All Answers 358098 Answers

$${\mathtt{AN}} = {\mathtt{6}}{cm}$$

no holly doesnt have enough money to buy what she wants to buy

Here's (b)....it's pretty much an extension of (a)

1000/7 = 142 numbers that have at least 1 factor of 7

1000/49 = 20 numbers that have at  an additional factor of 7

1000/343 =  2 numbers that have a third factor of 7

So.........    142 + 20 + 1 =  163

And 7^163 will divide 1000!   ...so n = 163

99^2   = 9801.....this is the greatest four digit square

cos(300) is equal to sin(?)

$$\small{
\cos{(300\ensurement{^{\circ}})} \text{ is equal to } \sin{(90\ensurement{^{\circ}}-300\ensurement{^{\circ}} )}
= \sin{(-210\ensurement{^{\circ}} )}
= \sin{(-210\ensurement{^{\circ}} + 360 \ensurement{^{\circ}} )}
=\sin{(150 \ensurement{^{\circ}} } )}
}$$

1996^2 - (1995)(1997)

1996^2  - (1996 - 1)(1996 + 1)

1996^2  - [ 1996^2  - 1]  =

1

Note that, in general...

a^2 - (a - 1) (a + 1)  =

a^2 - [a^2 - 1]  =  

1

Need more info....

cos (300) = 0.5      Sin (x) = 0.5    x= Sin-1 (0.5) =  30 degrees

(gcd = greatest common divisor)

1. There is only a solution when gcd( 1027, 712 ) = 1 or in other words 1027 and 712 are relatively prime.

2. Using the Euclidian algorithm to calcutate gcd(1027,712)

$$\small{
\begin{array}{|r|r|r|r|}
\hline
&&&\\
a & b & q& r \\
\hline
&&&\\
1027 & 712 & 1 & 315 \\
712 & 315 & 2 & 82 \\
315 & 82 & 3 & 69 \\
82 & 69 & 1 & 13 \\
69 & 13 & 5 &4 \\
13 & 4 &3 & \textcolor[rgb]{1,0,0}{1} \\
4 & 1 & 4 & 0 \\
&&&\\
\hline
\end{array}
}$$

The greatest common divisor gcd(1027,712) $$\textcolor[rgb]{1,0,0}{=1}$$

and we can go on

3. Using Extended Euclidean algorithm to calculate the first solution

$$\small{
\begin{array}{|r|r|r|r||r|r|}
\hline
&&&&&\\
a & b & q& r &x&y\\
\hline
&&&&&\\
1027 & 712 & 1 & 315 & \textcolor[rgb]{1,0,0}{-165} & 73 - 1 (-165) = \textcolor[rgb]{1,0,0}{238}\\
712 & 315 & 2 & 82 & 73 & -19 - 2\cdot 73 = -165\\
315 & 82 & 3 & 69 & -19 & 16 -3(-19) = 73\\
82 & 69 & 1 & 13 & 16 & -3 -1\cdot 16 = -19\\
69 & 13 & 5 &4 & - 3 & 1 -5(-3) = 16\\
13 & 4 &3 & 1 & 1 & 0 - 3\cdot 1 = -3 \\
4 & 1 & 4 & 0 & 0 & 0\cdot 4 + 1 = 1\\
&&&&&\\
\hline
\end{array}
}$$

$$\small{\text{
The first solution is $ (-165,238) \qquad 1027 \cdot (\textcolor[rgb]{1,0,0}{ -165} ) + 712 \cdot \textcolor[rgb]{1,0,0}{238} = 1
$}}\\\\$$

4. All Solutions:

$$\small{\text{$
\begin{array}{lcl}
\boxed{
a\cdot x + b \cdot y = 1 }\\\\
\mathrm{First~ solution~~} (x_0,~ y_0)\\
\mathrm{All~ solutions~~} \left(x_0+\dfrac{z\cdot b}{ gcd(a,b) } ,~ y_0 - \dfrac{z\cdot a}{ gcd(a,b) }\right)\\
\end{array}
$}}\\\\$$

$$\small{\text{$
\begin{array}{lcl}
\boxed{
1027\cdot x + 712 \cdot y = 1 }\\\\
\mathrm{First~ solution~~} (x_0=-165,~ y_0=238)\\
\mathrm{All~ solutions~~} \left(-165+\dfrac{z\cdot 712}{ 1 } ,~ 238 - \dfrac{z\cdot 1027}{ 1 }\right)\\
\end{array}
$}}\\\\$$

we have:

$$\small{\text{$
\begin{array}{lcl}
\left\{~\left(~ -165 + 712\cdot z,~ 238 - 1027\cdot z\right)~|~z \in Z ~
\right\} \\
\end{array}
$}}\\\\$$

