Use a calculator efficiently
Devise a strategy to solve a number problem involving remainders.
Devise and use problem solving strategies to explore situations mathematically (be systematic)
This problem explores patterns of divisibility in numbers. As such it is similar to the problem Powers of 7, Level 5 Number. In both these problems, as the numbers increase in some way, the pattern of the remainder goes round and round. The students must find where the pattern starts, and how it continues.
It would be an advantage if students have a calculator with a fraction facility.
The theory underpinning the problem begins to have extensive applications in university number theory courses.
What is the remainder when 1111 is divided by 7?
What is the remainder when 1111111 is divided by 7?
What is the remainder when 111.. (100 ones)...111 is divided by 7?
What is the remainder when a number with n digits, all of which are 1, is divided by 7?
Is there some number made up of n 1s which is divisible by 2, 3, 4, 5, 6, 8, or 9?
Since 1111 = 7 x 158 + 5, the remainder is 5.
Since 11111111 = 7 x 15973 + 4, the remainder is 4.
This last number gives us a clue as to how to proceed. Clearly 111111 = 7 x 15873 is divisible by 7. In that case 11111100000 is divisible by 7. So 111111111111 = 111111000000 + 111111 is divisible by 7 and so on. This means that we should break down the hundred ones into groups of 6.
How many are left at the end? 100 = 6 x 16 + 4, so there are four 1s left over. We know that the remainder on dividing 1111 by 7 is 5. So the remainder on dividing a hundred 1s by 7 is also 5.
(1) Let N be the number with n 1s. To find the answer in general we just need to find the possible numbers of 1s left after dividing by 6. Hence
n = 6k Here N has no remainder on dividing by 7.
n = 6k + 1 Since 1 divided by 7 has a remainder of 1, then the remainder here is 1.
n = 6k + 2 11 has a remainder of 4 and so does n.
n = 6k + 3 111 has a remainder of 6 and so does n.
n = 6k + 4 1111 has a remainder of 5 and so does n.
n = 6k + 5 11111 has a remainder of 2 and so does n.
In table form this can be summarised as follows:
form of n
6k + 1
6k + 2
6k + 3
6k + 4
6k + 5
The same applies to 4, 6, or 8.
What about 3? Now N is divisible by 3 if and only if the sum of the digits of N is divisible by 3. The smallest N that will work is therefore 111 because 1 + 1 + 1 = 3 is divisible by 3.
What about 5? If N is divisible by 5 then the last digit of N has to be 5 or 0. So 5 won't work either.
That leaves 9. As for 3, N is divisible by 9 if and only if all the digits of N sum to a number that is divisible by 9. We get such a number with nine 1s. So N = 111111111 is divisible by 9.
Printed from https://nzmaths.co.nz/resource/all-ones at 5:45pm on the 20th January 2021