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Question

How many integers are less than 1000 have the property that the sum of the digit of each such number divisible by 7 and the number itself is divisible by 3.


Solution

A number divisible by 3 has sum of idgits divisible by 3.
We are given sum of digits is divisible by 7 also.
So sum of digits is divisible by 7x3 = 21
Maximum sum of 3 digit number (<1000) = 9+9+9 = 27.
Only number divisible by 21 and < 28 = 21 itself.
Hence sum of digits = exactly 21 and it has to be of 3 digits since maximum sum of 2 digit number = 9+9 = 18.
Let us find all 3 digit numbers having sum of digits = 21
Starting with 9: other 2 digits add to 12 so numbers are: 939.948.957,966,975,984,993 (7 nos.)
Starting with 8: other 2 digits add to 13 so numbers are: 849,858,867,876,885,894 (6 nos.)
Starting with 7: other 2 digits add to 14 so numbers are: 759,768,777,786,795 ( 5nos)
Starting with 6: other 2 digits add to 15 so numbers are: 669,678,687,696  (4 nos)
Starting with 5: other 2 digits add to 16 so numbers are: 579,588,597 (3 nos)
Starting with 4: other 2 digits add to 17 so numbers are: 489,498 (2 nos)
Starting with 3: other 2 digits add to 18 so numbers are: 399 (1 number)
Total number = 7+6+5+4+3+2+1 = 28 Answer



 

Mathematics

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