0
999 is divisible by 9, but not by six; the next lower number divisible by 9 is 990, which is also divisible by 6, so that's the answer.
Some shortcuts for divisibility:
0 is divisible by any number.
If the last digit of a number is divisible by 2, the number itself is divisible by 2.
If the sum of the digits of a number is divisible by 3, the number itself is divisible by 3.
If the last TWO digits of a number are divisible by 4, the number itself is divisible by 4.
If the last digit of a number is divisible by 5, the number itself is divisible by 5.
If a number is divisible by both 2 and 3, it is divisible by 6.
If the last THREE digits of a number are divisible by 8, the number itself is divisible by 8.
If the sum of the digits of a number is divisible by 9, the number itself is divisible by 9.
990: 9+9+0=18, which is divisible by 9, so 990 is divisible by 9.
18 is also divisible by 3, so 990 is divisible by 3, and since 990 ends in 0 it's also divisible by 2, meaning that it's divisible by 6 as well.
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4-digit number divisible by 5 and 7?
To find a 4-digit number that is divisible by both 5 and 7, we need to find the least common multiple of 5 and 7, which is 35. The smallest 4-digit number divisible by 35 is 1005 (35 x 29 = 1005). The largest 4-digit number divisible by 35 is 9945 (35 x 284 = 9945). Therefore, the 4-digit numbers divisible by both 5 and 7 range from 1005 to 9945.
What is a 3 digit number that is divisible by both 3 and 5?
If it's to be divisible by 5, it must end in 5 or 0. And if it's to be divisible by 3, adding the digits together must result in a multiple of 3.The smallest such number is 105; the largest is 990.Since 3 x 5 = 15, all three-digit multiples of 15 would qualify. Note that 105 = 15 x 7, and 990 = 15 x 66.
Three digit number that is divisible by 4 and 9?
To be divisible by both 4 and 9 a number must be a multiple of the Least Common Factor of 4 and 9. The LCM is 36. The first 3 digit number in each hundred that is a multiple of 36 are as follows :- 108, 216, 324, 432, 504, 612, 720, 828 and 900.
Which digits should come in place of and if the no 62684 is divisible by both 8 and 5?
For a number to be divisible by both 8 and 5 then : 1) the final digit must be zero (as a multiple of 5 ending in 5 is not divisible by 8) 2) As 1000 is divisible by 8 then only the last 3 digits of the number need to be checked to confirm if it is divisible by 8. 680 ÷ 8 = 85. Therefore the number has to be changed to 62680 to be divisible by both 8 and 5. Therefore, replace the digit 4 in 62684 with 0.
What is the divisibility rule for 15?
The number is divisible by both 3 and 5. A number is divisible by 3 if the sum of the digits are and by 5 if the last digit is 0 or 5. Ex: 863,145 Non Ex: 93,460
