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Question 565182: Find all values that satisfy ALL the following conditions:
1. Positive whole numbers less than 100
2. Four more than each number is a multiple of 6.
3. The sum of the digits of each number is a multiple of 4.
4. Two digit numbers where the ten’s digit is greater than the one’s digit
Found 2 solutions by KMST, richard1234:
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! Positive whole numbers less than 100 means at most two digits, but
Two digit numbers where the ten’s digit is greater than the one’s digit implies exactly 2 digits.
Four more than each number is a multiple of 6 means even numbers, because an odd number plus 4 would still be odd and could not be a multiple of 6. So the ones number could be 0, 2, 4, 6, or 8.
The last two conditions
3. The sum of the digits of each number is a multiple of 4.
4. Two digit numbers where the ten’s digit is greater than the one’s digit
limit choices even further.
If the ten’s digit is greater than the ones digit, the ones digit cannot be 8. (The tens digit would have to be 9, and then the sum would be odd.
The ones digit has to be 0, 2, 4, or 6.
If the ones digit is 6, adding a greater tens digit (could only be 7, 8, or 9) would make the sum of digits 13, 14, or 15, and it would not be a multiple of 4.
If the ones digit is 4, we can make a multiple of 4 by adding 8 as the tens digit (5, 6, 7, and 9 do not work). The number would be 84, which is a multiple of 6, so that four more than 84 (88) would NOT be a multiple of 6.
Maybe the ones digit could be 2. Adding 6 as the tens digit would make the sum a multiple of 4 (2+6=8). Other numbers (3, 4, 5, 7, 8, or 9) do not work as tens digit. The number would be 62, and adding 4 we would get 66, which is a multiple of 6.
Our first answer is  .
Or maybe the ones digit is zero. Then the tens digit could be 4 or 8, and the digits would add up to a multiple of 4. The numbers would be 40 or 80. Adding 4 to each we get 44 (not a multiple of 6) and 84 (6 times 14).
So our second and last answer is  .
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! Modular arithmetic solution:
From (2) we know that the number is even, and it is congruent to -4 mod 6 (which is 2 mod 6 or 2 mod 3). Therefore the sum of the digits is 2 mod 3.
From (3), the sum of the digits is a multiple of 4. Since 8 is the only (small) multiple of 4 congruent to 2 mod 3, the sum of the digits is 8.
From (4), the tens digit is greater than the ones digit. Out of the numbers 17, 26, ..., 62, 71, 80, only the numbers 53, 62, 71, 80 satisfy (4). However we already know the number is even, so 62 and 80 are the only possible numbers.
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