How many 3-digit natural numbers have no 5's but at least one 6 ?
FINITE people HELP me out pleaz!
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well, the way i see it....
106
116
126
136
146
166
176
186
196
206
216
226
236
246
266
276
286
296
306
316
326
336
346
366
376
386
396
406
416
426
436
446
466
476
486
496
606
616
626
636
646
666
676
686
696
706
716
726
736
746
766
776
786
796
806
816
826
836
846
866
876
886
896
906
916
926
936
946
966
976
986
996
add em upComment
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It seems that you're missing the obvious instances ofOriginally posted by stanton
106 also can be 160 601 610 and so on for each of those.....
116
126
136
146
166
176
186
196
160
161
162
163
164
166
167
168
169
260
261
262
etc...
SpaceAceComment
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100 through 200 there are 17 instances.
200 through 300 there are 17 instances.
300 through 400 there are 17 instances.
400 through 499 there are 17 instances.
600 through 700 there are 81 instances.
700 through 800 there are 17 instances.
800 through 900 there are 17 instances.
900 through 999 there are 17 instances.
200 is right
final answer be dammned!!!Comment
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67510, sorry had too much free time I guess, the mirc script below will save you time though, even if it does take close to a minute.
alias weird.thing {
set %repeats 0
set %start.from 0
while (%start.from <= 100000) {
echo -a progress %start.from
if ($count(%start.from,0) >= 2) { inc %repeats | goto end }
if ($count(%start.from,1) >= 2) { inc %repeats | goto end }
if ($count(%start.from,2) >= 2) { inc %repeats | goto end }
if ($count(%start.from,3) >= 2) { inc %repeats | goto end }
if ($count(%start.from,4) >= 2) { inc %repeats | goto end }
if ($count(%start.from,5) >= 2) { inc %repeats | goto end }
if ($count(%start.from,6) >= 2) { inc %repeats | goto end }
if ($count(%start.from,7) >= 2) { inc %repeats | goto end }
if ($count(%start.from,8) >= 2) { inc %repeats | goto end }
if ($count(%start.from,9) >= 2) { inc %repeats | goto end }
:end
inc %start.from
}
}Comment

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