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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.

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Q: Which digits should come in place of and if the no 62684 is divisible by both 8 and 5?
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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 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 712 divisible by 2 3 4 5 6 10?

712 is divisible by 2 because the digit at ones place is an even number. 712 is not divisible by 3 because the sum of the numbers(7+1+2=10) is not divisible by 3. 712 is divisible by 4 as the last two digits(12) are divisible by 4. 712 is not divisible by 5 because the last digit is neither 5 nor 0. 712 is not divisible by 6 because it is not divisible by 3. 712 is not divisible by 10 because the last digit is not 0.


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


Place value of 0.92?

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Related questions

Which digits should come in place of and if the number 62684 is divisible by both 8 and 5?

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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?

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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


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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?

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What is a number with three digits has a two in the hunderds place its divisible by 3 4 5?

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