5, or .8 with a three repeating.
That's not possible. The largest single-digit number by which you might divide is 9. And, by definition, the remainder is always LESS than the number by which you divide. Thus, the largest remainder you can get, when you divide by 9, is 8.
81 is.
mathematics
Divide the two-digit number by the one-digit number. If the remainder is zero then the 2-digit number is a multiple and if not, it is not.
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.
im sorry but i can t figure it out
You can't have a remainder of 6 when you divide by 2! JHC!
20 if you divide by 17. 19 if you divide by 16. 18 if you divide by 15, 17 if you divide by 14. And so on. In fact any number from 10 to 99. That is, every two digit number.
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
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)
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.
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