n = (10^10 x 9^9 x 8^8…3^3 + 2^2 + 1^1)^2
How many zeroes does n^2 contain?
First off, you need to recognize the pattern: every number has an exponent equal to itself, i.e. 10^10, 8^8 , etc. That means, that the (….) represents 7^7 x 6^6 x 5^5 x 4^4.
Now that we can survey the entire series (and I do recommend writing out the remaining numbers), notice is that 10^10 will yield 10 zeroes. The next step, though, is a little trickier. If you pair 5 and 2, you get 10. Therefore, each pair of 5 and 2 will yield a zero. There are five 5’s, yet only two 2’s. But notice how both 4^4 and 8^8 contain a bounty of 2’s (4^4 = 2^8 and 8^8 = 2^24), so we will have more than enough 2’s to pair with those five 5’s. So (5 x 2)^5 will yield a total of 5 more zeroes to ‘n’, giving us a total of 15.
Now to the very last part. The question is looking for n^2. You want to make sure not to square 15. See, when you have some number to an exponent and you are looking for the number of zeroes, you want to multiply by 2, since you are doubling the number of zeroes. Therefore, the answer is 30.
Now watch all of that explained in video form here:
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