Any two digit number that fits the formula 3n+1
For example if n=4, 3x4+1=13 which fits your question.
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.
103
121
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.
0.0003
Every 2 digit number, when divided by the number one less than it, will result in a remainder of 1.
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.
17
103
121
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.
27.2222
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
506
I think you're wanting a number of two digits, one of which is 3, that when divided by 7 gives a quotient and a remainder of 1 and when that quotient is divided by 2 it gives a remainder of 1: Answer: 36 36 ÷ 7 = 5 r 1 5 ÷ 2 = 2 r 1 If you want the number to be such that if it is divided by 7 the remainder is 1 and if it is divided by 2 the remainder is 1, then: Answer: 43 43 ÷ 7 = 6 r 1 43 ÷ 2 = 21 r 1
The remainder is 8. To find the remainder when a number is divided by 9, add the digits together; if this sum has more than 1 digit, repeat until one digit remains. This digit is the remainder, unless it is 9 in which case the remainder is 0. examples: remainder 53 → 5 + 3 = 8 → remainder is 8 when 53 is divided by 9. remainder 126 → 1 + 2 + 6 = 9 → remainder is 0, ie no remainder, 126 is divisible by 9. remainder 258 → 2 + 5 + 8 = 15 → 1 + 5 = 6 → remainder is 6 when 258 is divided by 9.
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).