Largest 3-Digit Number Divisible By 5 & 9

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Hey guys! Today, we're diving into a cool math problem: What's the largest three-digit number that has all different digits and can be divided evenly by both 5 and 9? Let's break it down step by step so you can totally nail it. This is a classic number theory problem, so understanding the principles here will help you tackle similar questions with confidence. We'll explore the divisibility rules, look at how to construct the number, and then confirm our answer. Ready to get started?

Understanding the Divisibility Rules

Before we jump into finding the number, let's refresh our memory on the divisibility rules for 5 and 9. These rules are super handy because they let us quickly check if a number can be divided evenly without actually doing the division. Knowing these rules is key to solving our problem efficiently.

Divisibility by 5

A number is divisible by 5 if its last digit is either 0 or 5. This is the first rule we'll use to narrow down our options. Think about it – all multiples of 5 end in either 0 or 5 (like 5, 10, 15, 20, and so on). So, when we're looking for our three-digit number, we know immediately that it has to end in either 0 or 5. Keep this in mind as we move forward!

Divisibility by 9

A number is divisible by 9 if the sum of its digits is divisible by 9. This is another crucial rule. For example, the number 81 is divisible by 9 because 8 + 1 = 9, and 9 is divisible by 9. Similarly, 126 is divisible by 9 because 1 + 2 + 6 = 9, which is also divisible by 9. This rule helps us ensure that when we add up the digits of our three-digit number, the total can be divided evenly by 9. This significantly narrows down our possibilities and helps us find the correct answer faster.

Constructing the Number

Now that we know the divisibility rules for 5 and 9, let's start building our three-digit number. Remember, we want the largest number possible with distinct digits. We'll start by considering the hundreds place, then the tens, and finally the units place, keeping in mind our divisibility rules.

Starting with the Hundreds Place

To get the largest three-digit number, we want the highest possible digit in the hundreds place. Let's try 9. This is a good starting point because it maximizes the overall value of the number. So, for now, our number looks like 9 _ _.

Considering the Units Place

Since our number needs to be divisible by 5, the last digit must be either 0 or 5. Let's first consider 5 as the last digit. This gives us 9 _ 5. Now we need to find a digit for the tens place that satisfies the divisibility rule for 9. The sum of the digits should be divisible by 9.

Finding the Tens Digit

We have 9 _ 5, so the sum of the digits we have so far is 9 + 5 = 14. To make the total sum divisible by 9, we need to add a number that will bring the sum to the nearest multiple of 9. The next multiple of 9 after 14 is 18. So, we need 18 - 14 = 4 as the tens digit. This gives us the number 945. Now, let's check if this number meets all our criteria:

  • It's a three-digit number.
  • It has distinct digits (9, 4, and 5 are all different).
  • It ends in 5, so it's divisible by 5.
  • The sum of its digits is 9 + 4 + 5 = 18, which is divisible by 9, so the number is divisible by 9.

So, 945 seems like a strong contender! But, just to be sure we've found the absolute largest, let's consider the other possibility for the units digit.

Alternative Units Digit: 0

If we use 0 as the last digit, our number looks like 9 _ 0. Now we need to find the tens digit. The sum of the digits we have so far is 9 + 0 = 9. To make the total sum divisible by 9, we need to add another multiple of 9. We could add 0, but we need distinct digits, so that won’t work. The next multiple of 9 we can add is 9 itself, but we already have 9 in the hundreds place. So, we need to find the largest digit less than 9 that, when added to 9, results in a multiple of 9.

If we try 8 as the tens digit, our number would be 980. Let's check if this works:

  • It's a three-digit number.
  • It has distinct digits (9, 8, and 0 are all different).
  • It ends in 0, so it's divisible by 5.
  • The sum of its digits is 9 + 8 + 0 = 17. Whoops! 17 isn't divisible by 9, so 980 doesn't work.

Let's try 7 as the tens digit. Our number would be 970. Now let's check:

  • It's a three-digit number.
  • It has distinct digits (9, 7, and 0 are all different).
  • It ends in 0, so it's divisible by 5.
  • The sum of its digits is 9 + 7 + 0 = 16. Nope, 16 isn't divisible by 9 either.

Let's keep going. Try 6 as the tens digit. Our number is 960:

  • It's a three-digit number.
  • It has distinct digits (9, 6, and 0 are all different).
  • It ends in 0, so it's divisible by 5.
  • The sum of its digits is 9 + 6 + 0 = 15. Still not divisible by 9.

How about 5? But wait, we can’t use 5 because we already tried 5 as a units digit and got 945. Let’s try 4 as the tens digit. The number would be 940:

  • It's a three-digit number.
  • It has distinct digits (9, 4, and 0 are all different).
  • It ends in 0, so it's divisible by 5.
  • The sum of its digits is 9 + 4 + 0 = 13. Nope, not divisible by 9.

Let’s try 3 as the tens digit. The number is 930:

  • It’s a three-digit number.
  • It has distinct digits (9, 3, and 0 are all different).
  • It ends in 0, so it’s divisible by 5.
  • The sum of its digits is 9 + 3 + 0 = 12. Still not divisible by 9.

Finally, let's try 2. Our number is 920:

  • It's a three-digit number.
  • It has distinct digits (9, 2, and 0 are all different).
  • It ends in 0, so it's divisible by 5.
  • The sum of its digits is 9 + 2 + 0 = 11. Nope, not divisible by 9.

And then, let's try 1. Our number is 910:

  • It's a three-digit number.
  • It has distinct digits (9, 1, and 0 are all different).
  • It ends in 0, so it's divisible by 5.
  • The sum of its digits is 9 + 1 + 0 = 10. This also is not divisible by 9.

But what about 900? The sum of the digit is 9+0+0=9. So, it is divisible by 9, but the digits are not distinct.

So, we found a working number and tried every possibility and it failed. Let's ensure this is true.

Confirming the Answer

We've found that 945 fits all the requirements. It's a three-digit number with distinct digits, and it's divisible by both 5 and 9. We also tried 9 _ 0 but failed because none of the numbers that meet the requirements.

Therefore, the largest three-digit natural number with distinct digits that is divisible by both 5 and 9 is 945. You nailed it!

Wrapping Up

So, there you have it! Finding the largest three-digit number divisible by 5 and 9 involves understanding divisibility rules and systematically constructing the number. It's like a puzzle, and you've just solved it. Keep practicing these kinds of problems, and you'll become a math whiz in no time. Keep up the great work, and see you in the next math adventure!