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Which digits can be in the ones place of a number that is didisible by 2?
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.
Is a number has a 9 in the ones place it is always divisible by 3?
No, a number with a 9 in the ones place is not always divisible by 3. For a number to be divisible by 3, the sum of its digits must be divisible by 3, regardless of the digit in the ones place. For example, the number 29 has a 9 in the ones place, but its digit sum (2 + 9 = 11) is not divisible by 3.
If a number has a 9 in the ones place is it always divisible by 3?
No, a number with a 9 in the ones place is not always divisible by 3. For a number to be divisible by 3, the sum of its digits must be divisible by 3, not just the last digit. For example, the number 39 (3 + 9 = 12, which is divisible by 3) is divisible by 3, but the number 29 (2 + 9 = 11, which is not divisible by 3) is not.
What is 102 divisable by?
102 is divisible by: 1 2 3 6 17 34 51 102.
How do you test to see if the number is divisible by 4?
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.
How is a number divisible by 8?
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.
Which digits can be in the ones place of a number that is didisible by 2?
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.
Which digits should come in place of and if the number 62684 is divisible by both 8 and 5?
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.
How do you explain that even without listing all the factors of a number it is easy to tell that they are composite numbers?
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.
Is a number has a 9 in the ones place it is always divisible by 3?
No, a number with a 9 in the ones place is not always divisible by 3. For a number to be divisible by 3, the sum of its digits must be divisible by 3, regardless of the digit in the ones place. For example, the number 29 has a 9 in the ones place, but its digit sum (2 + 9 = 11) is not divisible by 3.
If a number has a 9 in the ones place is it always divisible by 3?
No, a number with a 9 in the ones place is not always divisible by 3. For a number to be divisible by 3, the sum of its digits must be divisible by 3, not just the last digit. For example, the number 39 (3 + 9 = 12, which is divisible by 3) is divisible by 3, but the number 29 (2 + 9 = 11, which is not divisible by 3) is not.
Is 268 divisible to any of these numbers 2 3 4 5 6 9 10?
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
Is 87 divisible by 7?
No, 87 is not divisible by 7. To determine if a number is divisible by 7, you can use the rule that states a number is divisible by 7 if the difference between twice the digit in the ones place and the number formed by the other digits is either 0 or a multiple of 7. In this case, the number formed by the other digits is 8, and twice the digit in the ones place is 14. The difference between 14 and 8 is 6, which is not a multiple of 7, so 87 is not divisible by 7.
What is 102 divisable by?
102 is divisible by: 1 2 3 6 17 34 51 102.
Which digits should come in place of and if the no 62684 is divisible by both 8 and 5?
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.
What digits can be in the ones place of a number that has 5 as a factor?
0 or 5
Can 1197 be divided by six?
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)
