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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: 2,778 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: 2,778 Thanks: 361 
Yep, that's it !


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digit, divided, leave, number, reminder, sum 
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