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Q: When divide ding a 3 digit number by a 1 digit number for what divisors can you get a remainder of 8?

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9. The divisor must be greater than the remainder. A 1 digit divisor that is greater than 8 can only be 9.

23 divide 3974

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.

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.

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.

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

im sorry but i can t figure it out

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

0.28235294117647

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.

5, or .8 with a three repeating.

109 is a prime number. Therefore, the only divisors are 1and 109.

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.

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)

62

I think you mean, the largest four digit number that has exactly three divisors. 2X3X1663=9,978

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.

The single-digit divisors of 7,000 are 1, 2, 4, 5, 7, 8.

No.

Yes- A remainder can be any number less than the dividend (the number by which the divisor is divided). An example of a 2 digit number is: 131/11=11 remainder 10.

The single digit divisors of 420 are: 1,2,3,4,5,6, and 7

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