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 Neither 3 nor 6 (Posted on 2011-03-11)
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.

 See The Solution Submitted by Ady TZIDON No Rating

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 part b (not very elegant) | Comment 4 of 6 |
Well, I am not clear what it means to augment every digit by 1.
Consider 89.  Does augmenting every digit by 1 turn 89 into 90 or into 100?  I'll assume no carryover, and thus 89 augmented is 90. That's the way my bicycle cylinder lock works.

Having said that, all squares > 9 end in one of the following pairs of ending digits.  Only one of the augmented ending digits is itself a possible ending pair of digits (36), which means that all squarish numbers greater than 9 must end in 25.  The only squarish number less than 9 is 0, which is a multiple of 25, so all squarish numbers are multiples of 25,

ending  augments to
00      11
01      12
04      15
09      10
16      27
21      32
24      35
25      36
29      30
36      47
41      52
44      55
49      50
56      67
61      72
64      75
69      70
76      87
81      92
84      95
89      90
96      07

Note that this proof needs some more work if something ending in 89 augments to something ending in 00, so my interpretation of the problem was a little self-serving.

It looks like OEIS sequence A061843 (cited by Charlie) gets around the whole issue by saying that a squarish number is not allowed to have any 9's in it.  If this is Ady's definition also, then my proof is complete.

Edited on March 13, 2011, 4:01 pm

Edited on March 13, 2011, 4:04 pm
 Posted by Steve Herman on 2011-03-13 15:59:37

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