p, q, r, s, t, u, v, w are all positive integers. If p=2^r5^27^t and q

Question

p, q, r, s, t, u, v, w are all positive integers. If   p=2^r5^s7^t  and  q=2^u5^v7^w , is  \frac{p}{q}  a terminating decimal?

1)  r > u ,  s < v 
2)  t > w

kyloxylo · Accepted Answer

Explanation by Expert's Global:

p=2^r5^s7^t    q=2^u5^v7^w

\frac{p}{q} = 2^{(r-u)}5^{(s-v)}7^{(t-w)}

This number will terminate if we have only 2s and 5s in the denominator. That means, we need to get rid of the  7^{(t-w)}

1) This does not help us get rid of the 7 in the denominator. Insufficient.
2) Since  (t-w)  is a positive number, we know that there is no 7 in the denominator. p/q will terminate.

Sufficient.

kyloxylo · Answer

Hi guys, I posted the question and the explanation. Now, I don't really understand the question stem.

What does it mean "is p/q a terminating decimal?" 
What does it mean "this number will terminate if"?

freedom128 · Answer

Non-terminating decimal means you have an infinite number of decimals in a number, i.e.
1/3= O.333333...
1/6= 0.166666... 
1/7= 0.142857...
1/11=0.090909... etc.

Terminating decimal means you have an finite number of decimals in a number, i.e.
1/2= O.5
1/4= 1/(2^2)= 0.25
1/5= 0.2
1/8= 1/(2^3)= 0.125
1/10=1/(2*5)= 0.1 etc. 
See that a number will terminate if we have "only" 2s and 5s in the denominator. Any presence of 3s, 6s, 7s, or 11s in the denominator will cause the number to be non-terminating. We need to get rid of 3s, 6s, 7s or 11s to ensure that a number has terminating decimal.

So,
1) This does not help us get rid of the 7 in the denominator. Insufficient.
2) Since is a positive number, we know that there is no 7 in the denominator. p/q will terminate. Sufficient

Posted from my mobile device

QuantMadeEasy · Answer

p/q will be a terminating decimal when q only is multiple of 2 or 5 or both, otherwise p/q will be non terminating decimal

(1) q may or may not be a multiple of 7. insufficient

(2) q is not a multiple of 7. Sufficient

B is correct.

mokiburnsred · Answer

I have a question on statement (2). We figure out that there is no 7 in the denominator. But, don't we still have to determine the powers of 2^(r-s) and 5^(s-v).

If (r-s) and (s-v) is greater than 0, then the number will be a WHOLE number and NOT a decimal at all.

My question is: Can this whole number be considered as a TERMINATING DECIMAL?

If not, I feel (C) is the answer.
Could any math expert comment on this please..  Bunuel   chetan2u

chetan2u · Answer

If you have only 2 or 5 or both in the denominator, the fraction will be terminating decimal irrespective of th epower..
So  \frac{1}{2^{154}}  or  \frac{1}{5^{112}}  or  \frac{1}{2^{154}5^2} , all will be terminating decimal..
Reason..  whatever be the power of say only 2, then if you multiply both numerator and denominator by same power but of 5, will give you 10 to that power, and similarly for power of 5
 
 \frac{1}{2^{154}*5^2} = \frac{1*5^{154-2}}{2^{154}*5^2*5^{154-2}}=\frac{5^{152}}{10^{154}}  which will be terminating decimal

\frac{p}{q}=\frac{2^r5^s7^t}{2^u5^v7^w} 
 Now for a fraction to be TERMINATING decimal, the denominator should contain only 2 or 5 or both.
So we are concerned with ONLY power of 7. If power of 7 in numerator is > than that in denominator, p/q will have only 2 and 5 in denominator..
Thus we are looking for whether 7^t>7^w or t>w

1)  r > u ,  s < v 
Nothing about t and w

2)  t > w 
Yes, this what we were looking for. answer is YES
suff

B

mokiburnsred · Answer

Thanks for the detailed explanation,  chetan2u .

If in statement (2): we plug some random values like r=5,u=3 and s=3,v=1 and t=2,w=1 we will get a whole number (2^2)(5^2)(7^1).
So, can this number be considered a 'terminating decimal' as well?

If yes, then the answer must be B. If not, I feel answer might be (C).

chetan2u · Answer

Integers are also part of terminating decimals. 
12.00, 12.123 all are terminating decimals

ccooley · Answer

A whole number is also considered to be a terminating decimal. A terminating decimal is just any number that has an 'end' - as opposed to a number like 1/3 = 0.33333...

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p, q, r, s, t, u, v, w are all positive integers. If p=2^r5^27^t and q

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p, q, r, s, t, u, v, w are all positive integers. If p=2^r5^27^t and q

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