pmenon wrote:
If x is a positive integer, is the remainder 0 when (3x + 1)/10?
(1) x = 3n + 2, where n is a positive integer.
(2) x > 4
n=7; x=23; 3x + 1=70; is the remainder 0 when 70 is divided by 10? Yes.
n=5; x=17; 3x + 1=52; is the remainder 0 when 52 is divided by 10? No.
Not Sufficient.
Ans: "E"
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Arriving at "n=5" or any number that x may be above 4 that will lead to some remainder other than 0 is easy.
Question is: How do we take care of the condition to rule out whether the expression may be evenly divisible by 10. Here's what I've done:
We need to find out whether:
\(\frac{3x+1}{10}\), upon division by 10, will lead to a 0 remainder.
Let's also consider the other condition from statement 1:
\(x = 3n + 2\)
Merging the above two,
\(\frac{3(3n + 2)+1}{10}=\frac{9n+7}{10}\)
What do we know now?
\(\frac{7}{10}\) always gives us a remainder of 7. So, that's constant.
If we somehow make \(\frac{9n}{10}\) to give us a remainder of 3, then earlier 7+3=10 will definitely give us a remainder of 0. Remainders of individual term is the remainder of the entire expression.
And how do we get a number like that?
What if the units digit of 9n=3; that'll solve our problem, will it not? All integers that end with 3 will give a remainder of 3 upon division by 10.
So, n's units digit must be 7, because 7*9=6
3; we get our 3 as the unit digit. In fact, any positive integer whose units digit is 7 will fit in well as n to make the entire expression give us 0 as remainder.
Put, n=7; find x; x>4. Conditions satisfy both statements and give us an affirmative answer.
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~fluke
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