Q: What is the smallest whole number which 5 8 9 can divide and leave a remainder 1?

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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.

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The smallest whole number that can be divided by both 3 and 6 and leaves a remainder of 1 is 7.

24

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.

There are positive integers which divide 2004 to leave a remainder of 24: 30, 33, 36, 44, 45, 55, 60, 66, 90, 99, 110, 132, 165, 180, 198, 220, 330, 396, 495, 660, 990, and 1980

Divide top number by bottom number leave remainder on top and move other number to sideex: x/c= a Rb Key: R=remainderThen, put your answer here:ex: ab/cPut your remainder here:ex: ab/c

It is LCM(3, 4, 5) + 2 = 62