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.
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.
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.
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.
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.
The smallest one is 1,008. The largest one is 9,990. In all, there are 500 of them, all of them even numbers divisible by 9. (To find them all, multiply 18 by all integers from 56 to 555.)
99,600 is the largest 5-digit number that meets both requirements.
99,996
990.
99990
The largest such number is 9990. 9990/2 = 4995, and 9990/5 = 1998.
The least three-digit number that is divisible by both 5 and 9 is 135.
9800 The LCM of 25 & 40 is 200, so need the largest multiple of 200 less than 10000. 10000 ÷ 200 = 50 so 49 x 200 = 9800 is the largest 4 digit number divisible by both 25 and 40.
Oh, dude, you're hitting me with the math questions now? Alright, let me think for a sec... So, the largest 5-digit number is 99999, but that's not divisible by 4. The next largest one that works is 99996, which is divisible by both 4 and 9. So, there you go, 99996 is your winner. Math can be fun, right?
There are at least 2. One is 2657205 the other is 3515625.You're right!There are 200,000 of them.The smallest one is 1,000,035 .The largest one is 9,999,990.You're welcome.Here are some more:4999995999999010555202944485etc.Any 7 digit number divisible by 45 works.
Any number whose last digit is 0 would be divisible by both these numbers. 1000, 1010, 1020...
54
54.