<p>at a certain hospital, 89 children were born in the month of June. If the children born on the 15th are more than other children born on any other day in June. What is the least number of children that could have been born on the 15th?</p>
<p>I need an explanation, guys! </p>
<p>90n+23p=4523</p>
<p>what is one possible value of n+p.. so the first numbers I tried were p =1 and n=50 and it worked.. but that is absolute luck. how to answer such question??</p>
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<p>This question relies on a mathematical principle called the Pigeonhole Principle. (OK, technically, I think it might not quite, but it’s at least similar.) You’ve got 30 dates available in June, and 89 babies to distribute among them. Since you want to know the smallest possible number of babies born on the 15th, distribute the babies as evenly as possible, to keep the number of babies per day low. You could have two babies born on every day; that’d be 60. Now you’ve got to distribute those other 29 babies; that’ll make 29 days with 3 babies, and one day with 2 babies. However, this won’t meet the requirement that “the children born on the 15th are more than other children born on any other day in June.” But if there were 4 babies born on the 15th, you could have the whole rest of the month filled with days on which 3 or fewer babies were born. It won’t necessarily fall out that way, but it could happen, so it answers the question, “What is the least number of children that could have been born on the 15th?”</p>
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<p>You’d be completely right if you’d found your answer through a little bit of intuition instead of blind luck. Note the similarity between the 23 in 23p and the final two digits of 4523. Based on that, you’re supposed to think, “If p is 1, then that will leave 90n = 4500, and…oh, hey, that works if n = 50.”</p>
<p>Because the coefficient of n is 90, 90n can have only zero in the ones place. That means you’ll have to use the 23p to make that 3 in the ones place of 4523. The only way that can happen is if p has 1 in its ones place; that is, if p is 1 or 11 or 21 or 31… You could play around with 11, 21, 31, etc., to see whether any of them work out, but I’d start by testing p = 1, because the arithmetic is easiest.</p>
<p>thanks for ur help! :)</p>