what is the greatest integer m for which ____ is an integer
number
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neelgandham
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Hi rupsk, I am sure that this is not the complete question. Can you please check and edit the post with the correct question?
Anil Gandham
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neelgandham
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N/10^m = Integerrupsk wrote:sorry my mistake it says what is the greatest integer m for which N/10^m is an integer?
N =3*9*12*15*...30*..45*..60*..75*..90. So N contains 5^7 and 2^n where n > 7
N/10^m = X*5^7*2^n/10^m
The maximum value of m = 7
IMO C
Anil Gandham
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Right approach, but I don't see how you get 7 5'sneelgandham wrote:N/10^m = Integerrupsk wrote:sorry my mistake it says what is the greatest integer m for which N/10^m is an integer?
N =3*9*12*15*...30*..45*..60*..75*..90. So N contains 5^7 and 2^n where n > 7
N/10^m = X*5^7*2^n/10^m
The maximum value of m = 7
IMO C
Basically, you need to know how many 10s are in the product of all the multiples of 3.
From what we know, the only 10s that can be factored out are 30, 60 and 90
BUT
We also have 15, 45, and 75, which all have 5s in them and there are enough 2s in the problem to multiply and create 10s.
So, from this we get 6 10s, which means that the product of all multiples of 3 that are less than 100 equal to 10^6
therefore the choice should be B
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shankar.ashwin
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N = 3*6*9*12*15*18*..... * 99.
For N to be divisible by 10^m, we need to find number of 0's N has.
Number of 0's can be found by finding number of factors of 5 in the series.
Factors of 5;
15 - 5*3 - 1
30 - 5*6 - 1
45 - 5*9 - 1
60 - 5*12 - 1
75 - 5*15 = 5*5*3 - 2
90 - 5*18 - 1
Adding we have 7 factors of 5. So 'm' would be 7 here. C
For N to be divisible by 10^m, we need to find number of 0's N has.
Number of 0's can be found by finding number of factors of 5 in the series.
Factors of 5;
15 - 5*3 - 1
30 - 5*6 - 1
45 - 5*9 - 1
60 - 5*12 - 1
75 - 5*15 = 5*5*3 - 2
90 - 5*18 - 1
Adding we have 7 factors of 5. So 'm' would be 7 here. C
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neelgandham
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As Shankar already explainedrupsk wrote:even if I try to find number of 5 I get value as 6. Please explain how you get 5^7?
N = 3*...*15*...30*...45*....60*....75(5*5*3)*...90*...99, So, the highest power of 3 is 7
Anil Gandham
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