im sorry but i can t figure it out
mathematics
Oh, dude, yeah, totally! A remainder can definitely be a 2-digit number. It's just whatever is left over after you divide one number by another. So, like, if you divide 100 by 3, you get a remainder of 1, which is a 1-digit number. But if you divide 100 by 7, you get a remainder of 2 digits, which is totally cool too.
You repetitively divide the number by two, taking the remainder as the digit (in binary). When you divide by 2, the remainder will either be 0 or 1.Example: convert 23 (base 10) to binary:23/2 = 11, remainder 1 (this is the ones digit)11/2 = 5, remainder 1 (this is the twos digit)5/2 = 2, remainder 1, (this is the fours digit)2/1 = 1, remainder 0, (this is the eights digit)1/2 = 0, remainder 1, (this is the sixteens digit). So now combine the digits (sixteens is the highest digit in this number):23 (base 10) = 10111 (base 2)
You divide when there is a remainder the same as you divide when there is none. The only difference is that when you divide the last digit in the dividend, you will wither add a decimal point and 0 to the right of the digit and keep dividing, designate the leftover number as a remainder, or you will put the remainder over the divisor to show the remainder as a fraction. For example: 761 divided by 10 is 76 with a remainder of 1. You can write 76 R1, 76 1/10 or 76.1
No.
A 2 digit number divided by a four digit number, such as 2345, will leave the whole 2-digit number as a remainder. It cannot leave a remainder of 1.
9. The divisor must be greater than the remainder. A 1 digit divisor that is greater than 8 can only be 9.
The process of multiplication doesn't produce remainders.The process of division does.If you want to divide a 3-digit number by a one-digit numberand get a remainder of 8, try these:107 divided by 9116 divided by 9125 divided by 9134 divided by 9143 divided by 9..Add as many 9s to 107 as you want to, and then divide the result by 9.The remainder will always be 8.
Every 2 digit number, when divided by the number one less than it, will result in a remainder of 1.
Divide the number by 5, and the remainder in that division is the 'ones' (50) digit. Take that quotient (without the remainder) and divide by 5. The remainder is the 'fives' (51) digit. Continue dividing until you have zero, with a remainder and that will be the leftmost digit. Example 27 (base ten) to base 5: 27 / 5 = 5, remainder 2 5 / 5 = 1, remainder 0 1 / 5 = 0, remainder 1 so 102 (base 5) is the same as 27 (base 10). You can check: 1 is in the (52=25) place, and the 2 is in the 'ones' place. So (1*25) + (0*5) + (2*1) = 27
2
the biggest remainder you can have when dividing by 2 is 1. Reason is ......if you divide an even number by 2 you will always get a round answer. However if you divide an odd number by 2 you will always get a full number with 1 reamining...i.e divide 3 by 2.........2 remainder 1......................divide 19 by 2 this will give you 9 remainder 1. Hope this helps