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Difference sum of ages between present and 7 years ago is:
96  69 = 27
Since \(27\div7=3\) remainder 6, so Alex should be 6 year old (and seven years ago he was not born yet).
Morgan is: 6 + 4 = 10 years old,
Total ages of Tim and Nancy is: 96  10  6 = 80 (years)
Tim is 2 years older than Namcy.
=> Age of Nancy: \(\left(802\right)\div2=39\)
Age of Tim : 39 +2 = 41

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\(AB^2=\left(62\right)^2+\left(1+3\right)^22.\left(62\right)\left(1+3\right).\cos120^0\)
\(=48\)
\(\Rightarrow AB=4\sqrt{3}\)
Similarity, \(AC=BC=4\sqrt{7}\),
=> perimeter and area of the triangle.

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\(f\left(x\right)=x\)
\(\Leftrightarrow\lefta+1ax\right=x\)
\(\Leftrightarrow\left(a+1ax\right)^2=x^2\)
\(\Leftrightarrow\left(a^21\right)x^2+2a\left(a+1\right)x+\left(a+1\right)^2=0\) (*)
if (a = 1) the equation becomes \(4x+4=0\) \(\Leftrightarrow x=2\) (not satisfied: sum of roots is 5/2)
if (a = 1) the equation becomes 0x = 0 (not satisfied sum of roots is 5/2).
Because sum of the roots (*) is 5/2 we infer that:
\(\dfrac{2a\left(a+1\right)}{a^21}=\dfrac{5}{2}\)
\(\Rightarrow9a^2+4a5=0\)
\(a=1,a=\dfrac{5}{9}\)

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We have: \(\dfrac{x}{\sqrt{91^2x^2}}=\dfrac{60x}{60}\)
We find \(x=35\)

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Assume that 0 < a < b < c < d.
The greatest number m formed by using the four digits is \(\overline{dcba}\).
And the least number n formed by using the four digits is \(\overline{abcd}\).
We have: \(\overline{dcba}+\overline{abcd}=1330\)
since \(\overline{abcd}>1000\) and \(\overline{dcba}>1000\) => \(\overline{dcba}+\overline{abcd}>2000\).
Therefore, there is no a, b, c, d sastifies the problem.

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The illustration shows the supremely easy way of solving this puzzle. The central star is the officer, and the dots are the men.

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The amount must have been $31.63. He received $63.31. After he had spent a nickel there would remain the sum of $63.26, which is twice the amount of the check.

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Labelling each red counter:
\(R_1BR_2B....R_{50}B\)
Then \(R_i\) (i=l, ... , n) must move i1 places to the left. (Once this has happened, all the blues will automatically be in the correct position). Thus, 0 + 1 + .. + 49 = \(\dfrac{50\times49}{2}\) moves are needed altogether.

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Let \(f\left(x\right)=\dfrac{\left(xa\right)\left(xb\right)}{\left(ca\right)\left(cb\right)}+\dfrac{\left(xb\right)\left(xc\right)}{\left(ab\right)\left(ac\right)}+\dfrac{\left(xc\right)\left(xa\right)}{\left(bc\right)\left(ba\right)}1\)
We have:
\(f\left(a\right)=0+1+01=0\)
Similarity \(f\left(b\right)=0,f\left(c\right)=0\)
f(x) is a degree of 2 and have 3 different solotutions (a, b, c) .
=> f(x) = 0
=> \(\dfrac{\left(xa\right)\left(xb\right)}{\left(ca\right)\left(cb\right)}+\dfrac{\left(xb\right)\left(xc\right)}{\left(ab\right)\left(ac\right)}+\dfrac{\left(xc\right)\left(xa\right)}{\left(bc\right)\left(ba\right)}=1\)

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42 = 2 x 3 x 7
85! = 1 . 2 . 3 ... 85
85! has 12 factors \(⋮7\), including: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84.
Among these factors, 47 = 7 . 7. So 85! has 13 factors \(⋮\) 7. So k = 13

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Let O(0, 0), M(2, y), N(x, 2), P(4, 3) as in figure below:
\(f\left(x,y\right)=\sqrt{4+y^2}+\sqrt{\left(x2\right)^2+\left(2y\right)^2}+\sqrt{\left(4x\right)^2+1}\)
\(=OM+MN+NP\)
\(\ge OP\)
\(Min_f=OP\) when M(2,y), N(x,2) in OP.
The equation of OP is \(y=\dfrac{3}{4}x\).
=> \(y=\dfrac{3}{4}.2\) and \(2=\dfrac{3}{4}x\).
\(y=\dfrac{3}{2},x=\dfrac{8}{3}\)
And \(Min_f=OP=\sqrt{4^2+3^2}=5\)

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The messenger reaches the second army after : 10/(9 +1) = 1 hour.
At the time of the messenger leaving the second army, the distance between him and the first army is: 10  2.1 = 8 miles.
The messenger meets the first army after 8/(9 + 1) = 0.8 hours more.
When the messenger gets back to the first army, two armies take 1 + 0.8 = 1.8 hours.
And the distance between two armies is : 10  (1 + 1) 1.8 = 6.4 miles.

