An algebra problem by Christopher Boo
× 0 . 0 . 0 . 1 1 0 1 1 1 1 1 2 1 1 3 1 1 4 1 1 5 1 1 6 1 1 7 1 1 x … … …
The above is a long multiplication, where only the first few digits of each number are shown. (The first two numbers continue repeating 1s.)
What digit is x ?
The answer is 9.
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6 solutions
It's amazing how few people realise a recurring decimal such as 0.1111111......... is much easier dealt with as a vulgar fraction
The question does not give a recurring 1 - just a string of unknown decimal places. The numbers are therefore not 1/9 exactly (although they are clearly very close) and the analysis doesn't quite work.
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Since we're inquiring about a digit that occurs whether or not the 1 repeats, the digit can be solved for by assuming the 1 repeats.
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This has now been corrected but, of course, in the original framing, the answer digit depended on the subsequent digits in the question and was 9 for many terminations but 8 for, for example, terminating with repeating zeroes.
It seems to me that this question is poorly stated because it can have multiple solutions (perhaps that is why you are given multiple attempts). As only a portion of the decimal expansion is known of each number, one should expect that any other digits do not contribute to the decimal expansion in the resulting number. Specifically, the remaining digits of the each multiplicand can be chosen to give the correct first few digits of the result; however, this is not accurate, because one could choose the remaining digits in the decimal expansion of each multiplicand to be all zero (repeating) which makes this problem into a simple arithmetic problem (i.e. 0.111111111 x 0.111111111 = 0.1234567898765...). This implies that the solution to the problem is (x = 8) which is just as valid as any other assumption on the remaining digits in the multiplicands. In reference to the solution in which this comment is placed, the assumption of (0.111... (repeating decimal) = 1/9 (in the limit)) as both multiplicands is valid, and it yields a different result (i.e. (x = 9)). Therefore, this is no single solution to this problem; perhaps it could be modified to yield a definite solution.
Sums of the columns in the long multiplication would be
1, 2, 3 ,4, 5, 6, 7, 8, 9, 10, ….
The 10, will result in a 1 carried over, to make the 9 in front of it into a 10.
This in turn will result in a 1 carried over into where the 8 would have been, so that there will be a 9 instead.
Let's say there is no decimal (for easier calculations)
Make sure to look number of 1' we r gonna multiply
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
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111111111 x 111111111 = 12345678987654321 − − > Nine 1's, now see what will happen after increasing 1 more digit
1111111111 x 1111111111 = 1234567900987654321
11111111111 x 11111111111 = 123456790120987654321
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. So we can see clearly that digit after increasing number of 1's changes from 8 to 9 if number of are 1's more than nine.
Hence, x = 9
The ... indicate that the remainder of the decimal could take any value We therefore have to consider all possibilities from (0.111111111)^2 (where the omitted decimal is 0 in both cases) to (0.111111112)^2 (where the omitted decimal is 9 recurring in both cases) This gives us an answer between 0.012345678987654321 and 0.012345679209876544 It's therefore not possible to identify the digit after the 7 (although it is clear that if the answer is given to 8 d.p. it would be 0.012345679
The factors are equal to 1 / 9 , so the product represents 1 / 8 1 . This is a repeating fraction, since it contains prime factors that do not divide 10. The period of repetition is a divisor of ϕ ( 8 1 ) = ϕ ( 3 4 ) = 2 ⋅ 3 3 . From the given values we see that the period is at least 9, so that is what we try first. (After that, we might have to try 18, 27, 54 yet.)
Thus we try to find x such that 1 2 3 4 5 6 7 x ⋅ 8 1 = 9 9 9 9 9 9 9 9 9 , or 1 2 3 4 5 6 7 x ⋅ 9 = 1 1 1 1 1 1 1 1 1 . From the final digits we conclude that x should be 9 in this case. It is easy checked that this gives the correct value.
Each of the fractions that are being multiplied are 9 1 . Therefore their product is 8 1 1 and its decimal expansion is 0.012345679
No. 1/81 is not exactly equal to 0.012345679.
0 . 1 ˉ × 0 . 1 ˉ = 9 1 × 9 1 = 8 1 1 .
Since 8 1 1 = 0 . 0 1 2 3 4 5 6 7 9 , we conclude x = 9 .