First number in a series to be divisible is

What is the divisibility rule for 2?
What is the divisibility rule for 9?
What is the divisibility rule for 11?
Construct the number using the least number of digits.
 
BigBeachBanana said:
What is the divisibility rule for 2?
What is the divisibility rule for 9?
What is the divisibility rule for 11?
Construct the number using the least number of digits.
Click to expand...
for 2 - it has to end with an even number
for 9 - the sum of numbers has to be divisible by 9
for 11 - the difference of the sum of numbers in odd places from right side and the sum of numbers in even places from right side has to be divisible by 11.

I do not know, however, how to construct these facts into a number.
 
Loki123 said:
for 2 - it has to end with an even number
for 9 - the sum of numbers has to be divisible by 9
for 11 - the difference of the sum of numbers in odd places from right side and the sum of numbers in even places from right side has to be divisible by 11.

I do not know, however, how to construct these facts into a number.
Click to expand...
You've already correctly determined that it has to end in 4, i.e. that it has to have the same number of 3's and 4's. If your number has [imath]N[/imath] 3's and [imath]N[/imath] 4's what is the sum of all its digits? What about the sum of digits in odd places and even places?
 
blamocur said:
You've already correctly determined that it has to end in 4, i.e. that it has to have the same number of 3's and 4's
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I'm not sure how you concluded that it has to have the same number of 3's and 4's from the divisibility rule of 2 only? It could be n-1 and n. I'm not saying you're wrong, but just wondering about your logic.
EDIT: I concluded that n has to be the same from the divisibility rule of 11.
 
BigBeachBanana said:
I'm not sure how you concluded that it has to have the same number of 3's and 4's from the divisibility rule of 2 only? It could be n-1 and n. I'm not saying you're wrong, but just wondering about your logic.
EDIT: I concluded that n has to be the same from the divisibility rule of 11.
Click to expand...
If you have a sequence of alternating 3's and 4's which starts with 3 and ends with 4 would not you expect the number of 3's be the same as the number of 4's ?
 
blamocur said:
If you have a sequence of alternating 3's and 4's which starts with 3 and ends with 4 would not you expect the number of 3's be the same as the number of 4's ?
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Right. I was thinking it could start with either 3 or 4, but the OP clearly started with 3.
 
blamocur said:
If you have a sequence of alternating 3's and 4's which starts with 3 and ends with 4 would not you expect the number of 3's be the same as the number of 4's ?
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i would not, but i guess 44-33 makes it a bit clearer
so even number of 4s and 3s (4x, 3x)
4x-3x=11k
x=11k

ends with 4
the sum has to be divisible by 9,

3x+4x=9q
7x=9q
x=9q/7

11k=9q/7
77k=9q

k=9 77=q

x=11k =99
x=9q/7=99

so x=99
double that
198

this is correct
i don't know how i did tbh
 
Loki123 said:
i would not, but i guess 44-33 makes it a bit clearer
so even number of 4s and 3s (4x, 3x)
4x-3x=11k
x=11k

ends with 4
the sum has to be divisible by 9,

3x+4x=9q
7x=9q
x=9q/7

11k=9q/7
77k=9q

k=9 77=q

x=11k =99
x=9q/7=99

so x=99
double that
198

this is correct
i don't know how i did tbh
Click to expand...
[math] \begin{cases} 7n\mod 9=0 \\ n\mod11=0\\ \end{cases}\implies \begin{cases} n=9m \\ n=11m\\ \end{cases}\\ lcm(9,11)=99 \implies n=99m [/math][imath]m=1[/imath] yields the smallest positive integer, thus [imath]n=99[/imath]. Therefore, you need 99 of each 3's and 4's giving a total of 198.
 
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