Divisibility if a number by 3 is not determined by its last digits: instead it is determined by the number's digital root.
You get the digital root of a number by adding together all its digits. If the answer is a big number, then find the digital root of the answer. Keep going until you have a number that is smaller than 10. If this number is 3, 6 or 9 (all divisible by 3) then the original number is divisible by 3. And it not, it is not.
All whole numbers are divisible by 1. Numbers are divisible by 2 if they end in 2, 4, 6, 8 or 0. Numbers are divisible by 3 if the sum of their digits is divisible by 3. Numbers are divisible by 4 if the last two digits of the number are divisible by 4. Numbers are divisible by 5 if the last digit of the number is either 5 or 0. Numbers are divisible by 6 if they are divisible by 2 and 3. Numbers are divisible by 9 if the sum of their digits is equal to 9 or a multiple of 9. Numbers are divisible by 10 if the last digit of the number is 0.
For 4:Are the last two digits in your number divisible by 4? If so, the number is too! For example: 358912 ends in 12 which is divisible by 4, thus so is 358912. For 6: The number must be divisible by 2 and 3. For 8: This one's not as easy, if the last 3 digits are divisible by 8, so is the entire number. Example: 6008 - The last 3 digits are divisible by 8, therefore, so is 6008.
Even numbers are divisible by 2. Even numbers that are divisible by 3 must be divisible by 2 and by 3, so they have to be divisible by 2 x 3.6, 12, 18, 24 …
For starters, the number 3 is divisible by 3 but not by 2. Next, 9, 27 etc. All the odd multiples of 3 are divisible by 3 but not 2. So there is an infinite number of numbers that are divisible by 3 and not 2.
All 4 digit numbers that are divisible by 9 are also divisible by 3. The first is 1008 and the last is 9999.
Last digit (0) is even, so it is divisible by 2 4 + 3 + 2 + 0 = 9 which is divisible by 3, so it is divisible by 3 last digit is 0 or 5, so it is divisible by 5 4 + 3 + 2 + 0 = 9 which is divisible by 9, so it is divisible by 9 last digit is 0, so it is divisible by 10 → 4320 is divisible by all the numbers 2, 3, 5, 9, 10
All whole numbers are divisible by 1. Numbers are divisible by 2 if they end in 2, 4, 6, 8 or 0. Numbers are divisible by 3 if the sum of their digits is divisible by 3. Numbers are divisible by 4 if the last two digits of the number are divisible by 4. Numbers are divisible by 5 if the last digit of the number is either 5 or 0. Numbers are divisible by 6 if they are divisible by 2 and 3. Numbers are divisible by 9 if the sum of their digits is equal to 9 or a multiple of 9. Numbers are divisible by 10 if the last digit of the number is 0.
Oh, dude, let me break it down for you. So, we start at 21 because it's the first number between 20 and 50 that's divisible by 3, and then we just keep adding 3 until we hit 48. So, that gives us 10 numbers in total. Easy peasy, right?
For 4:Are the last two digits in your number divisible by 4? If so, the number is too! For example: 358912 ends in 12 which is divisible by 4, thus so is 358912. For 6: The number must be divisible by 2 and 3. For 8: This one's not as easy, if the last 3 digits are divisible by 8, so is the entire number. Example: 6008 - The last 3 digits are divisible by 8, therefore, so is 6008.
Even numbers are divisible by 2. Even numbers that are divisible by 3 must be divisible by 2 and by 3, so they have to be divisible by 2 x 3.6, 12, 18, 24 …
Counting up from 1, there is a number divisible by 3 every 3 numbers. Thus to find out how many numbers are divisible by 3, we just have to divide the number we're counting to by 3. In this case, the top number is 50, so do: 50/3 = 16 with two remainder. Because the numbers divisible by 3 come last in every set of 3 numbers, we can discard the remainder. Therefore there are 16 numbers between 1 and 50 that are divisible by 3.
For starters, the number 3 is divisible by 3 but not by 2. Next, 9, 27 etc. All the odd multiples of 3 are divisible by 3 but not 2. So there is an infinite number of numbers that are divisible by 3 and not 2.
Oh, dude, let's break it down. So, first, you gotta figure out how many numbers are divisible by 3, right? Then, do the same for 7. But, here's the kicker - you gotta remember to subtract the numbers divisible by both 3 and 7. It's like a math puzzle, but with numbers. So, like, get your calculator out and have fun with it.
All 4 digit numbers that are divisible by 9 are also divisible by 3. The first is 1008 and the last is 9999.
No. 26 for instance the sum of the digits is 8 but not divisible by 4. 32 the sum of the digits is 5 but divisible by 4 The rules for some other numbers are 2 all even numbers are divisible by 2 3 The sum of the digits is divisible by 3 4 The last 2 numbers are divisible by 4 5 The number ends in a 0 or 5 6 The sum of the digits is divisible by 3 and is even 7 no easy method 8 The last 3 numbers are divisible by 8 9 The sum of the digits is divisible by 9
yEAH, i THINK SO
Oh, isn't that a happy little question! To find the two-digit numbers divisible by 3, we start by finding the first two-digit number divisible by 3, which is 12. Then, we find the last two-digit number divisible by 3, which is 99. Now, we can count how many numbers there are between 12 and 99 that are divisible by 3.