number

[rupsk]

If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which is an integer?

Explain this question
(1)3

(2)

6

(3)

7


(4)

8

(5)

10

[neelgandham]

Hi rupsk, I am sure that this is not the complete question. Can you please check and edit the post with the correct question?

what is the greatest integer m for which ____ is an integer

[rupsk]

sorry my mistake it says what is the greatest integer m for which N/10^m is an integer?

[neelgandham]

sorry my mistake it says what is the greatest integer m for which N/10^m is an integer?

N/10^m = 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

[rooster]

sorry my mistake it says what is the greatest integer m for which N/10^m is an integer?

N/10^m = 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

Right approach, but I don't see how you get 7 5's

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

[rupsk]

even if I try to find number of 5 I get value as 6. Please explain how you get 5^7?

[shankar.ashwin]

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

[neelgandham]

even if I try to find number of 5 I get value as 6. Please explain how you get 5^7?

As Shankar already explained

N = 3*...*15*...30*...45*....60*....75(5*5*3)*...90*...99, So, the highest power of 3 is 7