<p>Both from the May 08. What are the fastest ways to do these problems? I found the answer to both, but it took me longer than I would have liked.</p>
<li>90n + 23p = 4523
If n and p are positive integers in the equation above, what is one possible value of n + p?</li>
</ol>
<p>Any faster way than plugging in?</p>
<li>At a certain hospital, 89 children were born in the month of June. If more children were born on the fifteenth of June than on any other day in June, what is the least number of children that could have been born on the fifteenth of June?</li>
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<h1>15 is pretty straight forward. Don't feel like working it out but if you wan to be more efficient you could use a graphing calculator. Basically solve for n, go tot the table, scroll till you find a whole number for y and then add them together.</h1>
<p>First,
4523 = 15, 15 + 23, 15 + 23 + 23, mod 23 (ie the remainder when divided by 23 is 15, or 15+23, etc)
So now we have to find out which multiples of 90 mod 23 = 15, 15+23, etc mod 23.</p>
<p>90n mod 23 = 23 - 2n = 15, n = 4, p = 181
90n mod 23 = 23 - 2n = 15 - 23, n = not integer
90n mod 23 = 23 - 2n = 15 - 23 - 23, n = 27, p = 91
90n mod 23 = 23 - 2n = 15 - 23 - 23, n = not integer
90n mod 23 = 23 - 2n = 15 - 23 - 23 - 23, n = 50, p = 1</p>
<p>The trend is:
n = 4 + 23k, p = 181 - 90k where k is an integer, such that n>0, p>0</p>
<p>Second,
30 days in June. Divide 89 by 30 = 2.997 so you must have at least a few days with 3 babies [since babies can't be split]. For the 15th to have the most, it must have one more than three, so four.</p>
<p>Jesus!
All I did for 15 (and this took 5-6 seconds) was to realize 4500 = 90x 50
and 23 = 23 x1, therefore 4523 = 90(50) + 23(1) = 4523
so n + p = 51</p>
<p>Just want to add that more often then not the questions on the SAT will be written in such a way that a 'keen' eye will be able to solve them as quickly and as easily as 'IndiaRubber' did.
They don't get too adventurous on this exam. So be prepared to do exactly what what IR did. Another approach when you see a question like this (in case you're like most people and the answer doesn't just jump right out at you) is to divide the product by the larger of the two factors. In this scenario we'd take 4523 and divide by 90 (the larger of the two factors). Take the largest whole number that divides equally into the product, and then examine the remainder. Is it divisible by the smaller factor?
Yes? then you have your answer. No? Then take a smaller dividend and try again.
For example: 4523/90 = 50.25. Take the whole number 50 and determine the remainder. 90 * 50 = 4500 r 23. Now examine this remainder, is it divisible by the smaller factor, 23? Yes! This is your answer. If not, then go back and take a smaller dividend. You could try 49.</p>
<p>Because they aren't too adventurous it's a good idea to have a few exponents just memorized.
For example, if you see
2^x = 4^y = 16^z
you should know immediately what x, y and z equal.
Likewise, if you see
x^3 = y^2
you should be able to come up with potential values immediately.
Having just a few of these memorized can come in very handy.</p>
<p>
[quote]
90n + 23p = 4523
If n and p are positive integers in the equation above, what is one possible value of n + p?
Any faster way than plugging in?
[/quote]
90n + 23p = 4500 + 23
23p - 23 = 4500 - 90n
23(p-1) = (50-n)90
p-1 has to be a multiple of 90.
(p,n) ={ (1,50), (91,27), (181,4) }</p>