361
The smallest whole number that can be divided by both 3 and 6 and leaves a remainder of 1 is 7.
24
yes
5. To leave a remainder of 1 when divided by 2, the number must be odd. To leave a remainder of 2 when divided by 3, the number must also be two more than a multiple of 3. The multiples of 3 which are odd are 1 x 3, 3 x 3, 5 x 3, etc. The smallest odd multiple of three is 1 x 3 = 3 ⇒ required number is 3 + 2 = 5.
a number that can be divided and leave no remainder.
The smallest whole number that can be divided by both 3 and 6 and leaves a remainder of 1 is 7.
24
To find the greatest number that will divide 11296 and leave a remainder of 11, we need to use the concept of divisors. The number that satisfies this condition is called the Greatest Common Divisor (GCD). By using the Euclidean algorithm, we can find that the GCD of 11296 and 11 is 1. Therefore, the greatest number that will divide 11296 and leave a remainder of 11 is 1.
28 is one such number.
Any number greater than 199 will divide into both zero times and leave a remainder (of the original number); there is no upper limit to the numbers greater than 199. -------------------------------- The largest number to divide into both and leave the SAME remainder is 56 (leaving a remainder of 31 in both cases).
30/21 = 1 with a remainder of 9
yes
3 can't divide into anything and leave a remainder of 28. If you divide 43 into 157, the answer is 3, with a remainder of 28.
5. To leave a remainder of 1 when divided by 2, the number must be odd. To leave a remainder of 2 when divided by 3, the number must also be two more than a multiple of 3. The multiples of 3 which are odd are 1 x 3, 3 x 3, 5 x 3, etc. The smallest odd multiple of three is 1 x 3 = 3 ⇒ required number is 3 + 2 = 5.
a number that can be divided and leave no remainder.
Well, well, well, looks like we have a math problem on our hands. The number of positive integers that divide 2004 to leave a remainder of 24 is the same as the number of positive integers that divide 1980. Why? Because 2004 - 24 = 1980. So, the answer is the number of divisors of 1980, which can be calculated by prime factorizing 1980 and using the formula for the number of divisors. Voila!
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.