N = a = 0 ∑ 5 0 b = 0 ∑ a c = 0 ∑ a d = 0 ∑ 5 0 − a e = 0 ∑ c + d f = 0 ∑ 5 0 − a − d ( a 5 0 ) ( b a ) ( c a ) ( d 5 0 − a ) ( e c + d ) ( f 5 0 − a − d ) . How many digits does N have?
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For any a , d ≥ 0 we have b = 0 ∑ a c = a ∑ a e = 0 ∑ c + d ( b a ) ( c a ) ( e c + d ) = b = 0 ∑ a c = 0 ∑ a ( b a ) ( c a ) 2 c + d = 2 a × 3 a × 2 d = 6 a × 2 d and hence N = a + d + f ≤ 5 0 ∑ a ! d ! f ! ( 5 0 − a − d − f ) ! 5 0 ! × 6 a × 2 d = ( 6 + 2 + 1 + 1 ) 5 0 = 1 0 5 0 which has 5 1 digits.
By repeated use of the Binomial Theorem, a = 0 ∑ 5 0 b = 0 ∑ a c = 0 ∑ a d = 0 ∑ 5 0 − a e = 0 ∑ c + d f = 0 ∑ 5 0 − a − d ( a 5 0 ) ( b a ) ( c a ) ( d 5 0 − a ) ( e c + d ) ( f 5 0 − a − d ) = a = 0 ∑ 5 0 b = 0 ∑ a c = 0 ∑ a d = 0 ∑ 5 0 − a e = 0 ∑ c + d ( a 5 0 ) ( b a ) ( c a ) ( d 5 0 − a ) ( e c + d ) 2 5 0 − a − d = a = 0 ∑ 5 0 b = 0 ∑ a c = 0 ∑ a d = 0 ∑ 5 0 − a ( a 5 0 ) ( b a ) ( c a ) ( d 5 0 − a ) 2 c + d ⋅ 2 5 0 − a − d = a = 0 ∑ 5 0 b = 0 ∑ a c = 0 ∑ a d = 0 ∑ 5 0 − a ( a 5 0 ) ( b a ) ( c a ) ( d 5 0 − a ) 2 5 0 + c − a = a = 0 ∑ 5 0 b = 0 ∑ a c = 0 ∑ a ( a 5 0 ) ( b a ) ( c a ) 2 5 0 − a ⋅ 2 5 0 + c − a = a = 0 ∑ 5 0 b = 0 ∑ a c = 0 ∑ a ( a 5 0 ) ( b a ) ( c a ) 2 1 0 0 + c − 2 a = a = 0 ∑ 5 0 b = 0 ∑ a ( a 5 0 ) ( b a ) 2 1 0 0 − 2 a c = 0 ∑ a ( c a ) 2 c = a = 0 ∑ 5 0 b = 0 ∑ a ( a 5 0 ) ( b a ) 2 1 0 0 − 2 a ⋅ 3 a = a = 0 ∑ 5 0 ( a 5 0 ) 2 a ⋅ 2 1 0 0 − 2 a ⋅ 3 a = a = 0 ∑ 5 0 2 1 0 0 − a ⋅ 3 a = 2 5 0 a = 0 ∑ 5 0 2 5 0 − a ⋅ 3 a = 2 5 0 ⋅ 5 5 0 = 1 0 5 0 , which has 51 digits.
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We interpret each sum ∑ k = 0 n ( k n ) as choosing an arbitrary number k out of n items of a set, and counting in how many ways that may be done. Extending this idea gives an interpretation for the value of N : Consider a collection of 50 sheets of paper, which are initially blank.
First, choose a out of these 50 sheets and draw a circle on each of them.
Of the a sheets with a circle, pick b and draw a square on each of them.
Of the a sheets with a circle, pick c and draw a triangle on each of them.
Of the 5 0 − a sheets without a circle, pick d and draw an semicircle on each of them.
Of the c sheets with a triangle and the d sheets with a semicircle, pick e and draw a star on each of them.
Of the 5 0 − a − d sheets without circle or semicircle, pick f sheets and draw a dot on them.
After this process, each of the 50 sheets belongs to precisely one of the following categories:
only a circle
a circle and a square
a circle and a triangle
a circle, a triangle, and a star
a circle, a square, and a triangle
a circle, a square, a triangle, and a star
only a semicircle
a semicircle and a star
only a dot
blank
Since each of the 50 sheets belongs to precisely one out of 10 categories, there are N = 1 0 5 0 ways of making these choices. This number has 5 1 digits.