If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by

Another method can be
factors of 50 - factors of 100
1890-1620=270

Why can't 5 have 6 different powers. Because, you add 1 to the powers of 3 and 7, which I understand. Also, one 2 is understandable because more than one 2 will lead to divisibility by 100.
Thanks!

\(50 = 2*5^2\)

The power of 5 must be at least 2. It can be 3, 4, 5 or 6 as well. Here there are 5 different ways in which you can give a power to 5.

Similar question to practice: how-many-integers-from-1-to-200-inclusive-are-divisib-109333.html

Hope it helps.

here, we have to catch all factors for 50.

100=2^2*5^2
50=2*5^2

Now, to have number divisible by 50 and not by 100 we have to make sure that 2 is taken only once. As, if we pick 2 two's then it will form 100 with no. of 5 selected and since min 5 we have to select is 2, we cannot have 2 more than one.

Number= 2^7*3^5*5^6*7^8

So, there are 6 ways of picking 3, 9 ways of picking 7, 5 ways of picking 5 and only one way of picking 2 and that is 2^1.


Why only 5 ways for picking 5 when we have 7 ways(0,1,2,3,4,5,6)?

Reason: we need 50 so we have to pick 5^2 and greater than it. Also, we can pick only 2^1 because picking 2^2 will form 100 with 5 already selected. For example, 5^2*2^2=100. So, we cannot afford it.

Hence, total number of factors.

= 6(3's)*9(7's)*5(5's)*1(2's)=270

Hope it was helpful.

Thanks,
Gaurav

If N=2^7*3^5*5^6*7^8. How many factors of N are divisible by 50 but NOT by 100?

50=2*5^2
100=2^2*5^2
N=(2*2^6)*(5^2*5^4)*(3^5)*(7^8)
=(2*5^2)*(2^6*5^4*3^5*7^8)
We dont need any other power of 2 except (2*5^2)
no of factors (power of 5+1)*(power of 3+1)*(power of 7+1)= 5*6*9=270

Hence C

\(N=2^7*3^5*5^6*7^8\)

num factors of N/50: \(N=(2*5^2)[2^6*3^5*5^4*7^8]=[6+1][5+1][4+1][8+1]=1890\)
num factors of N/100: \(N=(2^2*5^2)[2^5*3^5*5^4*7^8]=[5+1][5+1][4+1][8+1]=1620\)

num factors of N/50 but not N/100: \(1890-1620=270\)

Answer (C)

Given

• N = \(2^7\) × \(3^5\) × \(5^6\) × \(7^8\)

To Find

• The number of factors of N that are divisible by 50 but NOT by 100.

Approach and Working Out

• We can calculate the factors that are multiples of 50 – the factors that are multiples of 100.

o N = \(2^7\) × \(3^5\) × \(5^6\) × \(7^8\)
o N = 2 × \(5^2\) × (\(2^6 \)× \(3^5\) × \(5^4\) × \(7^8\))

 Number of factors that are multiples of 50,
 = 7 × 6 × 5 × 9

o N = \(2^7\) × \(3^5\) × \(5^6\) × \(7^8\)
o N = \(2^2 \)× \(5^2\) × (\(2^5 \)× \(3^5\) × \(5^4\) × \(7^8\))

 Number of factors that are multiples of 100,
 = 6 × 6 × 5 × 9

• The answer = 7 × 6 × 5 × 9 - 6 × 6 × 5 × 9

o = 6 × 5 × 9
o = 270

Correct Answer: Option C

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