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I-C Finite Math 1.20
I-C Finite Math 1.20
Problem: Find the smallest number consisting of only 1's that is divisible by 9 and 11. How many digits does this number have?
My answer was the number has 10 digits, which is 1111111111. I calculated them and it is divisible by both 9 and 11. But the answer showed that the answer is 18. I checked my answer many times and it is true that 1111111111 is divisible by 9 and 11. Please help.
Re: I-C Finite Math 1.20
Recall the rules for divisibility by $9$ and $11$:
- A number is divisible by $9$ if and only if the sum of its digits is divisible by $9$. Since the number has only $1$'s, it must have a multiple of $9$ number of digits.
- A number is divisible by $11$ if and only if the alternating sum of its digits is divisible by $11$. Since the number has only $1$'s, the alternating sum of its digits will be $1$ if it has an odd number of digits, or $0$ if it has an even number of digits. Thus, we need the number to have an even number of digits.
Since we need the number to have a number of digits that is a multiple of $9$ and even, the smallest possible number of digits it may have is $18$.
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