"Regrouping" is a more modern word for "borrowing". When subtracting with decimals, if you are trying to subtract a larger digit from a smaller digit, you "regroup" the next digit to the left by taking one away from it and adding 10 to the number you are subtracting from. Example 84 - 19 _____ You can't subtract 9 from 4, so you take one away from the next digit over (the 8) and add 10 to the 4. 14 - 9 is 5 in the ones digits 7 - 1 is 6 in the tens digits Now if you are subtracting mixed numbers, the regrouping process is essentially the same, except that instead of always regrouping by tens, we regroup by the denominator size. 8 1/5 - 3 3/5 ______ We can't subtract 3/5 from 1/5, so we regroup one unit from the 8 into 5 fifths. 7 6/5 -3 3/5 _______ 4 3/5 It is very easy when you get some practice doing it.
Allowing repetitions, there are 9 combinations. Without repeated digits, there is only one combination of 3 digits from 3.
Adding the digits together gives 12, which is divisible by 3, so 273 is divisible by 3, and thus not prime.
Without repeating digits (not digets!) and without leading 0s, 600 of them.
22+2=
The problem of adding 23 and 40 is trivial. Since 2 + 4 = 6, we add the tens column and get 60; there is only 3 in the ones column so the answer is 63.
What is the answer for 8 1/3 - 5 2/6
"Regrouping" is a more modern word for "borrowing". When subtracting with decimals, if you are trying to subtract a larger digit from a smaller digit, you "regroup" the next digit to the left by taking one away from it and adding 10 to the number you are subtracting from. Example 84 - 19 _____ You can't subtract 9 from 4, so you take one away from the next digit over (the 8) and add 10 to the 4. 14 - 9 is 5 in the ones digits 7 - 1 is 6 in the tens digits Now if you are subtracting mixed numbers, the regrouping process is essentially the same, except that instead of always regrouping by tens, we regroup by the denominator size. 8 1/5 - 3 3/5 ______ We can't subtract 3/5 from 1/5, so we regroup one unit from the 8 into 5 fifths. 7 6/5 -3 3/5 _______ 4 3/5 It is very easy when you get some practice doing it.
Not sure what you meant, but 242 is not divisible by 3. By adding all digits, you can determine whether it is a multiple of 3 or not.
Yes, it is. Your answer is 406. You can determine whether a number is divisible by 3 by adding the digits. If the sum of the digits (in the above, 1+2+1+8 = 12) equals 3, the number is divisible by 3.
33 = 27 with repetition, 3! = 3*2*1 = 6 without repetition.
EXAMPLE: 24+2= 26 NO REGROUPING 56+3=59 NO REGROUPING 24+8=32 IS REGROUPING 56+4=60 IS REGROUPING TAKING THE ONES PLACE ONLY: FIRST EXAMPLE 4+2=6 HAS TO BE LESS THAN 9 4+8=12 YOU MAKE 10 IN THE ONES PLACE YOU CARRY OVER WHICH NOW THEY ARE CALLING REGROUPING. WE JUST CALLED IT CARRYING OVER AND BORROWING. HOPE THIS HELPS.
Allowing repetitions, there are 9 combinations. Without repeated digits, there is only one combination of 3 digits from 3.
Sum of digits is divisible by three so yes. You can check if a number is divisible by 3 (without actually doing the division), by adding up the digits. If that sum is divisible by 3 then the original number is divisible by 3: 6 + 8 + 4 = 18 (which is divisible by 3; if not sure add again: 1 + 8 = 9, yes) You'll have to 'memorize' that 3, 6 & 9 are divisible by 3.
Adding the digits together gives 12, which is divisible by 3, so 273 is divisible by 3, and thus not prime.
Yes. 33 = 3 * 11. You can tell it's divisible by 3 because adding the digits together gives six, which is divisible by 3.
With repeated digits, 43 = 64 Without repeated digits, 4*3*2 = 24