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A certain calculator is able to display at most 10 digits, so that any [#permalink]
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Bunuel wrote:
A certain calculator is able to display at most 10 digits, so that any number with a total of more than 10 digits before and after the decimal point cannot be displayed accurately. If x and y are positive integers less than 1,000, can the result of dividing x by y be displayed accurately on the calculator?

(1) 105 < x < 108

(2) 3 < y < 6


Interesting! It took me a minute to realize it, but this is a digits and decimals problem. It's basically asking whether x/y has no more than 10 digits. With a complex question stem like this, it helps to write out some examples quickly so you really understand what the different possible situations are.

For example, if x = 1 and y = 3, x/y = 1/3 = 0.333333..., which can't be displayed.

If x = 100 and y = 2, x/y = 50, which can be displayed.

If x = 2 and y = 100, x/y = 0.02, which can also be displayed.

Don't go further with this question unless you have a plan at this point!

The decimal form of a fraction "terminates" if its denominator has only 2s and 5s in its prime factorization, and not any other numbers. So, if y is a number like 2, 4, 10, 25, etc, then the decimal will terminate. Otherwise, the decimal will go on forever, so it won't be able to fit on the calculator.

Is it possible for a decimal to terminate, but still be too big for the calculator? For example, 123454543254354236262436.0 terminates, but won't fit. Same with 0.000000000000000000009. My intuition is that no, this probably won't happen here, since all of the numbers have to be less than 1,000. However, if I have time, I'm going to double check as I work with the statements to make sure this won't happen. I'm PRETTY sure the question is just asking whether the decimal terminates, which is the same as asking whether y has only 2s and 5s in its prime factorization.

Statement 1 105 < x < 108

In the analysis of the question, I decided that it doesn't really matter what x is, it only matters what y is. Let's check some specific examples to be sure. If x = 106 and y = 2, then the result is 53, which fits in the calculator.

If x = 106 and y = 7, then the result is 106/7, which can't be reduced further. The decimal will go on forever and won't fit in the calculator. So, we have a 'yes' and a 'no' case, and this statement is insufficient.

Statement 2 3 < y < 6

y is an integer, and the only integers between 3 and 6 are 4 or 5. So, y has to be either 4 or 5. Both of those numbers only have 2s and/or 5s in their prime factorization. So, any fraction with 4 or 5 in its denominator will give us a decimal that terminates eventually.

Now, is it possible that we'll get a decimal that's just too big to fit in the calculator, even though it terminates? The largest possible value for x is 999, and 999/4 isn't that big. It'll definitely fit. Also, any fraction over 4 will have a decimal that ends in either .0, .25, .5, or .75. Any fraction over 5 will end in either .0, .2, .4, .6, or .8. So, the decimal isn't going to have a ton of extra digits before terminating. I'll say that this decimal will definitely fit in the calculator. So, the statement is sufficient.

The answer is (B).
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Re: A certain calculator is able to display at most 10 digits, so that any [#permalink]
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Re: A certain calculator is able to display at most 10 digits, so that any [#permalink]
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