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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 else can I help you with?
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.
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.
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
How do you know a number is divisible by 4?
If a number can be divisible by 2 twice( ei, 48 in half is 24 and 24 can be divided in half. But you can check, without doing any dividing. If the last 2 digits (the tens and ones place digits) is divisible by 4, then the whole number is divisible by 4. How to tell if the last two digits are divisible by 4: add the ones digit to twice the tens digit: if this sum is divisible by 4, then so is the original. This sum can be repeated until a single digit remains. IF this digit is 4 or 8, then the original number is divisible by 4. So if you have a number 975239806723498658859303524, then the last two digits are 24, which is divisible by 4 and the entire number is divisible by 4 (4 + 2×2 = 8 which is 4 or 8, so 24 is divisible by 4). Why this works: Suppose you have a whole number M. Rewrite the number as M = 100*P + Q, where P & Q are whole numbers. Note that Q represents the last two digits, and P is the rest of the number. Now divide this by 4: M/4 = (100*P + Q)/4 = 100*P/4 + Q/4 = 25*P + Q/4. So since P is a whole number, then 25*P is a whole number, and if Q/4 is a whole number, then [M/4 = 25*P + Q/4], which is the sum of two whole numbers is also a whole number, so M is divisible by 4. Let M = 324, so P = 3, and Q = 24. 324 = 100*3 + 24, and 324/4 = 25*3 + 24/4 = 75 + 6 = 81. So 324 is divisible by 4. A number which is divisible by 4 is a multiple of 4
Is 712 divisible by 2 3 4 5 6 10?
Oh, dude, let's break it down. So, 712 can be divided by 2 because it's even. It can also be divided by 4 because if it's divisible by 2, it's also divisible by 4. But, like, it's not divisible by 3, 5, 6, or 10. So, you can totally divide it by 2 and 4, but the others are a no-go.
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.
Which digits should come in place of and if the no62684 id divisible by both 8 and 5?
The last digit should be 0.
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.
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 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.
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
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)
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.
Which of the following is divisible by 3 and by 5 but you not divisible by 10?
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.
When you round a number what should you do with the digits to the right of the place you are rounding to?
After determining whether to round up or down, the digits, to the right of the place, are discarded.
What is a number with three digits has a two in the hunderds place its divisible by 3 4 5?
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.
