I'm a 6th grader. Hopefully this won't be too tough for most of you. :]
Alrighty, here's the question:
Stephan's grandfather told him about a terrible year when the cicadas were so numerous that they ate all the crops on his farm. Stephan conjectured that both 13-year and 17-year locusts came out that year. Assume Stephan's conjecture was correct.
A. How many years pass between the years when both 13-year and 17-year locusts are out at the same time? Explain how you got your answer.
B. Suppose there were 12-year locusts, 14-year locusts, and 16-year locusts, and they all came out this year. How many years will it be before they all come out together again? Explain how you got your answer.
2006-10-05 13:40:05 · 7 answers · asked by GreenDay roxxx 1 in Science & Mathematics ➔ Mathematics
You need to figure the lowest common multiple of each number.
13 and 17 are both primes, so the lowest common multiple is just 13 x 17 = 221 years!
Now try 12, 14 and 16. The easiest way is to figure their LCM is to break them down into their prime factorizations:
12 = 2 x 2 x 3
14 = 2 x 7
16 = 2 x 2 x 2 x 2
So you need to take four 2s, one 3 and one 7.
LCM (12, 14, 16) = 2 x 2 x 2 x 2 x 3 x 7 = 336 years
The *long* way to do this would be to write down the list of multiples of each:
12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, ..., 324, 336...
14 = 14, 28, 42, 56, 70, 84, 98, 112, 126, 140, ..., 322, 336...
16 = 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, ..., 320, 336...
It will take awhile to see that the first time they all coincide is at year 336. That's why prime factorization is *much* easier.
2006-10-05 13:42:47 · answer #1 · answered by Puzzling 7 · 2⤊ 1⤋
We have to find a number with can be divided with both 17 an 13.
Both 17 and 13 is prim number, so the first number must be 13*17=221
Its the same situation we have to find a number there can be divided with all tre numbers.
the first number is 12=2*2*3 the sekund number is 14= 2*7
the third number is 16=2*2*2*2
the lowest commen number that can be divided wit all tre number must therfor be
2*2*3we are now sure the the number work for 12
2*2*3*7 we are now sure the number works for 12&14
2*2*3*7*2*2 we are now sure the number works for 12 &14&16
so the result will be 336
2006-10-05 21:40:59 · answer #2 · answered by Broden 4 · 0⤊ 0⤋
In both cases, you're looking for the least common multiple of these numbers. To find it with small numbers, decompose each of them into prime factors, and then look for the smallest set of prime factors that includes all of the prime factors from each one (with appropriate multiplicities - if one of youe numbers contains two copies of 2, then the LCM must also have at least 2 copies of 2). Thus for your second problem, we have 12=2²*3, 14=7*2, and 16=2^4. Thus the prime factors of the LCM are 2^4*3*7, and the LCM itself is 336. Thus, all the locusts will come out every 336 years in part B.
In part A, 13 and 17 are both prime, so the LCM must be 13*17, which is 221.
2006-10-05 20:48:11 · answer #3 · answered by Pascal 7 · 1⤊ 1⤋
hi im a 7th grader im sure i can help.....the answer ia 168
between 13 and 17 its 4.....and u add 12,14,16 and u get 42....so now u multiply 42 and 4 together and u get 168..its all about reading it carefully its really easy but hey i got your answer
2006-10-05 20:52:09 · answer #4 · answered by I want to believe 3 · 0⤊ 1⤋
ooooooo, youre CHEATING!! Nah, id do the same thing. But im out of school and thats all over for me! Yay!!
2006-10-05 20:41:31 · answer #5 · answered by Baby Jack born 4/5/09 4 · 0⤊ 2⤋
Yikes! I'm glad I'm not in school any more!
2006-10-05 20:41:45 · answer #6 · answered by ndtaya 6 · 0⤊ 2⤋
uhhhh yea no idea lol
2006-10-05 20:41:24 · answer #7 · answered by CB 5 · 0⤊ 2⤋