17
How many three digit numbers are divisible by 17?
Labels: mathschallenge
posted by Chris at
9:11 AM
Tuesday, September 8, 2009
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15 Comments:
53
17.Xmin>=100 and Xmin integer => Xmin=6
17.Xmax<=999 and Xmax integer => Xmax=58
Answer: Xmax-Xmin+1=53 multiples of 17 between 100 and 999
42!!!!!!!
53 is right.
öhm, what a question ^^
how many four digits numbers are which start with three and end with nine ??
lol
LOL. Thanks t:b:H. I had expected the (highly deserved) abuse quicker than that ;) Greetz.
53
53. The lowest number, when multiplied by 17 and ends with a 3 digit number is 6. 17 divided by 1000 gives you the highest(round down). 58 - 6+1= 53
No, you're wrong...
count'em
102
119
136
153
170
187
204
221
238
255
272
289
306
323
340
357
374
391
408
425
442
459
476
493
510
527
544
561
578
595
612
629
646
663
680
697
714
731
748
765
782
799
816
833
850
867
884
901
918
935
952
969
986
Ok, 53 is right, I just wanted to stir things up.
No, you are all wrong...
Count'em
n1
102
119
136
153
170
187
204
221
238
255
272
289
306
323
340
357
374
391
408
425
442
459
476
493
510
527
544
561
578
595
612
629
646
663
680
697
714
731
748
765
782
799
816
833
850
867
884
901
918
935
952
969
986
Ok, it was 53.
Just wanted to know how many would actually count them.
Okay, but i got one(someone can post if they think its worth):
What is worth more, a kilogram of ten-dollar gold coins or a half kilogram of twenty-dollar gold coins?
I seem to have started a silly season :).
I'll take the 1 kg, thanks.
... on second thoughts, if the $20 are dated 1933, I'll take those as they're supposedly worth $7.6 million each (but are illegal to own).
I'd take the 1 kg of $10 because $20 coins weigh more, therefore producing less money.
LOL
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