Largest palindrome product: Difference between revisions
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<pre>
906609
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=={{header|FreeBASIC}}==
<lang freebasic>function make_pal( n as ulongint ) as ulongint
'turn a number into a palindrom with twice as many digits
dim as string ns, ret
ns = str(n) : ret = ns
for i as uinteger = len(ns) to 1 step -1
ret += mid(ns, i, 1)
next i
return val(ret)
end function
function has_dig( n as ulongint, d as uinteger ) as boolean
'does the number n have d decimal digits?
if 10^(d-1)<=n and n<10^d then return true else return false
end function
dim as integer np
for d as uinteger = 2 to 7
for n as ulongint = 10^d - 1 to 10^(d-1) step -1 'count down from 999...
'since the first good number we encounter
'must be the highest
np = make_pal( n ) 'produce a 2d-digit palindrome from it
for f as ulongint = 10^d - 1 to 10^(d-1) step -1 'look for highest d-digit factor
if np mod f = 0 then
if has_dig( np/f, d ) then 'if np/f also has d digits we are done :)
print f;" *";np/f;" =";np
goto nextd
end if
end if
next f
next n
nextd: 'yes, I used a goto. sue me.
next d</lang>
{{out}}<pre>
99 * 91 = 9009
993 * 913 = 906609
9999 * 9901 = 99000099
99979 * 99681 = 9966006699
999999 * 999001 = 999000000999
9998017 * 9997647 = 99956644665999
</pre>
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