- **Elementary Math**
(*http://mymathforum.com/elementary-math/*)

- - **How many three digit number whose digit sum is 13 leave reminder 1 divided by 4**
(*http://mymathforum.com/elementary-math/44396-how-many-three-digit-number-whose-digit-sum-13-leave-reminder-1-divided-4-a.html*)

How many three digit number whose digit sum is 13 leave reminder 1 divided by 4Hi 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 |

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? |

Hi Hoempa Its easy now. It will be 18 numbers. Thanks again |

Yep, that's it :idea:! |

All times are GMT -8. The time now is 08:05 PM. |

Copyright © 2019 My Math Forum. All rights reserved.