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03/17/2021 (2011: Countdown Round, Problem 25)

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Q: A list of five positive integers has all of the following properties: the only integer that occurs more than once is 80, the median is 90 and the mean is 100. What is the largest possible integer that could appear in this list? A: 159 Using the first two bullets of information we have, we can write out the numbers we have, and we get [80, 80, 90, x, y], where x and y are unknown integers.  Next, we use the third bullet. The mean is just the sum of the numbers divided by the total amount of numbers. That means that the sum of the list is 5 ⋅ 100 = 500. 500 = 80 + 80 + 90 + x + y We can subtract (80 + 80 + 90) from both sides of the above equation to get: 250 = x + y. If we are trying to find the greatest number that can be on this list, we have to use the smallest value of x (to maximize the value of y). Because 90 isn't repeated (only 80 is repeated), and all the numbers are integers, that means 91 is the smallest x-value, and 250 - 91 = 159 is the greatest y-value.  That m...
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03/17/2021 (2011: Countdown Round, Problem 24)

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Q: What is the number of zeroes to the right of the right-most non-zero number in the product of the first 150 positive integers. A: 37 Basically, this question is asking for how many factors of 10 are in 150!. To do this, we must find the number of 5's in 150! (because the number of 2's will always outnumber the numbers of 5 in the product). We find the number of multiples of 5 to be: 150/5 = 30 We find the number of multiples of 25 to be: 150/25 = 6 We find the number of multiples of 125 to be: 150/125 = 1.2 → 1 We add these together to get 30 + 6 + 1 = 37 . Why does this work? When we divide by 5, we are finding the multiples of 5, but we aren't finding how many 5's 'fit' into each multiple. So then we have to add one for every multiple of 5² to find the additional 5's we missed. And we do this until we reach a 1 or a number that is smaller than 1. We then add the integers greater than 1 that we found.
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03/17/2021 (2011: Countdown Round, Problem 23)

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Q: If triangle PNM is a right triangle, as shown with PM = 5 units, PN = 4 units, and NQ ⊥ PM, what is the length of segment NQ, in units? Express your answer as a decimal to the nearest tenth. A: 12/5   First, label the side lengths :   Next, using the Pythagorean theorem, we find MN :   Let's now find the angle measures in terms of a , where a = angle P   Now, we can see that PMN ~ (is similar to)  PNQ. We can prove this using Angle-Angle Similarity. That means: PM/MN = PN/QN  We can plug in the values we know to get: 5/3 = 4/QN, which means that QN = 12/5
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