my query is:
1. Find the maximum value of n such that
570*60*30*90*100*500*700*343*720*81 is perfectly divisible by 30^n.
Its answer is 11.
my query is:
1. Find the maximum value of n such that
570*60*30*90*100*500*700*343*720*81 is perfectly divisible by 30^n.
Its answer is 11.
The maximum power of 30 here is determined by max p/w of 10 and not three since it is having more 3 than 10 max power of 10 over here is 11 and so the answer
I am doing this book currently.
So how does this thread work? I post my question, or I check whether its been answered already?? :lookround:
Hey Neha,
well,you post a query from the book,whichever you dint understand or think would be helpful to the people here.The brightest brains this side of the universe(thats what the description of the forum says) will reply your query.You can check the thread regularly if your query is answered or not.as far as repetition of questions is concerned,its fine if the questions are repeated
Cheers,
Shashank.
plz solve this:
n is a number such that 2n has 28 factors and 3n has 30 factors,then how many factors does 6n have?? answer given is 35
plz solve this:
n is a number such that 2n has 28 factors and 3n has 30 factors,then how many factors does 6n have?? answer given is 35
n=2^a*3^b*....=number of factors are=(a+1)(b+1).....
2n=2^(a+1)*3^b....=number of factors are=(a+2)(b+1)...=28
3n=2^a*3^(b+1)....=number of factors are=(a+1)(b+2)....=30
(a+2)(b+1)/(a+1)(b+2)=28/30
a=5,b=3
n=2^5*3^3
6n=2^6*3^4
number of factors is 7*5=35
n=2^a*3^b*....=number of factors are=(a+1)(b+1).....
2n=2^(a+1)*3^b....=number of factors are=(a+2)(b+1)...=28
3n=2^a*3^(b+1)....=number of factors are=(a+1)(b+2)....=30
(a+2)(b+1)/(a+1)(b+2)=28/30
a=5,b=3
n=2^5*3^3
6n=2^6*3^4
number of factors is 7*5=35
Why did u take N to consists only of 2's and 3's ???
I mean why the factors of N cant be (a+1)(b+1)(c+1)....?
coz 28 can be (7*4) or (7*2*2)
By (7*4) it means that N consists only of 2 and 3 but by (7*2*2),it means that there could also be a third prime number too building up N...
I hope u get my query..
kano_kyosuke SaysWhat is the answer.I guess even i am making some mistake
hmm even i am getting the same answer
i think none of thses is the answer
Why did u take N to consists only of 2's and 3's ???
I mean why the factors of N cant be (a+1)(b+1)(c+1)....?
coz 28 can be (7*4) or (7*2*2)
By (7*4) it means that N consists only of 2 and 3 but by (7*2*2),it means that there could also be a third prime number too building up N...
I hope u get my query..
the question says 28 factors for 2n.
when we take the ratios,it is see that,
(a+2)(b+1)k=28
(a+1)(b+2)k=30
as they all are integers,
a=5 and b=3 satisfy the condition.now,putting them in either of the equations,we get k=1.
well u see in the case "a" is not coming out to be same,the answer can be found out by this process.First find the LCM of all the numbers say,x and y, and then try to find out a number which when divided by x and y leaves the specified remainder.Let the number found be z.then to z add the lcm of x and y to find more such numbers.
Example,
Find a number which when divided by 6 leaves remainder 5 and when divided by 5 leaves remainder 1?
to solve,we can easily find that 11 will be the smallest number to satisfy the given conditions.LCM of 5 and 6 is 30.So by adding 30 to 11 we can get more numbers like 41,...
Hope this solves your query.
saru421 Saysbut i wanted to ask what if a question arises in which this difference "a" is not same always. How will we solve such question ?
well u see in the case "a" is not coming out to be same,the answer can be found out by this process.First find the LCM of all the numbers say,x and y, and then try to find out a number which when divided by x and y leaves the specified remainder.Let the number found be z.then to z add the lcm of x and y to find more such numbers.
Example,
Find a number which when divided by 6 leaves remainder 5 and when divided by 5 leaves remainder 1?
to solve,we can easily find that 11 will be the smallest number to satisfy the given conditions.LCM of 5 and 6 is 30.So by adding 30 to 11 we can get more numbers like 41,...
Hope this solves your query.
can you tell me which book is best to consult for higher level maths for cat.(other than Arun Sharma)
my query is:
Find the maximum value of n such that
77*42*37*57*30*90*70*2400*2402*243*343 is perfectly divisible by 21^n.
The answer given is 6.
my query is:
Find the maximum value of n such that
77*42*37*57*30*90*70*2400*2402*243*343 is perfectly divisible by 21^n.
The answer given is 6.
see the maximum power can be determined by taking minimum power available for 3 & 7 in this case it is for 7 which comes out to be 6 and so the answer
abhay15783 Sayssee the maximum power can be determined by taking minimum power available for 3 & 7 in this case it is for 7 which comes out to be 6 and so the answer
But how the power is 6 for 7. I think the power for 7 comes out to b 4.
my query is:
Find the maximum value of n such that
77*42*37*57*30*90*70*2400*2402*243*343 is perfectly divisible by 21^n.
The answer given is 6.
we have to find the number of 3's and the number of 7's
number of 7's are 1+1+1+3=6
number of 3's are 1+1+1+2+3+5=13
so,number of 21's possible are 6.
we have to find the number of 3's and the number of 7's
number of 7's are 1+1+1+3=6
number of 3's are 1+1+1+2+3+5=13
so,number of 21's possible are 6.
Can u plz explain how did u get
number of 7's as 1+1+1+3 in detail. I m not getting how to calculate this step.
Can u plz explain how did u get
number of 7's as 1+1+1+3 in detail. I m not getting how to calculate this step.
Is sum 1 trying to increase his posts by asking such questions. I don't think it is possible for anyone to give a more trivial solution until and unless you tell that 7 has one power of 7,70 has 1 power of 7 and so on....
abhay15783 SaysIs sum 1 trying to increase his posts by asking such questions. I don't think it is possible for anyone to give a more trivial solution until and unless you tell that 7 has one power of 7,70 has 1 power of 7 and so on....
No i m not at all trying to increase my number of posts. I got some confusion in d solution but now i have understood. I was calculating it using some other method, so got confused. And i thought this was a thread where we could ask our confusions.
Anyways thanks for clarifying my doubt.
hi everybody, can you please help me with this quest
Q.the number of positive integers not greater than 100,which are not divisible by 2,3, or 5 is?
Q.V1,V2,V3,V4,.....,,V999,V1000 are all natural numbers and Vn + V(n+1)=k,11.If V987=987, what is the value of V236?
2.If V100=100,what is the value of V10+V11+V12+V13+V14-V15-V16-V17-V18-V19?
hi everybody, can you please help me with this quest
Q.the number of positive integers not greater than 100,which are not divisible by 2,3, or 5 is?
Q.V1,V2,V3,V4,.....,,V999,V1000 are all natural numbers and Vn + V(n+1)=k,11.If V987=987, what is the value of V236?
2.If V100=100,what is the value of V10+V11+V12+V13+V14-V15-V16-V17-V18-V19?
there are 50 nos. divisable by 2
there are 33 divisable by 3
there are 20 divisable by 5
there are 16 divisable by lcm of2 and 3
there are 6 divisable by lcm of 3 and 5
there are 10 divisable by lcm of 2 and 5
there are 3 divisable by lcm of 2 3 and 5
hence total not divisable by these three are = 100 - (50 + 33+20 - 16 - 6 - 10 + 3) = 26