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June 5th, 2014, 02:31 PM   #1
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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
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June 5th, 2014, 04:04 PM   #2
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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?
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June 5th, 2014, 07:55 PM   #3
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Hi Hoempa

Its easy now. It will be 18 numbers.

Thanks again
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June 6th, 2014, 01:27 AM   #4
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Yep, that's it !
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