Grades 11-12 Video Solutions 2025-2025_11-12

2025_11-12_15

Video Transcription

This is problem 15 on the grades 11 and 12, 2025, Math Kangaroo.  So, what is the smallest positive integer n, such that the square root of 2 times the square root of 3 times the square root of n, all nested, is an integer?  And our answer choices are a, 2 to the 12 times 3 to the 6th, b, 2 to the 4 times 3 to the 14,  c, 2 to the 4 times 3 to the 6 times 5 to the 8th, d, 2 to the 4 times 3 to the 2, and e, none of the previous.  So, you're gonna attempt this one for yourself.  Maybe pause the video.  Okay, so  As I've said, you know, multiple times before, there are a lot of ways to go about this one. But to me, the most obvious is to really just stop working with these  just very nasty  nested roots. And we can do that by, let's expand this a little bit. So, we have root 2, give ourselves a little more space, times root 3  times root n. And that's, that's really all this question is asking. It's like, we can replace the blank spaces with time symbols.  And then we can express sort of everything as a fraction. So, we get 2 times, and put this in parentheses, put this next one in parentheses.  So, yeah, this whole thing, this whole parentheses should go here. The smaller one should go here. So, 3 times, and then finally,  this last square root should be in this parentheses.  And now, let's apply some exponents. So, we have n to the 1 half.  Then we have that this, like, 3 times n to the 1 half should also be to the 1 half.  And then finally, we have that this 2 times 3 times n to the 1 half to the 1 half  should also be to the 1 over 2.  And this sort of helps us out, because we can just, like, distribute all of these things inwards. So, this should be 2 to the 1 over 2 times, now, 3 times n to the 1 over 2  to the 1 fourth. And then we obtain  2 to the 1 over 2 times 3 to the 1 over 4. And, you know, finally distributing, we get n to the 1 half to the 1 fourth times n to the 1 over 8.  And critically, we want this thing to be an integer. So, we want each of these  sort of, like, fractions to be completed.  So, because we can only adjust n, we want n to have, like, some other copy of 2. And we also want it to have  some copy of 3.  So, then finally, we get that n to the 1 over 8 should, like, complete the 2 to the 1 half.  So, it should equal 2 to the 1 over 2, because, you know, 2 to the 1 over 2 times 2 to the 1 over 2 is equal to 2 to the 1.  Then times 3 to the 3 over 4 for, like, much the same reason. We need, like, the remaining part of this fraction to make this an integral amount.  We can raise this thing to the 8th, like, I guess, both sides to the 8th.  Then move back up here. And we obtain that n should equal 2 to the 1 over 2  to the 8 times 3 to the 3 over 4  to the 8. And expanding, you get that this should be 2 to the 4 times 3  to the 6th.  So, n equals 2 to the 4 times 3 to the 6th.  Which, unfortunately, is not included as an answer choice. 5 to the 8, certainly.  We get something of that form, but, like, this extra 5 to the 8 is tacked on. And we get, we know that 2 to the 4 times 3 to the 6 is  smaller than 2 to the 4 times 3 to the 6 times 5 to the 8.  So, our answer is actually E, none of the previous.  Because we don't have to, we don't want to deal with that exponent of 5.

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