If a and b are single-digit positive numbers and a/b is NOT a recurrin

Hello all

My attempt:

This looks like a very tricky question to me but lets give it a try

Point to note is that the fraction \(\frac{a}{b}\) is a terminating decimal meaning that \(b\) is of the form \(2^x * 5^y\) where \(x\) and \(y\) are whole numbers.
Also \(a\) and \(b\) are single digit \(+ve\) numbers meaning that \(a,b={1,2,3...9}\)

Statement 1:

\(\frac{-1}{3} > \frac{-a}{b} > \frac{-4}{5}\)

\(0.33<\frac{a}{b}<0.8\)
Working with \(b\) in the form of \(2^x * 5^y\) we have \(b\) in the set \({2,4,5,8}\) but we must not forget that this condition of terminating decimal is dependent on the reduced fraction and therefore we will check at the end.

So
for \(b=2\) we have \(\frac{a}{b} = \frac{1}{2}\) falls within \(0.33\) and \(0.8\)
for \(b=4\) we have \(\frac{a}{b} = \frac{2}{4}\) and \(\frac{3}{4}\) falls within \(0.33\) and \(0.8\)
for \(b=5\) we have \(\frac{a}{b} = \frac{2}{5}\) and \(\frac{3}{5}\) falls within \(0.33\) and \(0.8\)
for \(b=8\) we have \(\frac{a}{b} = \frac{3}{8}\), \(\frac{4}{8}\), \(\frac{5}{8}\) and \(\frac{6}{8}\) falls within \(0.33\) and \(0.8\)
the odd case here is \(b=6\)
for \(b=6\) we have \(\frac{a}{b} = \frac{3}{6}\) falls within \(0.33\) and \(0.8\)
Alternatively we can just multiply integers to the fractions to find the reducible forms of the fraction.
Thus \(a={1,2,3,4,5,6}\). Hence we cannot solve this from this statement alone.


Statement 2:

Now the given statement basically means that \(b\) is a perfect number.
Theory: A perfect number is a positive integer that is equal to the sum of its proper positive divisors, that is, the sum of its positive divisors excluding the number itself. Refer https://en.wikipedia.org/wiki/Perfect_number. First perfect number is \(6\) and the second is \(28\).

Now \(b=6\)
Thus the terminating fractions which can be formed are \(\frac{3}{3}\), \(\frac{6}{3}\), \(\frac{9}{3}\)
So \(a={3,6,9}\)
Hence we cannot find \(a\) from this statement alone.

Combining the two statements:
If we combine the two sets of \(a\) from statement 1 and 2 we have only one common solution which is \(a=3\).
Hence solvable.

I will go with option \(C\)

Phew, been a while since I looked at a problem and this one was a doozie.

I'm also going with answer C, however my approach involved using the number line and statement 2 before actually comparing "answers".

Statement (1) : plot a number line with -4/5 and -1/3 at each end. You know -a/b will fall somewhere in the range. Alternatively you can multiply by a negative, so you're dealing with positive fractions instead of negatives. Make sure you FLIP the signs if you choose to do this.

Tip - list out some fractions that fall within the range. This helps you get a "feel" for where the answer will lie.

This statement is clearly NS, as you will find multiple fractions that fall into this range.

Statement (2) : B is the sum of its positive factors excluding B itself. The only single digit, positive integer that B can be is 6. This still doesn't tell you A. NS.

St (1) St (2) : combined both statements provide a single, concrete answer for A = 3. With 6 as a denominator the only possible, non-repeating fraction is -3/6.

C. Both statements are sufficient.

My Approach:
Given: If a and b are single-digit positive numbers and a/b is NOT a recurring decimal.

1.) -1/3 > -a/b > -4/5
==> 1/3 < a/b < 4/5
==>0.3333 < a/b < 0.8
->Now there are many values of a and b satisfying this inequality, like a/b can be 1/2, 2/4, 3/6, 4/8,2/5, 3/5, 3/4, 2/8, 3/8, 5/8, 6/8.
and a can be 1,2,3,4,5 or 6.
So, Insufficient.

2.) b is equal to the sum of its positive divisors excluding b itself.
so, for b=1,2,3,4,5,7,8,9 this is not possible.
Only for b=6, we have divisors as 2,3,1 and 2+3+1=6.
So, we have b=6.
Again nothing is given about value of a. We can have a = 3 or a=6 or a = 9. [Only with these values we get terminating decimal values]
So, (2) is also insufficient.

Now using both 1 and 2 we get a = 3 answer.
So, C.

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