Q: Can a remainder be a 2 digit number?

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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.

2

Every 2 digit number, when divided by the number one less than it, will result in a remainder of 1.

You can't have a remainder of 6 when you divide by 2! JHC!

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)

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.

17

27.2222

62

99

It is 997.

If the number is even, and the last digit is 0 or 5, the remainder is 0 ,or there is no remainder. If the number is odd or even (where the last digit is not 0 or 5), there is a remainder.

5, or .8 with a three repeating.

-2

81 is.

2 x 6 + 0 = 12 2 x 1 + 2 = 4 4 is not [divisible by] 8, so 60 is not divisible by 8. (The remainder when 60 is divided by 8 is 4). To test divisibility by 8: Add together the hundreds digit multiplied by 4, the tens digit multiplied by 2 and the units (ones) digit. If this sum is divisible by 8 so is the original number. (Otherwise the remainder of this sum divided by 8 is the remainder when the original number is divided by 8.) If you repeat this sum on the sum until a single digit remains, then if that digit is 8, the original number is divisible by 8 otherwise it gives the remainder when the original number is divided by 8 (except if the single digit is 9, in which case the remainder is 9 - 8 = 1).

mathematics

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.

0.0003

Regardless of the dividend (the number being divided), no divisor can produce a remainder equal to, or greater than, itself..... dividing by 4 cannot result in a remainder of 5, for example, Therefore the only single-digit number which can return a remainder of 8 is 9. 35 ÷ 9 = 3 and remainder 8

This is impossible. A two digit number n divided by 345 has a remainder equal to n. If you meant to say divided by 3,4,5 or 6 then the answer is 62.

100

111

The simplest answer is 10124 divided by 125.

First note that when you divide any number by the 10, the remainder will be the same number as the units digit. For example, when 26 is divided by 10, the remainder is 6 and when 37832728 is divided by 10, the remainder is 8. For this problem, we just need to know what the units digit of the number 2^400. So let's see if we can see a pattern. 2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32, and 2^6 = 64. The pattern we are looking for is the value of the units digit. It is 2, 4, 8, 6, 2, 4, 8, 6, and it repeats. Every fourth remainder is 8 and since 400 is a multiple of 4, 2^400 will have 6 as its units whatever the number is and therefore the remainder will be 6. Hope that helps :)