The last digit should be 0.
2,4,6,8,and 0 are divisible by 2 in the ones digit. Zero is only divisible in a number with 2 digits or more. 0 itself is not divisible by 2.
You test if the last two digits are divisible by 4. If the digit in the tens' place is odd, the digit in the units place must be 2 or 6. If the digit in the tens' place is even, the digit in the units place must be 0, 4 or 8.
You did not provide the choices to pick from, but here's how you can tell: it must have a five in the ones place, and the sum of the digits must add up to a multiple of 3.
After determining whether to round up or down, the digits, to the right of the place, are discarded.
As these numbers have no common factors then any number divisible by 3, 4 & 5 is also divisible by (3 x 4 x 5) = 60. The answer is thus 240.
Sol. Since the given number is divisible by 5, so 0 or 5 must come in place of $. But, a number ending with 5 is never divisible by 8. So, 0 will replace $. Now, the number formed by the last three digits is 4*0, which becomes divisible by 8, if * is replaced by 4. Hence, digits in place of * and $ are 4 and 0 respectively.
If the number formed by the last three digits is divisible by 8. This requires that: if the digit in the hundreds place is even, the last two digits must form a number divisible by 8 and if the digit in the hundreds place is odd, the last two digits must form a number divisible by 4 but not by 8.
2,4,6,8,and 0 are divisible by 2 in the ones digit. Zero is only divisible in a number with 2 digits or more. 0 itself is not divisible by 2.
To know if a number is composite without listing its factors, you can use these rules:All numbers that end with 2, 4, 6, 8, and 0 (even numbers) are divisible by 2.If the sum of the digits in a number is divisible by 3, the number is divisible by 3.If the last two digits are divisible by 4, the number is divisible by 4.All numbers that have 5 in the one's place are divisible by 5.If a number is divisible by 2 AND is divisible by 3, it is divisible by 6.If the last 3 digits of a number are divisible by 8, the number is divisible by 8.If the sum of the digits of number is divisible by 9, the number is divisible by 9.If a number ends wuth 0, the number is divisible by 10.
For a number to be divisible by both 8 and 5 then : 1) the final digit must be zero (as a multiple of 5 ending in 5 is not divisible by 8) 2) As 1000 is divisible by 8 then only the last 3 digits of the number need to be checked to confirm if it is divisible by 8. 680 ÷ 8 = 85. Therefore the number has to be changed to 62680 to be divisible by both 8 and 5. Therefore, replace the digit 4 in 62684 with 0.
Yes, 268 is divisible by 2 and 4. To check if a number is divisible by 2, we just need to look at the ones place digit. If it is even, then the number is divisible by 2. In this case, the ones place digit of 268 is 8, which is even, so 268 is divisible by 2. To check if a number is divisible by 4, we need to look at the last two digits of the number. If the last two digits
Since 6 = 2 x 3, in order for a number to be evenly divisible by 6, it needs to be divisible by 3 and divisible by 2. 1197 is divisible by 3 (by inspection of sum of digits), but it is not divisible by 2 (the ones place digit must be 0,2,4,6, or 8)
You test if the last two digits are divisible by 4. If the digit in the tens' place is odd, the digit in the units place must be 2 or 6. If the digit in the tens' place is even, the digit in the units place must be 0, 4 or 8.
You did not provide the choices to pick from, but here's how you can tell: it must have a five in the ones place, and the sum of the digits must add up to a multiple of 3.
After determining whether to round up or down, the digits, to the right of the place, are discarded.
As these numbers have no common factors then any number divisible by 3, 4 & 5 is also divisible by (3 x 4 x 5) = 60. The answer is thus 240.
240 is three digits, and has a '2' in the hundreds column. 240/3 = 80 240/4 = 60 240/5 = 48