quickest ways to do these math problems

<p>1) There are 6 volunters who can work at the information desk of a hospital, and the volunteers are always assigned to work in pairs. How amyn differnet pairs of these 6 volunteers can be assigned?
a) 15
b) 18
c) 24
d) 30
e) 36</p>

<p>i know there are repetitions on this one, and that there are a total of 30 different possible ways of having pairs but half of them are repeats. so a)? any other way to think of it? </p>

<p>2) What is the smallest positive integer k for which 3^5 times 5^3 times 7^4 times k is the square of an integer? a) 3 b) 5 c) 15 d) 21 e) 105</p>

<p>i know the answer is c) 15 by using my ti-89 and doing it, but how do u do it without a calculator?</p>

<p>Quickest Way to do this problem:
9) The hypotenuse of a right triangle has length 10 and the legs have lengths x and y. If x is less than 6, which of hte following describes all possible values of y?
A) 0<y<4
B) 4<y<6
C) 6<y<8
D) 8<y<10
E) 10<y<16</p>

<p>Answer is D). The way i did it was that x^2+y^2=100, so if x^2+y^2 is going to be greater than 100, i plugged in different values for x (one high value like 5.999999 and one low value like .111111), to see what the limits were. Any better ways to do it, as it took me about 2 minutes to do it this way.</p>

<p>1, when the problem asks you for number of pairs, you don't have to care about order. So the combination of 6 things taken at 2 times is:
nCr(6,2)=15 </p>

<p>2, Do it this way:
take the square root of 3^5 * 5^3 * 7^4 * x and you get 7^2 * 3^2 * 5 * sqrt of (15*k), thus k would be 15</p>

<p>3, x<6 so x^2 <36 so y^2 > 64, take the sqrt of both sides and you get y>8, since y is a leg it must be smaller than the hypotenus, y<10</p>