Arithmetic cirtemoeG
The arithmetic mean and the geometric mean of two distinct 2 digit numbers
x
and
y
are two integers one of which can be obtained by reversing the digits of the other. Then
x
+
y
equals
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3 solutions
not really, just clearly explaining it for others takes a long explanation
I don't quite understand the problem, unless my logic is flawed, we know (x+y)/2=sqrt(yx)....Call yx=p, then x+y=2sqrtp. So the sum we are looking for is yx so 2sqrt(p) is our answer. So either 148=2sqrt(p), 130=2sqrt(p), 82=sqrt(p), or 116=sqrt(p). Thus: 74=sqrt(p), 65=sqrt(p), 41=sqrt(p), or 58=sqrt(p). None of which are perfect squares and the question asks for two two digit numbers which supposedly add up to a 'perfect sum'.
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Read the question. The equation is not 2 x + y = x y .
The equation is that if 2 x + y = 1 0 a + b , where 0 ≤ b ≤ 9 , then x y = 1 0 b + a . That is what "reversing the digits of the other" refers to.
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O haha, I looked at the picture and that threw me off. I just assumed that was the equation we were solving.
Thanks for giving it a picture !!!!
For the 4th step, you could simply say tha 47 is a prime no.so it couldn't be the multiple of two different numbers.
I used kind of a use-all-of-your-forces method. Let the numbers be x, y; x being the larger number. It is easy to show that AM and GM will be also 2 digited. First, I decided that summation of AM and GM will be divisible by 11, as both are reverse of each other. From this equation, we can see, ( x − y ) 2 = 0 m o d 4 4 Then, difference of AM and GM will be divisible by 9. From this equation, we find, ( x − y ) 2 = 0 m o d 3 6 Combining these, ( x − y ) 2 = 0 m o d ( 4 ∗ 1 1 ∗ 9 ) = 0 m o d ( 4 ∗ 1 2 1 ∗ 9 ) LHS being a square number, the power of 11 must be even. Hence, ( x − y ) = 0 m o d 6 6 = 6 6 , x and y are distinct. Taking the largest value for x ( = 9 9 ) and smallest value for y ( = 1 0 ) respectively, we can see, 1 0 < = y < = 3 3 , and AM will be between 43 and 66. Hence, the last digit of GM will be either 4, 5 or 6. We can say, the last digit of x y will have 5 or 6. As y ( y + 6 6 ) m o d 1 0 = y ( y + 6 ) m o d 1 0 , we can easily verify that, y m o d 1 0 = 2 , 5 , 9 We get 7 possible pairs of (y,y + 66) from the above logic: ( 1 2 , 7 8 ) , ( 1 5 , 8 1 ) , ( 1 9 , 8 5 ) , ( 2 2 , 8 8 ) , ( 2 5 , 9 1 ) , ( 2 9 , 9 5 ) , ( 3 2 , 9 8 ) As the product must be a square number, ( 2 2 , 8 8 ) , ( 3 2 , 9 8 ) remains only. The first pair is easily ruled out.
firstly i would like to state the key facts
(1) the geometric mean is an integer implying x ∗ y to be a perfect square {REMEMBER: geometric mean= s q r t x ∗ y }..hence both x &y should be perfect squares.
(2) secondly, x & y are two digit numbers.
integrating the two facts we approximate the set to which x &y belong as- {16,25,36,49,64,81} (all the 2-digit perfect squares). We see that none pair adds up to 148,82,116 [as given in the choices]....... but (81+49) gives 130, leaving us with only one possible choice.
if any anomaly be encountered, please comment!......thanks!! :)
Respect (1), your theorem isn't right (i.e. x y perfect square => x and y are perfect squares is wrong, counterexample x=2 y=98, x y=14^2, but they aren't squares) x&y perfect squares is a sufficient condition, but is not necessary. Your method is probably the shortest I know, but it has a kind of luck, or at least, not complete rigorous reasoning.
Anyway, it's a very interesting aproach
I solved this by mostly using logic. I used Y as the larger of the two.
A= arithmetic mean
G= geometric mean
"n" will be explained later
1) Knowing that the arithmetic mean is simply the sum of the 2 numbers in half, you know A. Since A and G have inverse digits, you know the values of each.
Sum | A | G |
148 | 74 | 47 |
130 | 65 | 56 |
82 | 41 | 14 |
116 | 58 | 85 |
2) Another way to use the geometric mean is to find the common ratio, or n. X n=G , G n=Y. This is why Geometric means are always closer to the smaller number. Using this, we conclude that 14*n must be greater that 41, as Y must be greater than 41. Therefore, n is at least 2.92, and if true, X would be around 4.8, and since X must be 2 digits, we have eliminated another choice.
Sum | A | G |
148 | 74 | 47 |
130 | 65 | 56 |
82 | 41 | 14 |
-------|-----|-----|
3) The arithmetic mean must be greater than the geometric mean, based off the binomial formula, and also if you think about it, since the geometric mean is proportionally closer to one the smaller value, it must be smaller.
Sum | A | G |
148 | 74 | 47 |
130 | 65 | 56 |
-------|-----|-----|
-------|-----|-----|
4) We will deal with the first of the 2 options left now. if G is 47, then X is less than 47. if x is 46, it is 28 away from A, so Y would be 102, but since they both must be 2 digits, we eliminate one more option.
Sum | A | G |
-------|-----|-----|
130 | 65 | 56 |
-------|-----|-----|
-------|-----|-----|
Voilà!