Determine the total number of concatenated 3-digit positive integers of the form ABC, such that each of their nonzero digits A, B, and C satisfies this relationship:
. . .
___ 0.A + 0.B + 0.C
0.ABC = ------------------
3Notes:
. ---
0.ABC = 0.ABCABCABC.......
.
0.A = 0.AAAA.......
Computer program/excel solver solutions are welcome- but a semi-analytic (p&p with a hand-calculator) methodology is preferred.
There are 15:
(Actually, see caveat at the end)
[ABC, lhs (matches rhs)]
[111, 1/9]
[222, 2/9]
[333, 1/3]
[370, 10/27]
[407, 11/27]
[444, 4/9]
[481, 13/27]
[518, 14/27]
[555, 5/9]
[592, 16/27]
[629, 17/27]
[666, 2/3]
[777, 7/9]
[888, 8/9]
[999, 1]
for abc=1:999
lhs=sym(abc/999);
rhs=sym(sod(abc)/27);
if lhs==rhs
disp([abc lhs])
endThere are 15:
[ABC, lhs (matches rhs)]
[111, 1/9]
[222, 2/9]
[333, 1/3]
[370, 10/27]
[407, 11/27]
[444, 4/9]
[481, 13/27]
[518, 14/27]
[555, 5/9]
[592, 16/27]
[629, 17/27]
[666, 2/3]
[777, 7/9]
[888, 8/9]
[999, 1]
for abc=1:999
lhs=sym(abc/999);
rhs=sym(sod(abc)/27);
if lhs==rhs
disp([abc lhs])
end
end
All the repeated-digit integers work, but also:
370, 407, 481, 518, 592, 629.
end
All the repeated-digit integers work, but also:
370, 407, 481, 518, 592, 629.
The caveat: The puzzle requires that all the digits must be positive, specifically not zero. My program only required that A not be zero. 370 and 407 should not have been allowed. Only realized this after reading Larry's comment.
Edited on August 1, 2023, 8:46 am
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Posted by Charlie
on 2023-08-01 08:41:30 |
computer solution
