Wiki User
∙ 14y agoBecause is shows 1020 is divisible by 10 and everything that divides by 10, also divides by 5 and 2, because 10 itself divides by 5 and 2.
Wiki User
∙ 14y agoBack in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." I think "regrouping" is a better term for all of it. These problems might be easier to visualize if you copy them vertically. Example: 56 - 39 Just looking at it, you might think there's a problem with subtracting nine from six until you realize that 56 is 5 tens and 6 ones which is the same thing as 4 tens and 16 ones. Now you can subtract 9 from 16, leaving 7 in the ones place and 3 from 4, (the regrouped 5) leaving 1 in the tens place. 56 - 39 = 17 Example: 45 + 28 5 + 8 is 13, which won't fit in the ones place, so we leave 3 of the ones there and regroup the ten other ones into one ten which we add in the tens column. 1 + 4 + 2 = 7 45 + 28 = 73
A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.
Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." I think "regrouping" is a better term for all of it. These problems might be easier to visualize if you copy them vertically. Example: 56 - 39 Just looking at it, you might think there's a problem with subtracting nine from six until you realize that 56 is 5 tens and 6 ones which is the same thing as 4 tens and 16 ones. Now you can subtract 9 from 16, leaving 7 in the ones place and 3 from 4, (the regrouped 5) leaving 1 in the tens place. 56 - 39 = 17 Example: 45 + 28 5 + 8 is 13, which won't fit in the ones place, so we leave 3 of the ones there and regroup the ten other ones into one ten which we add in the tens column. 1 + 4 + 2 = 7 45 + 28 = 73
Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." I think "regrouping" is a better term for all of it. These problems might be easier to visualize if you copy them vertically. Example: 45 + 28 5 + 8 is 13, which won't fit in the ones place, so we leave 3 of the ones there and regroup the ten other ones into one ten which we add in the tens column. 1 + 4 + 2 = 7 45 + 28 = 73
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.
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Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." These problems might be easier to visualize if you copy them vertically. Example: 56 - 39 Just looking at it, you might think there's a problem with subtracting nine from six until you realize that 56 is 5 tens and 6 ones which is the same thing as 4 tens and 16 ones. Now you can subtract 9 from 16, leaving 7 in the ones place and 3 from 4, (the regrouped 5) leaving 1 in the tens place. 56 - 39 = 17
If the last two digits are divisible by 4 then the number is divisible by 4. Thus, if the tens digit is even and the units digit is 0 or 4 or 8 OR if the tens digit is odd and the units digit is 2 or 6 then the number is divisible by 4.
Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." I think "regrouping" is a better term for all of it. These problems might be easier to visualize if you copy them vertically. Example: 56 - 39 Just looking at it, you might think there's a problem with subtracting nine from six until you realize that 56 is 5 tens and 6 ones which is the same thing as 4 tens and 16 ones. Now you can subtract 9 from 16, leaving 7 in the ones place and 3 from 4, (the regrouped 5) leaving 1 in the tens place. 56 - 39 = 17 Example: 45 + 28 5 + 8 is 13, which won't fit in the ones place, so we leave 3 of the ones there and regroup the ten other ones into one ten which we add in the tens column. 1 + 4 + 2 = 7 45 + 28 = 73
If the digit in the tens place is divisible by 2 and the digit in the units place is a 0, then the number is divisible by twenty.
261, 264, or 267
Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." I think "regrouping" is a better term for all of it. This problem might be easier to visualize if you copy it vertically. Example: 45 + 28 5 + 8 is 13, which won't fit in the ones place, so we leave 3 of the ones there and regroup the ten other ones into one ten which we add in the tens column. 1 + 4 + 2 = 7 45 + 28 = 73
A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.A number is divisible by 4 if the last two digits are divisible by 4.
Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." I think "regrouping" is a better term for all of it. These problems might be easier to visualize if you copy them horizontally. Example: 56 - 39 Just looking at it, you might think there's a problem with subtracting nine from six until you realize that 56 is 5 tens and 6 ones which is the same thing as 4 tens and 16 ones. Now you can subtract 9 from 16, leaving 7 in the ones place and 3 from 4, (the regrouped 5) leaving 1 in the tens place. 56 - 39 = 17 Example: 45 + 28 5 + 8 is 13, which won't fit in the ones place, so we leave 3 of the ones there and regroup the ten other ones into one ten which we add in the tens column. 1 + 4 + 2 = 7 45 + 28 = 73
Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." I think "regrouping" is a better term for all of it. These problems might be easier to visualize if you copy them vertically. Example: 56 - 39 Just looking at it, you might think there's a problem with subtracting nine from six until you realize that 56 is 5 tens and 6 ones which is the same thing as 4 tens and 16 ones. Now you can subtract 9 from 16, leaving 7 in the ones place and 3 from 4, (the regrouped 5) leaving 1 in the tens place. 56 - 39 = 17 Example: 45 + 28 5 + 8 is 13, which won't fit in the ones place, so we leave 3 of the ones there and regroup the ten other ones into one ten which we add in the tens column. 1 + 4 + 2 = 7 45 + 28 = 73
Back in the day, regrouping in addition was called "carrying" and regrouping in subtraction was called "borrowing." I think "regrouping" is a better term for all of it. These problems might be easier to visualize if you copy them vertically. Example: 45 + 28 5 + 8 is 13, which won't fit in the ones place, so we leave 3 of the ones there and regroup the ten other ones into one ten which we add in the tens column. 1 + 4 + 2 = 7 45 + 28 = 73
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