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Question: The number of ways to choose 7 distinct natural numbers from the first 100 natural numbers such that...

The number of ways to choose 7 distinct natural numbers from the first 100 natural numbers such that any two chosen numbers differ atleast by 7 can be expressed as nC7. Find the number of divisors of n.

A

The number of divisors of n is 7.

B

The number of divisors of n is 6.

C

The number of divisors of n is 8.

D

The number of divisors of n is 5.

Answer

The number of divisors of n is 7.

Explanation

Solution

Let the 7 distinct natural numbers chosen from {1,2,,100}\{1, 2, \dots, 100\} be x1,x2,,x7x_1, x_2, \dots, x_7, sorted in ascending order: 1x1<x2<<x71001 \le x_1 < x_2 < \dots < x_7 \le 100. The condition is that xi+1xi7x_{i+1} - x_i \ge 7 for i=1,2,,6i = 1, 2, \dots, 6.

We use a transformation to simplify this condition. Let y1=x1y_1 = x_1, y2=x26y_2 = x_2 - 6, y3=x312y_3 = x_3 - 12, ..., y7=x736y_7 = x_7 - 36. In general, yi=xi6(i1)y_i = x_i - 6(i-1) for i=1,,7i=1, \dots, 7.

Now, let's check the differences between consecutive yiy_i: yi+1yi=(xi+16i)(xi6(i1))=xi+1xi6i+6i6=(xi+1xi)6y_{i+1} - y_i = (x_{i+1} - 6i) - (x_i - 6(i-1)) = x_{i+1} - x_i - 6i + 6i - 6 = (x_{i+1} - x_i) - 6. Since xi+1xi7x_{i+1} - x_i \ge 7, we have yi+1yi76=1y_{i+1} - y_i \ge 7 - 6 = 1. This means yi+1>yiy_{i+1} > y_i, so the yiy_i are distinct and in increasing order: y1<y2<<y7y_1 < y_2 < \dots < y_7.

Now, let's find the range of yiy_i. The minimum value of y1y_1 is x11x_1 \ge 1, so y11y_1 \ge 1. The maximum value of y7y_7 is determined by the maximum value of x7x_7: y7=x76(71)=x736y_7 = x_7 - 6(7-1) = x_7 - 36. Since x7100x_7 \le 100, the maximum value of y7y_7 is 10036=64100 - 36 = 64.

So, we are choosing 7 distinct numbers y1,y2,,y7y_1, y_2, \dots, y_7 from the set {1,2,,64}\{1, 2, \dots, 64\}. The number of ways to do this is (647)\binom{64}{7}.

The problem states that this number of ways is equal to nC7nC7. Therefore, (n7)=(647)\binom{n}{7} = \binom{64}{7}, which implies n=64n = 64.

We need to find the number of divisors of n=64n=64. The prime factorization of 64 is 262^6. The number of divisors of a number pap^a is (a+1)(a+1). So, the number of divisors of 64=2664 = 2^6 is (6+1)=7(6+1) = 7. The divisors are 20,21,22,23,24,25,262^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6, which are 1,2,4,8,16,32,641, 2, 4, 8, 16, 32, 64. There are 7 divisors.