 My Math Forum How many three digit number whose digit sum is 13 leave reminder 1 divided by 4
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 June 5th, 2014, 01:31 PM #1 Member   Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1 How many three digit number whose digit sum is 13 leave reminder 1 divided by 4 Hi What is the way to do this in easy way? How many three digit natural number leave a reminder 1 when divided by 4 and the sum of the individual digits are 14. I did this by trial and error. But is there any better way? Thanks June 5th, 2014, 03:04 PM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 For digitsum 13, (from the topic title) The last digit is odd, so it's 1, 3, 5, 7 or 9. If it's 1, 5 or 9, the last digit is even, so 0, 2, 4, 6 or 8. If it's 3 or 7, the last digit is odd, so 1, 3, 5, 7 or 9. Since the digitsum is 13, If the last digit is 1 then the middle digit is at least 4. That's three possibilities; 4, 6 or 8. Now, for every such number there is one choice for the first digit. If the last digit is 5 then the middle digit is at most 6, since 085 has 2 digits. Gives 4 possibilities; 0, 2, 4 or 6. If the last digit is 9 then the middle digit is at most 2. That's two possibilies. If the last digit is 3 then the midde digit could be any such digit. That's 5 posibilities. If the last digit is 7 then the middle digit is at most 5. That's 3 possibilities. So in total, we have 3 + 4 + 2 + 5 + 3 = 16 such numbers. Can you do it for digitsum 14 now, as in your post? June 5th, 2014, 06:55 PM #3 Member   Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1 Hi Hoempa Its easy now. It will be 18 numbers. Thanks again June 6th, 2014, 12:27 AM #4 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Yep, that's it ! Tags digit, divided, leave, number, reminder, sum ,

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# The sum of the three digits is the sum of which is divided by 13

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