The LCM of 5 and 9 is 45 and so 45*22 = 990
Any number whose last digit is 0 would be divisible by both these numbers. 1000, 1010, 1020...
If n is divisible by both 5 and 6, then it should be divisible by 30 (5 * 6). Considering you are asking for only two-digit numbers, the answer(s) would be 30, 60, and 90. So, three numbers.
None. Extremely simple proof: To be divisible by 5 the units digit must be 5. But then it would not be divisible by 2.
The lowest common multiple of 5 and 9 is 45 and 11 times 45 is 495 which is the greatest 3 digit number divisible by 5 and 9
To find the greatest 3-digit number divisible by both 5 and 9, we need to find the highest common multiple of 5 and 9 within the range of 100 to 999. The highest common multiple of 5 and 9 is 45. To find the greatest 3-digit number divisible by 45, we divide 999 by 45, which equals 22 with a remainder. Therefore, the greatest 3-digit number divisible by both 5 and 9 is 45 x 22 = 990.
Any number whose last digit is 0 would be divisible by both these numbers. 1000, 1010, 1020...
If n is divisible by both 5 and 6, then it should be divisible by 30 (5 * 6). Considering you are asking for only two-digit numbers, the answer(s) would be 30, 60, and 90. So, three numbers.
None. Extremely simple proof: To be divisible by 5 the units digit must be 5. But then it would not be divisible by 2.
10977120 would be one. There are several.
The lowest common multiple of 5 and 9 is 45 and 11 times 45 is 495 which is the greatest 3 digit number divisible by 5 and 9
To find the greatest 3-digit number divisible by both 5 and 9, we need to find the highest common multiple of 5 and 9 within the range of 100 to 999. The highest common multiple of 5 and 9 is 45. To find the greatest 3-digit number divisible by 45, we divide 999 by 45, which equals 22 with a remainder. Therefore, the greatest 3-digit number divisible by both 5 and 9 is 45 x 22 = 990.
All numbers that are divisible by 8 and 12 would also be divisible by 4.
There is no such number. Once you have one n-digit number which is divisible by 4 then 10 times that number will be an n+1 digit number which is divisible by 4. And this process can continue ad infinitum.
For this kind of question, start from the left, and use the greatest possible digit for each position. Thus, in the first position you would have a 9, in the second an 8, etc.
The list of numbers that are divisible by three is infinite. The first four are: 3, 6, 9, 12 . . .
A Venn diagram for numbers divisible by both 4 and 5 would have two overlapping circles. One circle would represent numbers divisible by 4, while the other circle would represent numbers divisible by 5. The overlapping region where the two circles intersect would represent numbers divisible by both 4 and 5. This intersection would include numbers that are multiples of both 4 and 5, such as 20, 40, 60, and so on.
4,000 That's a 4 digit number, and it's not divisible by 9 (the implication in the question is probably "without remainder"). A 5 digit would be 90000. Another would be 18000 etc etc.