|
The only 4-digit perfect squares without duplicate digits, together with the sums of their digits, are:
1024 7
2401 7
2304 9
2601 9
3025 10
5041 10
2704 13
3721 13
6241 13
3481 16
9025 16
1089 18
1296 18
1764 18
2916 18
3249 18
4356 18
4761 18
5184 18
6084 18
7056 18
9216 18
9801 18
1369 19
1936 19
2809 19
4096 19
5329 19
6724 19
7921 19
9604 19
1849 22
5476 22
7396 25
7569 27
8649 27
Only the digital sums 18 and 19 have at least 4 representatives, and among those, only the following has the same digital sum for each column when arranged as columns:
1764
3249
5184
9801
The final version of the program is shown, after the determination of 18 and 19 was made using only the first part:
DECLARE FUNCTION sod# (x#)
DEFDBL A-Z
DIM sq(100)
DIM sd(100)
st = -INT(-SQR(1000))
fin = INT(SQR(10000))
FOR n = st TO fin
ns = n * n
sum = sod(ns)
IF sum > 0 THEN
noSq = noSq + 1
sq(noSq) = ns
sd(noSq) = sum
END IF
NEXT
DO
done = 1
FOR i = 1 TO noSq - 1
IF sd(i) > sd(i + 1) THEN
SWAP sd(i), sd(i + 1)
SWAP sq(i), sq(i + 1)
done = 0
END IF
NEXT
LOOP UNTIL done
FOR i = 1 TO noSq
PRINT sq(i); sd(i)
IF sd(i) = 18 AND first18 = 0 THEN first18 = i
IF sd(i) > 18 AND last18 = 0 THEN last18 = i - 1
IF sd(i) = 19 AND first19 = 0 THEN first19 = i
IF sd(i) > 19 AND last19 = 0 THEN last19 = i - 1
NEXT
FOR a = first18 TO last18 - 3
FOR b = a + 1 TO last18 - 2
FOR c = b + 1 TO last18 - 1
FOR d = c + 1 TO last18
t1 = 0: t2 = 0: t3 = 0: t4 = 0
n(1) = sq(a)
n(2) = sq(b)
n(3) = sq(c)
n(4) = sq(d)
FOR i = 1 TO 4
t1 = t1 + n(i) \ 1000
t2 = t2 + (n(i) \ 100) MOD 10
t3 = t3 + (n(i) \ 10) MOD 10
t4 = t4 + (n(i)) MOD 10
NEXT
IF t1 = t2 AND t2 = t3 AND t3 = t4 THEN
PRINT n(1)
PRINT n(2)
PRINT n(3)
PRINT n(4)
END IF
NEXT
NEXT
NEXT
NEXT
FOR a = first19 TO last19 - 3
FOR b = a + 1 TO last19 - 2
FOR c = b + 1 TO last19 - 1
FOR d = c + 1 TO last19
t1 = 0: t2 = 0: t3 = 0: t4 = 0
n(1) = sq(a)
n(2) = sq(b)
n(3) = sq(c)
n(4) = sq(d)
FOR i = 1 TO 4
t1 = t1 + n(i) \ 1000
t2 = t2 + (n(i) \ 100) MOD 10
t3 = t3 + (n(i) \ 10) MOD 10
t4 = t4 + (n(i)) MOD 10
NEXT
IF t1 = t2 AND t2 = t3 AND t3 = t4 THEN
PRINT n(1)
PRINT n(2)
PRINT n(3)
PRINT n(4)
END IF
NEXT
NEXT
NEXT
NEXT
FUNCTION sod (x)
s$ = LTRIM$(STR$(x))
t = 0
FOR i = 1 TO LEN(s$)
IF INSTR(i + 1, s$, MID$(s$, i, 1)) THEN sod = 0: EXIT FUNCTION
t = t + VAL(MID$(s$, i, 1))
NEXT
sod = t
END FUNCTION
From Enigma No. 1553, "Squares from squares", by Susan Denham, New Scientist, 11 July 2009, page 26.
|
