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4 is an even number and so all multiples of 4 must be even. A number ending in 3 is not even and so cannot be a multiple of 4.

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Q: Why a number ends in 3 cannot be multiple of 4?

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Check it out: 4,8,12,16,20,24,28,32,36. No threes.

A number that ends in 3 can't be a multiple of 4 because the 4 times table goes in a pattern, (e.g 4, 8, 12, 16, 20, 24.) If the number ends in either 0 2 4 6 8 then it couldbe a possible multiple of 4. It always ends in an even number. 3 is not an even number so it is not going to be the last number of a multiple of 4.

I think so. Because it ends in a 3.

There is no such number. Since if x were the largest multiple of 3 and 7 then what about 2x? 2x would be a multiple of 3 since x is a multiple of 3; 2x would be a multiple of 7 since x is a multiple of 7; and 2x is bigger than x. So x cannot be the largest.

Any multiple of 6 is a multiple of 3 and 6.

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Check it out: 4,8,12,16,20,24,28,32,36. No threes.

If a number is divisible by 4, it also means that the same number is divisible by 2. But if the number ends in a 3, it can't be divisible by 2 and, to a further extent, can't be divisible by 4.

No.3 is less than 5, thus cannot be a multiple of it; 3 is prime, therefore cannot be a multiple of any number besides 3 (which is one of its factors); by definition, 3 is ismply not a multiple of any number but 1 and 3.

A number that ends in 3 can't be a multiple of 4 because the 4 times table goes in a pattern, (e.g 4, 8, 12, 16, 20, 24.) If the number ends in either 0 2 4 6 8 then it couldbe a possible multiple of 4. It always ends in an even number. 3 is not an even number so it is not going to be the last number of a multiple of 4.

If the number is even, it is a multiple of 2 If the sum of the digits make a number divisible by 3, the number is a multiple of 3 If the number ends in 5 or 0, the number is a multiple of 5 If the number is divisible by 2 and 3, the number is a multiple of 6 If the sum of the digits make a number divisible by 9, the number is a multiple of 9

All multiples of 6 are even numbers.

No number that is a multiple of 3, can be a prime number. A prime number must only be divisible by itself and 1. It cannot be divisible by any other number. Therefore if it is a multiple of 3, then it must be divisible by 3 and hence, not a prime number.

Since 6 is an even number, its multiples are all even numbers as well. This means that each of its multiples ends in an even number (0, 2, 4, 6, 8). Since 3 is an odd number, any integer ending in a 3 is odd, thus proving that the number is not a multiple of 6. cause 6 is even and any number multiply by an even number is even Obviously any number ending with a 3 is odd 6 = 3 x 2 6 x Z = 3 x 2 x Z so the number is even and therefore cannot ends in 1, 3, 5, 7 or 9

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There are different rules for different number 2: ends in 2,4,6,8,0 3: digits add up to multiple of 3 4: last two digits are divisible by 4 5: ends in 5,0 6: follows the rules of both 2 and 3 9: digits add up to a multiple of 9 10: ends in 0

I think so. Because it ends in a 3.

It would help to have that other number. If the other number is a multiple of 3, the LCM is that number. If it is not a multiple of 3, the LCM is 3 times that number.

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