Four distinct positive integers a , b , c , and d are such that a + b + c + d = 1 0 0 0 . Find the minimum value of lcm ( a , b , c , d ) .
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Simple and elegant. Nice solution.
Suppose that lcm ( a , b , c , d ) = k a = l b = m c = n d where k , l , m and n are distinct. Rewriting b , c and d in terms of a , gives the equation a + l k a + m k a + n k a = 1 0 0 0 This rearranges to lcm ( a , b , c , d ) = k a = k 1 + l 1 + m 1 + n 1 1 0 0 0 We can see that this is minimized when k = 1 , l = 2 , m = 3 , n = 4 giving the answer 4 8 0
This is exactly a replica of my own solution.
Just awesome
Wow!Keep up the good work man!
If a is the largest number of the four, it will be over 2 5 0 . This means any lcm greater than a will be significantly greater than a . Best case scenario, we want lcm ( a , b , c , d ) = a .
To minimise a , we need to make b , c & d as large as possible, so we make these the largest distinct fractions of a : 2 a , 3 a & 4 a . Now we check to see if this gives us integer solutions for a , b , c & d :
a + 2 a + 3 a + 4 a = 1 0 0 0 a × 1 2 2 5 = 1 0 0 0 a = 4 8 0 b = 2 4 8 0 = 2 4 0 c = 3 4 8 0 = 1 6 0 d = 4 4 8 0 = 1 2 0
As these are all integer solutions, we know that the minimum possible lcm is 4 8 0 .
We can't get a lower lcm than this which is a multiple of a , since the lower limit of a is 2 5 0 , meaning any potential multiples of a are greater than 5 0 0 . We know we can't get a lower lcm that is equal to a , as the other numbers are the highest possible distinct fractions of a .
this is written so simply that I feel stupid now, good problem
I still don't understand the first line where you said " If a is the largest number of the four, it will be over 250 . This means any greater than a will be significantly greater than a . Best case scenario, we want lcm (a, b , c ,d ) = a . Please explain this to me by giving any example or something i can't understand this
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I'm labelling the largest of the four numbers 'a'. The largest number must be greater than 250, because if it is not, the four numbers cannot add up to 1000. The lowest common multiple of the numbers must be a multiple of the largest number. If the lowest common multiple is 2x the largest number, it will be greater than 500. If we assume the lowest common multiple can be exactly 1x the largest number, we may find a solution which is less than 500. (this assumption turns out to be correct, as it gives us a solution of 480).
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that's best solution so far
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tx Aaditya..appreciate that!! Please don't hesitate to give me an Upvote
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If a, b, c, d are integers and if d < c < b < a and if a+b+c+d = p, a given number,
lcm (a,b,c,d) has the lowest possible value if b = 1/2 a, c = 1/3 a and d = 1/4 a, and then this value (for lcm) equals a.
If p = 1000, a + 1/2a + 1/3a + 1/4a = a + 13/12 a = 25/12 a = 1000, hence a = 12 x 1000/25 = 12 x 40 = 480, hence lcm (a,b,c,d) = 480.
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Suppose that a , b , c , d ∈ N , with a < b < c < d and a + b + c + d = 1 0 0 0 , and write N = l c m ( a , b , c , d ) . Then 1 0 0 0 < 4 d , and so d > 2 5 0 . If N > d , then N = k d for some k > 1 , and hence N ≥ 2 d > 5 0 0 . Thus values of N ≤ 5 0 0 are only obtained when N = d .
Suppose now that N ≤ 5 0 0 , and that N = d . Then a , b , c all divide d , and hence a ≤ 4 1 d , b ≤ 3 1 d , c ≤ 2 1 d , and hence 1 0 0 0 = a + b + c + d ≤ ( 4 1 + 3 1 + 2 1 + 1 ) d = 1 2 2 5 d so that d ≥ 4 8 0 . Thus we deduce that N ≥ 4 8 0 in all cases.
The case a = 1 2 0 , b = 1 6 0 , c = 2 4 0 , d = 4 8 0 shows that N = 4 8 0 is possible. The smallest possible value of N is 4 8 0 .