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OA' = 1.
\(\Delta ABC\) has edges: 2x, 2x, 2x (with x be the radius of a smaller circle).
\(OA=\dfrac{2}{3}\sqrt{\left(2x\right)^2x^2}=\dfrac{2\sqrt{3}x}{3}\)
We have: OA' = OA + AA'
=> \(1=\dfrac{2\sqrt{3}x}{3}+x\)
=> \(x=\dfrac{3}{2\sqrt{3}+3}\)

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By the Dirichlet principle, one of the subsets, say X; must contain at least 49/3 + 1=17 elements, say \(x_1< x_2< ...< x_{17}\). Form the differences \(x_2x_1,x_3x_1,...,x_{17}x_1\) and remove \(x_1\) (because a, b, c are to be different) if it appears on the list. If one of the remaining differences belongs to X, then we are done.
Otherwise, by the Dirichlet principle again, one of the subsets, say Y (\(\ne\) X); must contain at least 15/2 + 1 = 8 elements from these differences \(y_j=x_{i_j}x_1\), say \(y_1< y_2< ...< y_8\). Consider the differences \(y_2y_1,y_3y_1,...,y_8y_1\) and remove \(y_1\) and \(x_{i_1}\) if they appear on the list. If one of these differences belong to Y, then we are done. If one of them, say \(y_jy_1=x_{i_j}x_{i_1}\left(\ne x_{i_1},x_{i_j}\right)\), belong to X; then let \(x_{i_1},x_{i_j},x_{i_j}x_{i_1}\) are different elements of X and \(\left(x_{i_j}x_{i_1}\right)+x_{i_1}=x_{i_j}\) and we are done.
Thus, we may assume 5 of these differences \(z_k=y_{jk}y_1\) belong to the remainging subset Z and say \(z_1< z_2< ...< z_5\) Form the difference \(z_2z_1,z_3z_1,z_4z_1,z_5z_1\) and remove \(z_1,y_{j_1},x_{i_{j_1}}\) if they appear on the list. The remaining difference \(z_kz_1=y_{j_k}y_{j_1}=x_{i_{j_k}}y_{i_{j_1}}\) must belong to one of X, Y or Z: As above, we get three distinct elements a,b,c in one of X, Y or Z such that a + b = c

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Let \(S_n\) be the number of n letter words with even number of a's and \(T_n\) be the number of n letter words with odd number of a's. Then \(S_n+T_n=3^n\). Among the \(S_n\) words, there are \(T_{n1}\) words ended in a and \(2S_{n1}\) words ended in b or c: So we get \(S_n=T_{n1}+2S_{n1}\) . Similarly \(T_n=S_{n1}+2T_{n1}\).
Subtracting these, we get \(S_nT_n=S_{n1}T_{n1}\) . So \(S_nT_n=S_1T_1=21=1\).
Therefore, \(S_n=\dfrac{\left(3^n+1\right)}{2}\) (because \(S_n+T_n=3^n,S_nT_n=1\))

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f(a) = f(b) <=> \(a^2+a^2+b=b^2+ab+b\)
\(\Leftrightarrow a^2b^2+a^2ab=\Leftrightarrow0\)
\(\Leftrightarrow\left(ab\right)\left(a+b\right)+a\left(ab\right)=0\)
\(\Leftrightarrow\left(ab\right)\left(2a+b\right)=0\)
\(\Leftrightarrow2a=b\) (since \(a\ne b\))
=> \(f\left(x\right)=x^2+ax2a\)
=> \(f\left(2\right)=4+2a2a=4\)

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The shortest length of the first s would be F(1) = 1.
The shortest length of the second s would be F(2) = 1.
The shortest length of the third s would be F(3) = 2. (= 1 + 1).
The smallese length of the fourth s would be F(4) = 3 (=2 + 1).
....
And the shortest length of the tenth would be F(10) = F(9) + F(8).
Here F is the Fibonaci sequence: 1, 1, 2, 3 , 5, 8, 13, 21, 34, 55, ...
F(10) = 55.
So the shortest length of the longest s is 55.

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Let the size of the field square be 1. Assume each son has the area x, we have:
\(4x=\dfrac{3}{4}\) \(\Rightarrow x=\dfrac{3}{16}\)
So it suggest that we should divide the square into 16 small squares. And we can group into 4 parts as following:

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Write \(n=\overline{abcd}\), so that \(9n=\overline{dcba}\). Since n ≥ 1000, we know 9n ≥ 9000 and so d = 9. Since 9n < 10000, we know n ≤ 1111; since \(a\ne b\), this means a = 1 and b = 0. Finally, because \(\overline{dcba}\) is a multiple of 9, so is d + c + b + a, therefore c = 8 and n = 1089 . (Indeed, 9 · 1089 = 9801.)

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Since an odd number raised to any power is odd, our number is 999^{k} for some odd k. When a number ending in 9 is raised to an odd power, the units digit is always 9 .