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| Neither 3 nor 6 (Posted on 2011-03-11) |
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0, 25, 2025, 13225…are squares that remain squares
if every digit in the number defining them is augmented by 1.
Let's call them squarish numbers.
a. List two more samples of squarish numbers.
b. Prove that all such numbers are evenly divisible by 25.
c. Why are there neither 3-digit nor 6-digit squarish numbers?
d. Prove that between 10^k and 10^(k+1) there is at most one squarish number.
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Submitted by Ady TZIDON
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Solution:
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a),b),d) are covered by the solvers, either explicitly or implicitly.
To address c) note that :
" Incrementing each digit means b^2-a^2 = R_n, the n-digit repunit (10^n-1)/9; so solutions must be of the form a = (u-v)/2, b = (u+v)/2, where u * v = R_n.
It remains to check that this is in the right range and a has no 9's. "(Sloane)"
Consider the length of 3 : u*v=111=3*37 , so b=40/2=20 and a 34/2=17 and indeed 20^2-17^2=400-289=111, but 289 has the disturbing "9".
Show to yourself the same for all possible solutions for the length of 6: 111111=3*37*7*11*13.
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