Divisibility Rules

Question

It is less than 100,000. It is an even number. It is divisible by 2 3 and 6. None of the digits are less than 4. None of the digits are used more than once. The sum of the digits is 30. The digit in the hundreds place is 2 less than the digit in the thousands place. The digit in the ten-thousands place is one more than the digit in the thousands place. The digit in the tens place is 1 less than the digit in the hundreds place. If the number were increased by 6, it would be divisible by 2, 3, 4, 6, and 9.

Answer 1

Neat-looking problem!

Okay, so we are choosing digits from this list: {4, 5, 6, 7, 8, 9}

5 unique digits which sum to 30.

The numbers in the list sum to 39, so we must omit 9.

Our number must include the digits 4, 5, 6, 7, 8 in some order.

Even, so cannot end in 5 or 7.

Let's see what else we can figure out.

"The digit in the hundreds place is 2 less than the digit in the thousands place."

We have one of these:

A75DE or

A86D4 or

A64D8

What does that allow for the tens place?

A6438 <-- omitted!

A754E or

A8654 or

Three final options to select from:

87546

67548 <-- omitted; n is divis. by 4, so n+6 must not be.

78654 <-- omitted; digit in ten-thousands is not 1 more than adjacent digit.

Confiming that 87546 + 6 is divisible by 9 and 4:

87552 <-- Sum of digits is 17+10 = 27; Last 2 digits = 52 = (4)(13) --> DONE!

87546 <-- Answer!

Did you or someone you know make this problem up yourself? If so, it's pretty neat!