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What is the smallest integer with exactly 100 factors including 100?

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  • What is the smallest integer with exactly 100 factors including 100?


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mathonanapkin | 20/01 2018 04:04
45360 is the smallest 2^4*3^4*5*7
Positive: 60 %
Answer #2 | 08/05 2015 07:18
Seems to me if we break down 100, we would arrive at factors 2, 2, 5, 5. So that is a total of '4' factors required to get 100. So there are '4' factors that get us to 100. We will need another '96' factors beyond this...since we need to have a total of 100 factors (4+96=100). So for the second part of this problem, remember that the 'smallest integer' multiplier we can use is '2'. So we will want the remaining 96 factors to be 2's (or 2x2x2...x2). Since we will need 96 of these...it will be 2^96. Putting this all together...it will be 2x2x5x5x(2^96). That should yield the smallest number with 100 factors that includes the number 100. You can play with this and figure it out.
Positive: 54.545454545455 %
Answer #3 | 08/05 2015 00:18
Seems to me if we break down 100, we would arrive at factors 2, 2, 5, 5. So that is a total of '4' factors required to get 100. So there are '4' factors that get us to 100. We will need another '96' factors beyond this...since we need to have a total of 100 factors (4+96=100). So for the second part of this problem, remember that the 'smallest integer' multiplier we can use is '2'. So we will want the remaining 96 factors to be 2's (or 2x2x2...x2). Since we will need 96 of these...it will be 2^96. Putting this all together...it will be 2x2x5x5x(2^96). That should yield the smallest number with 100 factors that includes the number 100. You can play with this and figure it out.
Positive: 54.545454545455 %
Answer #4 | 08/05 2015 00:48
The number of factors a number has is driven by its breakdown into prime powers. Theorem: If n = the Product of p_i^a_i, (for i = 1 to k, which I'll omit going forward), then n has exactly Product (a_i + 1) factors. E.g., to find the factors of 360 = 2^3 3^2 5, you have to choose one factor from each of the following rows: 1 2 4 8 1 3 9 1 5 Do you see how every row has a number of choices = the power of the prime, plus 1? At this point, the truth of the Theorem should be obvious, so I don't think any more proof is needed. So 100 = 2^2 5^2. How can you get that? p^99 works but we can find a smaller number. We also need to have 100 = 2=^2 5^2 as a factor. Let n be the number we seek. n must = 2^2 5^2 x other factors (possibly including 2 and 5). The powers + 1 must multiply to 2x2x5x5 also. So n's power of 2 and 5 can't just be 2 because 2+1 = 3 is not a factor of 100. The powers can be 4, since 4+1 = 5. So what if we have n = 2^4 5^4 X other factors. We can either increase the powers on 2 and 5, or add other factors 2^4 5^4 p q, where p and q are primes not = 2 or 5, will give us a total of 5x5x2x2 = 100 factors. What are the smallest factors we can use? The smallest primes not used are 3 and 7, so 2^4 5^4 3 7 = 210000 is a possible solution. Can we get a smaller number by increasing the powers of 2 or 5, instead of using 3 or 7? We'd have to go up to a power of 9. If the power is 5, we get a factor of 6, which doesn't divide 100. If the power is 6, we get 7 if the power is 7, we get 8 If the power is 8, we get 9. None of those divide 100 So the next power we can have is 9, getting 10, which is a factor of 100. If the smaller of 2 and 5, which is 2, had a power of 9, that would increase the number by a factor of 2^5 = 32, which is larger than what we had in the possible solution above, 3x7 = 21. (And we would still need more factors.) So it's more effective (in keeping n small) to use 3x7 than to use 2^9. That means that 210000 is the smallest number.
Positive: 50 %
Answer #5 | 08/05 2015 07:48
The number of factors a number has is driven by its breakdown into prime powers. Theorem: If n = the Product of p_i^a_i, (for i = 1 to k, which I'll omit going forward), then n has exactly Product (a_i + 1) factors. E.g., to find the factors of 360 = 2^3 3^2 5, you have to choose one factor from each of the following rows: 1 2 4 8 1 3 9 1 5 Do you see how every row has a number of choices = the power of the prime, plus 1? At this point, the truth of the Theorem should be obvious, so I don't think any more proof is needed. So 100 = 2^2 5^2. How can you get that? p^99 works but we can find a smaller number. We also need to have 100 = 2=^2 5^2 as a factor. Let n be the number we seek. n must = 2^2 5^2 x other factors (possibly including 2 and 5). The powers + 1 must multiply to 2x2x5x5 also. So n's power of 2 and 5 can't just be 2 because 2+1 = 3 is not a factor of 100. The powers can be 4, since 4+1 = 5. So what if we have n = 2^4 5^4 X other factors. We can either increase the powers on 2 and 5, or add other factors 2^4 5^4 p q, where p and q are primes not = 2 or 5, will give us a total of 5x5x2x2 = 100 factors. What are the smallest factors we can use? The smallest primes not used are 3 and 7, so 2^4 5^4 3 7 = 210000 is a possible solution. Can we get a smaller number by increasing the powers of 2 or 5, instead of using 3 or 7? We'd have to go up to a power of 9. If the power is 5, we get a factor of 6, which doesn't divide 100. If the power is 6, we get 7 if the power is 7, we get 8 If the power is 8, we get 9. None of those divide 100 So the next power we can have is 9, getting 10, which is a factor of 100. If the smaller of 2 and 5, which is 2, had a power of 9, that would increase the number by a factor of 2^5 = 32, which is larger than what we had in the possible solution above, 3x7 = 21. (And we would still need more factors.) So it's more effective (in keeping n small) to use 3x7 than to use 2^9. That means that 210000 is the smallest number.
Positive: 50 %
mathonanapkin | 20/01 2018 04:04
45360
Positive: 50 %

Possible answer