Question 1200481: Winnie is counting some after-dinner mints. Whenever she divides all of her mints into groups of either 8 or 9 or 10 or 11 or 12, she always has 3 mints left over. The smallest number of mints that she could have (greater than 3) is
a) 363
b) 1983
c) 3963
d) 7923
e) 95 043
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: 3963 (choice C)
Explanation:
Let's list the prime factorization of 8 through 12.
8 = 2*2*2
9 = 3*3
10 = 2*5
11 = 1*11
12 = 2*2*3
The unique primes mentioned were: 2, 3, 5, 11- The prime 2 shows up at most 3 times, making 2^3 part of the LCM.
- The prime 3 shows up at most 2 times, making 3^2 part of the LCM.
- The prime 5 shows up at most 1 time, making 5^1 part of the LCM.
- The prime 11 shows up at most 1 time, making 11^1 part of the LCM.
The LCM of {8,9,10,11,12} is 2^3*3^2*5^1*11^1 = 8*9*5*11 = 3960
In short,
LCM = 3960
Many calculators specialize in finding the LCM very quickly based on an input list of values.
For example, the command LCM({8,9,10,11,12}) in GeoGebra will produce the output 3960.
This is what it would look like if you used WolframAlpha
https://www.wolframalpha.com/input?i=LCM+of+%7B8%2C9%2C10%2C11%2C12%7D
The key takeaway is that
LCM = 3960
If Winnie had 3960 mints, then she could divide it in piles of 8, piles of 9, etc all the way up to piles of 12.
There wouldn't be any leftovers or remainders if she had 3960 mints.
Add on 3 to this LCM and we'll guarantee 100% that she'll have this amount of leftovers (since 3 is smaller than 8 through 12)
That's how we go from 3960 to 3960+3 = 3963
This is guaranteed to be the smallest value we're after because it's closely tied to the LCM which is the smallest multiple of 8 through 12.
Check:
The table below shows the remainder for each answer choice when we divide the mints into groups of 8, groups of 9, etc all the way up to groups of 12.
| Choice A | Choice B | Choice C | Choice D | Choice E | | Groups of 8 | 3 | 7 | 3 | 3 | 3 | | Groups of 9 | 3 | 3 | 3 | 3 | 3 | | Groups of 10 | 3 | 3 | 3 | 3 | 3 | | Groups of 11 | 0 | 3 | 3 | 3 | 3 | | Groups of 12 | 3 | 3 | 3 | 3 | 3 |
Choice A is very close, but 363/11 = 33 remainder 0, meaning there aren't any leftovers here. Therefore, we must eliminate choice A.
Choice B is close also, but 1983/8 = 247 remainder 7. We must eliminate choice B.
If Winnie had 1 more candy then 1984/8 = 248 remainder 0.
I'll let you check each case of 3963.
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