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If k is an integer, what is the smallest possible value of k such that 1040*k is the square of an integer?
A. 2
B. 5
C. 10
D. 15
E. 65
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A. 2
B. 5
C. 10
D. 15
E. 65
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Bunuel
1040=2^4*13*5
We need 13*5 to be multiplied with 1040 to make is a perfect square.
Therefore, k=13*5=65
Answer : E
Posts: 1,776
Bunuel
1040 = (2^4)*(13)*(5)
We need one more factor of 13 as well as 5 so as to make square of an integer.
So, k = 13*5 = 65
Hence option (E).
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Bunuel
1040K = 13*8*10*K = 13*2^4*5*K
so K should have one 13 and one 5 i.e. 65 Answer E
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Bunuel
We need to find a number k such that 1040k = n^2 for some number n.
The prime factorization of 1040 is \(2^4 5^1 13^1\).
In order for 1040k to be a square number, we need every prime factor in 1040k to be at least raised to the second power. The only numbers that are not are 5 and 13 - thus, we need one more factor of 5 and one more factor of 13. Therefore, k = 5*13 = 65.
Answer: E
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Answer = E. 65
We require to find factors of 1040 which are perfect squares
1040 = 4 * 4 * 65
To make 1040 a perfect square, it has to be multiplied by 65
k = 65
We require to find factors of 1040 which are perfect squares
1040 = 4 * 4 * 65
To make 1040 a perfect square, it has to be multiplied by 65
k = 65
Bunuel
1040k - k needs to be replaced.
__1_02________
1 |1040K
1
-------------
202 | *040k
404
K should be 4
102^2 = 10404
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Factorizing 1040 = (2^4).5.13
Thus we need 5 and 13 to make it a perfect sqaure.
Thus 65. Ans E.
Thus we need 5 and 13 to make it a perfect sqaure.
Thus 65. Ans E.
Bunuel
MAGOOSH OFFICIAL SOLUTION:
Attachment:
Making_A_Square.png [ 17.87 KiB | Viewed 30152 times ]
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Posts: 385
Factorize 1040
1040 = 2 * 2 * 2 * 2 * 5 * 13
as we know that square of any number should have even powers of factors, the only number missing is 5 * 13 = 65
Correct answer - E
1040 = 2 * 2 * 2 * 2 * 5 * 13
as we know that square of any number should have even powers of factors, the only number missing is 5 * 13 = 65
Correct answer - E
Posts: 5
1040
^
10*104
^ ^
2*5 52*2
^
26*2
^
13*2
2*2*2*2*5*13=1040
2*2*2*2*5*13*K=1040k
(2*2*5*13)=(2*2*K) Since its a square, distribute the 2's and set the sides equal
5*13 = K (Cancel out the 2's)
65=K
^
10*104
^ ^
2*5 52*2
^
26*2
^
13*2
2*2*2*2*5*13=1040
2*2*2*2*5*13*K=1040k
(2*2*5*13)=(2*2*K) Since its a square, distribute the 2's and set the sides equal
5*13 = K (Cancel out the 2's)
65=K
Answer E and here is why,
First we need to factorize 1040, or 2^4*5*13*k=a perfect square. To have a perfect square, we must have squared variables before equal sign or 5 and 13. So 5*13 is 65
First we need to factorize 1040, or 2^4*5*13*k=a perfect square. To have a perfect square, we must have squared variables before equal sign or 5 and 13. So 5*13 is 65
Bunuel
Since you need the 'square', so start by taking the square root (which will be equal to the integer we want to find since int*int=1040k)
Now,
(1040k)^1/2
= (2^4 * 13 * 5k)^1/2
= 4 (13 * 5k)^1/2
To make this value an integer,
you need to multiply it with a 13 and a 5 inside the root
therefore, k= 13* 5 = 65
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rhine29388
"as we know that square of any number should have even powers of factors"
Can someone give me a link where this rule is discussed in more detail?
(if there even is such a thing)
For example what would I need to do if the number was "1080k"
Prime factorization would be: 3^3 * 2^3 * 5
So I have to add one 3, one 2 and one 5 that all factors are even? Is my thought approach right?
Thank you very much for any input
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Let's get some
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Bunuel
1040=2*2*2*2*5*13
So we need 5 and 13 to make it a perfect square. Which is 65
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Asked: If k is an integer, what is the smallest possible value of k such that 1040*k is the square of an integer?
1040 = 2^4*5*13
Smallest integer value of k = 5*13 = 65
1040*65 = 2^4*5^2*13^2 =( 2^2*5*13)^2
IMO E
1040 = 2^4*5*13
Smallest integer value of k = 5*13 = 65
1040*65 = 2^4*5^2*13^2 =( 2^2*5*13)^2
IMO E
Posts: 13
Bunuel
How this factor came out??
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