(27 - 19 +31) / 34  =  39/34  = 1 5/34 

Find out how many eagles per ton hay costs:

(4/5 eagle)/ 1/6 ton  =  6/1  *  4/5   =   24/5  =  4.8 eagles per ton

THEN  2/3 ton would cost:

2/3 ton  *  24/5  eagles/ton =  (24*2) /(5*3) = 48/15 = 16/5 = 3.2 eagles

That is a really good answer Mellie :))

b )   8x + 11y = 97      ....I'm with Melody, here......"trial and error" might be the shortest distance between two points, since the numbers are small

Note that  x can't be greater than 12 and y can't be greater than 8 ....  and y must be odd because 8x is always even.....and the only way to get an odd (97, in this case) is to add an odd to an even....

When y = 3, x = 8

And y can't be 1, 5 or 7

So.... 8 "8" confusions and 3 "11" confusions seem to work.......

1/6 of a ton of hay costs 4/5 of an eagle

2/3 or  4/6  of a ton of hay costs 4 times as much as  1/6 of a ton of hay, which is 3 1/5 of an eagles!

1.32

3 m  =  300 cm

24cm x 7 = 168 cm      300cm - 168cm = 132 cm   1.32m

A: I can only come up with two:    Twelve   8  Confusions bills

                                               One 8 Confusions bill and eight 11 Confusions bills

Take 96 and remove multiples of 11 from it succsessively    96  85  74  63  52  41  30  19  8     only one of which is divisible by 8......That is the only combo of 11s and 8s that will work...

Likewise with the 8s   :  96 88 80 72 64 56 48  40  32 24  16  8    only 88 is divisible by 11 .....

B:  Do the same thing for 97 :  97  86  75  64  53  42  31  20  9  1       I find one way of buying something for 97 confusions with only 8s and 11s....    eight 8s  and three 11s

C.  For 69     69  58  47  36  25  14  3  (nothing divisible by 8)   I find no way to pay 69 using only 8s and 11s

                   69  61  53  45  37  29  21  13  5   (nothing evenly divisible by 11)       I think you will get some 10kg bitcoins!!

The other way is to solve the diophantine equation more formally.

I have done the first one on a peice of paper in front of me but it takes even longer to get to your answer than the easier way that I just used, so do it the easy way :))

The Smith family has 13 children, all born exactly 2 years apart. How old will the second-youngest child be when the oldest child is five times the age of the youngest child?

When the youngest child turns 6, the oldest will be 30, which makes him/her 5X older than the youngest one.

The second-youngest child will be 8 years old.

Hi Mellie,

Now, we can do this the easy way .... or we can do it the hard way.

I'll start with the easy way

11x+8y=96

if x=0     8y=96    y=12          So   12 * 8 confusion notes will be good

if x=1    

11+8y=96

8y=85     No integer solution

if x=2          22+8y=96        8y=74        No integer solution

if x=3           33+8y=96        no integer solution

Just keep doing this and you will find that the only other solution is     

8*11confusions+1*8confusion      88+8=96

So there are only 2 ways.

You can do the other two similarly. :)    

This method is time consuming but the numbers are all quite small so it is quite doable :)

Well....Ms. SmartyPants......I even provided a general solution.....so.....take that  !!!   

Well aren't you just so clever.  I worked it out Too :))

The oldest child is 24 years older than the youngest

Let y be the time elapsed from the birth of the youngest to the time when the oldest is 5 times that age......so we have....

5y    = 24 + y     subtract y from each side

4y = 24     then y = 6

So, six years from the birth of the youngest, the oldest will be 30 and the youngest will be 6....thus.....the oldest is five times as old as the youngest at that time......

Then, the second to the youngest will be 8  at that time....two years older than the youngest....!!!

I'm going to employ maximum cheating on this one.......

Using the Euclidian algorithm, we have

1027  = 1(712) + 315

712 = 2(315) + 82

315 = 3(82) + 69

82 = 1(69) + 13

69 = 5(13) + 4

13 = 3(4) + 1

4 = 4(1) + 0

                              1        2        3         1        5          3          4

1  0                        1        2        7         9        52     165       712     

0  1                        1        3       10       13       75      238     1027

Take the  determinant of          165     712

                                                 238   1027        =   -1

So.....

1027(165) - 712(238) = - 1   but...we need a  "+1 "   .....so......multiplying through by -1, we have

1027(-165) + 712(238) = 1

So....x = -165   and y = 238....however.....this is only one solution.....the general solution is given by

x , y   =  [ (-165 + 712k), (238 - 1027k) ]

I'll concede you are right......the question does NOT specify the divisors multiplied together must equal the result, just that they must be divisors.