It's SAT math question- Algebra lvl. i was wondering if there's an easy and fast way to answer this question instead of testing one by one. Anyway, how do i solve this?
The integer n is equal to k^2 for some integer k. if n is divisible by 24 and by 10, what is the smallest possible positive value of n?
Thanks.
The integer n is equal to k^2 for some integer k. if n is divisible by 24 and by 10, what is the smallest possible positive value of n?
Thanks.
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Prime factors and least common multiple (LCM).
24 = 2•2•2•3
10 = 2•5
The LCM is the number which has the minimum prime factors to be divisible by 24 and 10. Start with 24:
2•2•2•3
Now, consider the 10. Since you already have a 2 from 24's prime factors, you just need a factor of 5:
2•2•2•3•5 = 120
This is the smallest number divisible by both 24 & 10, BUT it is NOT a perfect square so the associated k is not an integer. To be a perfect square you need to be able to pair up like factors.
(2•2)•2•3•5
As you can see, one of the twos, the three and the five aren't paired. Thus you need additional factors of 2•3•5 so that now you have an n of:
(2•2)•(2•2)•(3•3)•(5•5) = 3600
So 3600 is the smallest possible positive value of n.
[As a check: k = sqrt(3600) = 60 and 60 is an integer.]
24 = 2•2•2•3
10 = 2•5
The LCM is the number which has the minimum prime factors to be divisible by 24 and 10. Start with 24:
2•2•2•3
Now, consider the 10. Since you already have a 2 from 24's prime factors, you just need a factor of 5:
2•2•2•3•5 = 120
This is the smallest number divisible by both 24 & 10, BUT it is NOT a perfect square so the associated k is not an integer. To be a perfect square you need to be able to pair up like factors.
(2•2)•2•3•5
As you can see, one of the twos, the three and the five aren't paired. Thus you need additional factors of 2•3•5 so that now you have an n of:
(2•2)•(2•2)•(3•3)•(5•5) = 3600
So 3600 is the smallest possible positive value of n.
[As a check: k = sqrt(3600) = 60 and 60 is an integer.]
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n is divisible by 24 & 10.
24=2*2*2*3
10=2*5
Smallest number that divides n could be 2*2*2*3*5 = 120.
120 is not the answer because n is equal to k^2 for some integer k.
120 = 2*2*2*3*5, now each number must be repeated evenly so we can find a square root for n.
therefore, n is 120*2*3*5 = 120*30 = 3600. and k is 60.
Answer: 3600
24=2*2*2*3
10=2*5
Smallest number that divides n could be 2*2*2*3*5 = 120.
120 is not the answer because n is equal to k^2 for some integer k.
120 = 2*2*2*3*5, now each number must be repeated evenly so we can find a square root for n.
therefore, n is 120*2*3*5 = 120*30 = 3600. and k is 60.
Answer: 3600
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Make factor trees to get prime factors for each number.
24 factors into 2*2*2*3 and 10 factors into 2*5.
2 is common to both, but in different quantities. How many 2s are needed? Always use the largest number or 2*2*2 (2^3 if using exponents). 3 is unique to 24, so it is needed. 5 is unique to 10, so it is needed.
Multiplying these together gives the lowest common multiple 2*2*2*3*5, but notice we do not have perfect square values - there are an odd number of 2s, 3s, and 5s. To get a perfect square, we need multiples of 2 of each, so we need another 2, another 3, and another 5 to get 2*2*2*2*3*3*5*5, which is 3600.
24 factors into 2*2*2*3 and 10 factors into 2*5.
2 is common to both, but in different quantities. How many 2s are needed? Always use the largest number or 2*2*2 (2^3 if using exponents). 3 is unique to 24, so it is needed. 5 is unique to 10, so it is needed.
Multiplying these together gives the lowest common multiple 2*2*2*3*5, but notice we do not have perfect square values - there are an odd number of 2s, 3s, and 5s. To get a perfect square, we need multiples of 2 of each, so we need another 2, another 3, and another 5 to get 2*2*2*2*3*3*5*5, which is 3600.
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factorise 2*2*2*3 , 2*5,
have even #s of 3,4 & 5 = 9*16*25 = 3600 <--------
have even #s of 3,4 & 5 = 9*16*25 = 3600 <--------