find the least no. which is divisble by all the no. from 1 to 10
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2520 is your answer
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Same answer with only a slightly different thought process:
2*3*4*5*6*7*8*9*10
eliminate 2, it is present in three other evens (4,6,8,10)
eliminate 8, it is in 6*4, 24
eliminate 9, it is in 3*6, 18
eliminate 10, it is in 4*5, 20
All the others stand, therefore:
3*4*5*6*7=2520
Weird thinking, correct answer.
What about 3*4=12--doesn't that eliminate 6? Sure but we need 6*4 to eliminate 8, so gotta keep the 6.
Also, the 3 is in 6*4 as well, but take it away and we can't eliminate 9.
Seems interesting that the product of five consecutive numbers produces the correct answer.
So now am wondering, are there always a set of consecutive numbers that will produce the LCM for any given factorial?
Suppose it was the numbers 1 through 11 instead...
27,720 is the answer (cheated--used the highest multiplicity of primes technique) . Nope, 11*10*9*8*7=55,440 (obviously, the highest prime, 11, must be included). That's 2*27,720. Just a lucky coincidence for 1 through 10's product.
2*3*4*5*6*7*8*9*10
eliminate 2, it is present in three other evens (4,6,8,10)
eliminate 8, it is in 6*4, 24
eliminate 9, it is in 3*6, 18
eliminate 10, it is in 4*5, 20
All the others stand, therefore:
3*4*5*6*7=2520
Weird thinking, correct answer.
What about 3*4=12--doesn't that eliminate 6? Sure but we need 6*4 to eliminate 8, so gotta keep the 6.
Also, the 3 is in 6*4 as well, but take it away and we can't eliminate 9.
Seems interesting that the product of five consecutive numbers produces the correct answer.
So now am wondering, are there always a set of consecutive numbers that will produce the LCM for any given factorial?
Suppose it was the numbers 1 through 11 instead...
27,720 is the answer (cheated--used the highest multiplicity of primes technique) . Nope, 11*10*9*8*7=55,440 (obviously, the highest prime, 11, must be included). That's 2*27,720. Just a lucky coincidence for 1 through 10's product.
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