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How to find the $n$th number i m sorry is divisible by every the number from $1$ come $10$?
I think $1cdot 2cdot 3cdot 4cdot 5cdot 6cdot 7cdot 8cdot 9cdot 10 = 3628800$ is the answer because that $n = 1$
Can a formula be offered for the $n$th such number?
Hint: note that $362880$ has actually a divisor that $2^7$, while you only require a divisor that $2^3$. There are a variety of smaller number which space divisible by every the number from $1$ to $10$. You require to discover the smallest one. As soon as you uncover that, any multiple of it will also be divisible by all the numbers from $1$ with $10$, so multiply it through $n$
I don"t think you have that fairly right. If other is divisible by 8, the is immediately divisible through 2 and also 4, and also the same goes because that 9 and also 3. You want to think about the prime factorisations of every one of the number from 1-10 and find the lowest usual divisor. It would be:
$$ 5 * 7 * 8 * 9 = 2520 qquad n=1$$
I"m not sure that this would certainly be the most efficient means to execute it. But I"d imagine the following smallest number could be discovered by multiply 2520 by any type of integer you like. So because that example, the following number the would accomplish this would be:
$$ 5 * 7 * 8 * 9 * 2 = 5040 qquad n=2$$$$ 5 * 7 * 8 * 9 * 3 = 7560 qquad n=3$$$$ 5 * 7 * 8 * 9 * 4 = 10080 qquad n=4$$etc...
So through this logic, I"d speak the formula is just:
$$ T_n = 2520n $$
I"m not certain if this gets every number. But every number here is guarantee to be divisible through 10. I"d it is in interested to check out if the answer is various :)
edited Apr 30 "19 at 10:23
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answer Apr 30 "19 in ~ 5:32
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With a comptuer scientific research approach, I introduced this code in Julia the tries every numbers between one to $362880$.
for ns in 1:362880 uncovered = false for j in 2:10 if i%j != 0 break end if j == 10 discovered = true finish end if found
show i break endendThis code outputs $2520$ i beg your pardon is the the smallest number corresponding the question.Using
Ross Millikan note in his answer, us may have actually the correct solution.
edited Apr 30 "19 at 11:40
answer Apr 30 "19 in ~ 9:16
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As someone currently said, you have to look in ~ the prime determinants for every number 1 come 10. The cheat is to have the smallest collection that will certainly do all of them. Right here are the prime factors:
2 = 23 = 34 = 2 x 25 = 56 = 2 x 37 = 78 = 2 x 2 x 29 = 3 x 310 = 2 x 5So now we need a set of primes that will certainly cover the above. The variety of 2s we require is three to obtain to 8 (which is additionally enough for the other even numbers), we require two 3s to do nine, but we only require one each of 5 and also 7.
This is the group:
2, 2, 2, 3, 3, 5, 7
Multiple them together to gain your value, 2520.
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There cannot be a smaller number, any tiny integer will be deficient the one prime. If we look in ~ 120, because that example, you deserve to see why. The prime factors of 120 are:
120 = 2 x 2 x 2 x 3 x 5It has enough 2s, however not enough 3s and is lacking the 7. Because of this we deserve to see the it is no divisible through 9 (i.e., 3 x 3) or 7. As a generalisation, any number that is absent one the those primes is going come fail; it will not be divisible through the number that needs that prime.