<p>90n + 23p= 4523</p>
<p>If n and p are positive integers in the equation above, what is one possible vale of n + p ?</p>
<p>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?</p>
<p>What I just did for the first one was get it to a number which was divisible by 90. I subtracted 23, by subbing in 1 for p. Then 4500 was divisible by 90 to get 50. So n=50 and p=1. N+P=51. There’s probably a better way but that’s what I did off the top of my head.</p>
<p>I’m pretty sure the second one is 4 but I’m not sure how I figured it out.</p>
<ol>
<li><p>I’m not sure which mathematical you should use to solve this, but just try plugging in numbers! One possible value is n=4, p=181. So, one possible value of n+p=185.</p></li>
<li><p>In this case, since there are 30 days in June, you have 29 days and the fifteenth day of June. To solve this question, you have to find the max value of kids that are born on 29 days in order to get the least number of kids born on 15th of June. So, the max value is 28<em>3 + 1.
In words, 3 kids are born every day for 28 days, and 1 kid is born on the 29th day. 28</em>3+1=85. So, the min value of kids born on the 15th is 4.</p></li>
</ol>
<p>@Desafinado nice, i didn’t you can sub numbers randomly</p>
<p>@hyehwangkim thank you :)</p>
<p>1.
90n + 23p = 4523
90n + 23p = 4500 + 23
90n + 23p = (90)50 + (23)1
n=50 and p=1
n+p=51</p>
<p>2.
Not the most efficient way to solve, but a bit more intuitive.</p>
<p>We’ll use this fact:
the greatest number in a set of numbers is greater than their average.</p>
<p>Average number of children/day = 89/30 = 2.9666…</p>
<p>The number of children born on June 15th must be greater than 2.97.</p>
<p>Let’s try 3 children for June 15th, 2 children for any other day.
3 + (2)29 = 61. Not enough children (remember, 89 children were born in the month of June).</p>
<p>Let’s try 4 children for June 15th, 3 children for any other day.
4 + (3)29 = 91.
Two too many, so on one of 29 days 1 child was born instead of 3, and 3 were born on each of the remaining 28 days:
4 + 1 + (3)28 = 89.
4 is the answer.</p>
<p>
Thought of a couple more approaches - one is no brainer and the second one is kind of 'brainy".
1.
90n + 23p = 4523
Rewrite it as
90x + 23y = 4523
Solve for y
y = (4523 - 90x)/23
Type this into your trustworthy TI-8* Y-editor, go to Table and scroll down until you see an integer in the Y1 column (TBLSET: Table Start = 1, Table Step = 1, Auto, Auto).
The first time it happens is when x=4: y=181.
Going back to n and p
n=4 and p=181; 4+181=185 is one possible value of n+p.</p>
<p>2.
This solution is for those who are academically inclined and bent on finding all positive pairs (n,p).
90n + 23p= 4523
90n + 23p= 4500 + 23
23p - 23 = 4500 - 90n
23(p-1) = 90(50-n)
If p=1 then n=50.
<a href=“n,p”>b</a> = (50,1).**</p>
<p>If p not=1 then
(50-n)/(p-1) = 23/90 = 46/180
<a href=“n,p”>b</a> = (27,91) or (n,p) = (4,181)**</p>
<p>There is one more possible scenario in the hospital question (just wanted to collect’em all).</p>
<p>
</p>
<p>4 children were born on June 15th; 2 children per day were born on two days; 3 children were born on each of the remaining 27 days.
4 + 2 + 2 + (3)27 = 89</p>
<p>===================
Another method here:
<a href=“http://talk.collegeconfidential.com/sat-preparation/759703-math-help-center-2.html#post12423647[/url]”>http://talk.collegeconfidential.com/sat-preparation/759703-math-help-center-2.html#post12423647</a></p>