Aman Bansal
Last Activity: 14 Years ago
dear rajat,
Lets look at a sample set of 1001 thru 1100, every number in here contains a zero and as such is eliminated based on the question. Same applies to 2001 thru 2100, 3001 thru 3100 ...... and finally 9001 thru 9100
Now look at any of the remaining sets of 100 numbers for ex. 1101 thru 1200. We have (100/4) 25 multiples of 4 in here and of those 25, 5 contain zero and of the remaining 18, 9 are divisible by 8 and 4. That leaves us 9 numbers which are divisible only by 4 but not by 8.
This trend is similar for any set of 100 numbers. Now we know that we have 9 numbers which do not contain zero and are divisible by 4 but not by 8 in a set of 100 numbers.
Alll we have to do now is see how many sets of 100 numbers we have i.e. 1000-1099, 1100-1199................. 9899 thru 9999 or we can simply say (9999-999)/100 = 90.
Do not forget to eliminate the sets of 1001 thru 1100, 2001 thru 2100 ....... 9001 thru 9100, altogether 9 sets.
That leaves us 90-9 = 81 sets
Each of these 81 sets have 9 numbers that are relevant for us, so the answer is
81 * 9 = 729
Don’t forget to approve the answer if it is beneficial to you… !!
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AMAN BANSAL