the daily average wages of 20 illiterate employees is decreased from Rs 25 to Rs 10, thus the average salary of all the literate and illiterate employee is decreased by Rs 10 per day. The number of literate employees working in the organisation are:
a) 15 b) 20 c)10 d)25
please show the working.. thanx in advance 😃 if possible please also answer(with working) the question number 19 n 20 in Pg-92 from the Quant book by arun sharma..
Initial total wages of illiterate emp = 25*20 = 500
Total wages after deduction = 10*20 = 200
Difference = 500-200 = 300
i.e. rs. 300 got deducted from total wages of literate+illiterate employees.
Now given, this deacrese in total avegare is Rs.10
=> Total employees = 300/10 = 30
Literate emp = 30-20 = 10(c)
List the other ques...
In an organization, the daily average wages of 20 illiterate employees is decreased from Rs 25 to Rs 10, thus the average salary of all the literate and illiterate employee is decreased by Rs 10 per day. The number of literate employees working in the organisation are:
a) 15 b) 20 c)10 d)25
please show the working.. thanx in advance 😃 if possible please also answer(with working) the question number 19 n 20 in Pg-92 from the Quant book by arun sharma..
ans is 10
sol-
let literate men b x and let y be his sal so
xy+20x25/x+20=z where z b the avg of dere per day sal
now acoording to ques
xy+20x10/x+20=z-10
calculate u"ll get x=10
find two three digit numbers whose sum is a multiple of 504 and the quotient is a multiple of 6
parthabar Saysfind two three digit numbers whose sum is a multiple of 504 and the quotient is a multiple of 6
Let the ,3-digit numbers be n1 and n2
n1+n2= 504a
n2/n1=6b
where a,b are integers.....
so,n2=6b(n1)
n2+n1=504a
n1+6b(n1)
n1(6b+1)=504a-----------(1)
use smallest values,put a=1 and b=1
so 7n1=504
or n1=72
which is not a 3 digit number.
so take a=2,b=1
then put it in (1)
7n1=504.2
n1=144, therefore n2=864
Hence, the 3-digit numbers are 144 and 864
Define a number K such that it is sum of the squares of the first M natural numbers (ie K =1^2+2^2+3^2......+M^2) where M
thanks in advance
Define a number K such that it is sum of the squares of the first M natural numbers (ie K =1^2+2^2+3^2......+M^2) where M
Well i tried very naive approach
k= (m)(m+1)(2m+1)/(24) .. (6*4 as we need a no divisible by 4)
put the values in m 1,2,....
values of k will b 7,8,15 at this pt observe this relation ie no = sum of previous 2 nos
Then as per above deductions numbers will b 7,8,15,23,38
Well i tried very naive approach
k= (m)(m+1)(2m+1)/(24) .. (6*4 as we need a no divisible by 4)
put the values in m 1,2,....
values of k will b 7,8,15 at this pt observe this relation ie no = sum of previous 2 nos
Then as per above deductions numbers will b 7,8,15,23,38
same process by me also
it takes time
any one hav a more shrt cut
Define a number K such that it is sum of the squares of the first M natural numbers (ie K =1^2+2^2+3^2......+M^2) where M
thanks in advance
k = M(M + 1)(2M + 1)/6
For k to be divisible by 4, one of M or (M + 1) should be divisible by 8 (as we have an extra two in denominator), as (2M + 1) will always be odd.
For M = 8, 16, 24, 32, 40, 48; M will be divisible by 8 and
for M = 7, 15, 23, 31, 39, 47, 55; (M + 1) will be divisible by 8
So, for total 13 values of M 55, k will be divisible by 4.
k = M(M + 1)(2M + 1)/6
For k to be divisible by 4, one of M or (M + 1) should be divisible by 8 (as we have an extra two in denominator), as (2M + 1) will always be odd.
For M = 8, 16, 24, 32, 40, 48; M will be divisible by 8 and
for M = 7, 15, 23, 31, 39, 47, 55; (M + 1) will be divisible by 8
So, for total 13 values of M 55, k will be divisible by 4.
How can we be sure that in denominator will always be cancelled by some of the values of numerator.The 3 part is disturbing me.
plz help:drinking:
Define a number K such that it is sum of the squares of the first M natural numbers (ie K =1^2+2^2+3^2......+M^2) where M
thanks in advance
got it
for k=m(m+1)(2m+1)/6 to be divisible by 4
there are just 13 values for which m=4,8,12,16......52 (this is itself taking care of the rest of the conditions.
Is there any catch?
Hi..
here's my question -
Q. Find the sum of the series
1/(2^0.5 + 1^0.5) + 1/(2^0.5 + 3^0.5) + ......... + 1(120^0.5 + 121^0.5)
I didnt find its answer was posted on pg 212.
So is the answer to this is 10.
rationalising each fraction individually we ll get 1 in their denominator.
in numerator only 121^0.5 and -(1)^.5 remains. i.e. 11-1=10
Am i right?
How can we be sure that in denominator will always be cancelled by some of the values of numerator.The 3 part is disturbing me.
plz help:drinking:
The same thing is bothering me also. While finding the values, we didn't bother about the 3 in denominator.
deepak_pgi SaysThe same thing is bothering me also. While finding the values, we didn't bother about the 3 in denominator. :lookround:
6=2*3.....So we need to find values of M for which M,(M+1) and (2m+1) will cancel 6 entirely.For even values of M,either (M+1) or (2M+1) will cancel the 3 from denominator.For odd values of M other than multiples of 3,(2M+1) cancels the 3..and for multiples of 3 as well,3 of denominator gets cancelled.Thats why we aint bothered abt the 3 in denominator
spectramind07 Says6=2*3.....So we need to find values of M for which M,(M+1) and (2m+1) will cancel 6 entirely.For even values of M,either (M+1) or (2M+1) will cancel the 3 from denominator.For odd values of M other than multiples of 3,(2M+1) cancels the 3..and for multiples of 3 as well,3 of denominator gets cancelled.Thats why we aint bothered abt the 3 in denominator
As per you, for odd values of M other than multiples of 3,(2M+1) cancels the 3. But this is not the case with 23 as 2*23+1=47 , which is not divisible by 3. (M+1=24) will be responsible for the cancellation of 3 in this case.
PS:- Just wanted to bring this case to your notice.
spectramind07 Says6=2*3.....So we need to find values of M for which M,(M+1) and (2m+1) will cancel 6 entirely.For even values of M,either (M+1) or (2M+1) will cancel the 3 from denominator.For odd values of M other than multiples of 3,(2M+1) cancels the 3..and for multiples of 3 as well,3 of denominator gets cancelled.Thats why we aint bothered abt the 3 in denominator
As per you, for odd values of M other than multiples of 3,(2M+1) cancels the 3. But this is not the case with 23 as 2*23+1=47 , which is not divisible by 3. (M+1=24) will be responsible for the cancellation of 3 in this case.
PS:- Just wanted to bring this case to your notice.
Actually any ways it does. wat spectramind07 is saying may not be perfectly true(I wish it is). But when you work out to find values of m,m+1 or 2m+1; you will find each among one of the three part is divisible by 3.and the sequence will repeat after m=4.
So i guess we all are answered now.
