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 Celebrity Squares (Posted on 2011-03-29)

Three different 5-digit perfect squares between them use five different digits. No two of these squares share a common first digit, nor a common last digit. Each of the five digits is used a different number of times, the five numbers of times being the same as the five digits of the perfect squares. At least 3 digits are used their own number of times. What are the three squares? (Theme suggested by New Scientist, March 2011)

 See The Solution Submitted by broll No Rating

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 computer solution (spoiler) | Comment 1 of 2

This program finds those 5-digit squares which use only the digits 1 through 5, as the sum of the integers 1 through 5 is 15, which is precisely the number of digits needed for three 5-digit numbers.

CLS
FOR n = 100 TO 316
nsq = n * n
s\$ = LTRIM\$(STR\$(nsq))
good = 1
FOR i = 1 TO 5
IF MID\$(s\$, i, 1) > "5" OR MID\$(s\$, i, 1) < "1" THEN good = 0
NEXT
IF good THEN PRINT n, nsq
NEXT

The 5-digit squares appear to the right of their square roots:

111           12321
112           12544
115           13225
182           33124
185           34225
188           35344
211           44521
229           52441
235           55225

Making the program go farther (with the hindsight that there are only a manageable nine such squares):

CLS
FOR n = 100 TO 316
nsq = n * n
s\$ = LTRIM\$(STR\$(nsq))
good = 1
FOR i = 1 TO 5
IF MID\$(s\$, i, 1) > "5" OR MID\$(s\$, i, 1) < "1" THEN good = 0
NEXT
IF good THEN PRINT n, nsq: ct = ct + 1: sq\$(ct) = LTRIM\$(STR\$(nsq))
NEXT
PRINT

FOR a = 1 TO ct - 2
as\$ = sq\$(a)
FOR b = a + 1 TO ct - 1
bs\$ = sq\$(b)
FOR c = b + 1 TO ct
good = 1
cs\$ = sq\$(c)
IF LEFT\$(as\$, 1) = LEFT\$(bs\$, 1) OR LEFT\$(as\$, 1) = LEFT\$(cs\$, 1) OR LEFT\$(bs\$, 1) = LEFT\$(cs\$, 1) THEN good = 0
IF RIGHT\$(as\$, 1) = RIGHT\$(bs\$, 1) OR RIGHT\$(as\$, 1) = RIGHT\$(cs\$, 1) OR RIGHT\$(bs\$, 1) = RIGHT\$(cs\$, 1) THEN good = 0
REDIM dct(5)
FOR i = 1 TO 15
dused = VAL(MID\$(as\$ + bs\$ + cs\$, i, 1))
dct(dused) = dct(dused) + 1
NEXT
FOR i = 1 TO 4
FOR j = i + 1 TO 5
IF dct(i) = dct(j) THEN good = 0
NEXT
NEXT
IF good THEN PRINT as\$: PRINT bs\$: PRINT cs\$: PRINT
NEXT
NEXT
NEXT

adds to the output the candidate sets of three squares based on non-repeated first digits and non-repeated last digits, as well as having a different number of each digit:

12544
34225
44521

12544
34225
52441

35344
44521
55225

Only in the last of these, seen by inspection, do three of the digits appear their own number of times: one 1, four 4's and five 5's.

 Posted by Charlie on 2011-03-29 15:44:16

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