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Part of 1995 Putnam

Problems(2)

Putnam 1995 A3

Source:

7/1/2014
The number d1d2⋯d9d_1d_2\cdots d_9 has nine (not necessarily distinct) decimal digits. The number e1e2⋯e9e_1e_2\cdots e_9 is such that each of the nine 99-digit numbers formed by replacing just one of the digits did_i in d1d2⋯d9d_1d_2\cdots d_9 by the corresponding digit ei    (1≤i≤9)e_i \;\;(1 \le i \le 9) is divisible by 77. The number f1f2⋯f9f_1f_2\cdots f_9 is related to e1e2⋯e9e_1e_2\cdots e_9 is the same way: that is, each of the nine numbers formed by replacing one of the eie_i by the corresponding fif_i is divisible by 77. Show that, for each ii, di−fid_i-f_i is divisible by 77. [For example, if d1d2⋯d9=199501996d_1d_2\cdots d_9 = 199501996, then e6e_6 may be 22 or 99, since 199502996199502996 and 199509996199509996 are multiples of 77.]
Putnamcollege contests
Putnam 1995 B3

Source:

7/1/2014
To each number with n2n^2 digits, we associate the n×nn\times n determinant of the matrix obtained by writing the digits of the number in order along the rows. For example : 8617↦det⁡(  86    17  )=508617\mapsto \det \left(\begin{matrix}{\;8}& 6\;\\ \;1 &{ 7\;}\end{matrix}\right)=50. Find, as a function of nn, the sum of all the determinants associated with n2n^2-digit integers. (Leading digits are assumed to be nonzero; for example, for n=2n = 2, there are 90009000 determinants.)
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