Short answer: 1.
Long anwer: The prime factorization of 8 is 2 * 2 * 2 and the prime factorization of 11 is 11. By the Fundamental Theorem of Arithmetic, the factorization is unique and all common prime factors must appear in both lists (2, 2, 2 and 11). Since no prime divides both 8 and 11, then the only other possible answers are 1 and -1.
Because 1 > -1, the answer is 1.
The largest 5 digit number is 99999. If this is divided by 11 it leaves a remainder of 9. Therefore the largest 5 digit number divisible by 11 is 99999 - 9 = 99990.
If a number is divisible by both 8 and 11, it will be divisible by 88.
No. It is divisible by 11.No. It is divisible by 11.No. It is divisible by 11.No. It is divisible by 11.
1870 is divisible by both 17 and 11 1870 / 17 = 110 1870 / 11 = 170
The LCM of both numbers is 198
-11
56
Divide it by 11. If the answer is a whole number, it's divisible.
To find a 4-digit number divisible by both 11 and 17, we need to find the least common multiple of the two numbers, which is 187. Therefore, the smallest 4-digit number divisible by both 11 and 17 is 187*11 = 2057.
11
The number 143 is divisible by 11. This can be determined by performing the division operation 143 ÷ 11, which results in a quotient of 13 with no remainder. Therefore, 11 is a factor of 143, making it divisible by 11.
88 = 8x11 For the number to be divisible by both 8 and 11 it must be a multiple of their product (since 11 is prime 8, being smaller, shares no factors with it). You can't multiply 8x11 by any integer smaller than 2 (not counting 1 of course). Multiplying by the next integer - 2 - produces a number bigger than 120