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.
Every multiple of 4 is an even number.
Every even number must end in 0, 2, 4, 6, or 8.
'3' is not on that list.
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Check it out: 4,8,12,16,20,24,28,32,36. No threes.
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.
I think so. Because it ends in a 3.
If the number is even, it's divisible by 2. If the sum of the digits is a multiple of 3, the whole number is divisible by 3. If the last two digits are a multiple of 4, the whole number is divisible by 4. If the last digit is a 0 or a 5, the whole number is divisible by 5. If the number is even and divisible by 3, it's divisible by 6. If the sum of the digits is a multiple of 9, the whole number is divisible by 9. If the number ends in 0, it's divisible by 10.
Any multiple of 6 is a multiple of 3 and 6.
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
Check it out: 4,8,12,16,20,24,28,32,36. No threes.
All multiples of 6 are even numbers.
Any multiple of four must be even; any number ending in three is odd. Look at the multiples of four from 1 to 9. They are in the order : 4,8,12,16,20,24,28,32,36. Now, any number when multiplied by 4, will have the one's digit as the one's digit of the above multiples In the above multiples, no multiple has 3 as the one's digit, i.e., no multiple ends in three. Take an example. 15x4 will have 0 in the one's digit because 5x4 = 20 and has 0 in the one's place.
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.
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.
Because the multiples of 4 are even numbers and 3 is an odd number
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
A prime number has no factors other than 1 and itself. Each multiple of fifteen includes 3, 5, and 15 among its factors.
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.
The digit with which a multiple of 4 ends depends on the last digit of the other factor. If the last digit is a zero, the product ends with zero; if the last digit is a 1, the product ends with 4; etc. The only options for the last digit of the product are 0, 2, 4, 6, 8.