Question
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.
The number of divisors of n is 7.
The number of divisors of n is 6.
The number of divisors of n is 8.
The number of divisors of n is 5.
The number of divisors of n is 7.
Solution
Let the 7 distinct natural numbers chosen from {1,2,…,100} be x1,x2,…,x7, sorted in ascending order: 1≤x1<x2<⋯<x7≤100. The condition is that xi+1−xi≥7 for i=1,2,…,6.
We use a transformation to simplify this condition. Let y1=x1, y2=x2−6, y3=x3−12, ..., y7=x7−36. In general, yi=xi−6(i−1) for i=1,…,7.
Now, let's check the differences between consecutive yi: yi+1−yi=(xi+1−6i)−(xi−6(i−1))=xi+1−xi−6i+6i−6=(xi+1−xi)−6. Since xi+1−xi≥7, we have yi+1−yi≥7−6=1. This means yi+1>yi, so the yi are distinct and in increasing order: y1<y2<⋯<y7.
Now, let's find the range of yi. The minimum value of y1 is x1≥1, so y1≥1. The maximum value of y7 is determined by the maximum value of x7: y7=x7−6(7−1)=x7−36. Since x7≤100, the maximum value of y7 is 100−36=64.
So, we are choosing 7 distinct numbers y1,y2,…,y7 from the set {1,2,…,64}. The number of ways to do this is (764).
The problem states that this number of ways is equal to nC7. Therefore, (7n)=(764), which implies n=64.
We need to find the number of divisors of n=64. The prime factorization of 64 is 26. The number of divisors of a number pa is (a+1). So, the number of divisors of 64=26 is (6+1)=7. The divisors are 20,21,22,23,24,25,26, which are 1,2,4,8,16,32,64. There are 7 divisors.