Vampire fan club?

Consider the second column written as $A + F = 1 .$ Since there are no nonzero digits, this must involve a carry; that is, the actual sum must be 11. Therefore the first column has a 1 carried from the second column.

This implies $1 + F = 3 ,$ so $F = 2 .$

$\begin{array} {ccccc} \large & & \color{#D61F06}2 & \color{#3D99F6}A & \color{#20A900}N & \color{#69047E}G\\ \large + & & & \color{#D61F06}2& \color{#3D99F6}A & \color{#20A900}N \\ \hline \large & & 3 & 1 & 2 & 2 \end{array}$

$N$ and $A$ are different in the third column, and no digits are zero, so the sum in this column must be 12; this means there will be a 1 carrying into the second column. The second column is $1 + A + 2 = 11 ,$ so $A = 8 .$

$\begin{array} {ccccc} \large & & \color{#D61F06}2 & \color{#3D99F6}8 & \color{#20A900}N & \color{#69047E}G\\ \large + & & & \color{#D61F06}2& \color{#3D99F6}8 & \color{#20A900}N \\ \hline \large & & 3 & 1 & 2 & 2 \end{array}$

By the same argument as with $N$ and $A$ in the third column, with $N$ and $G$ in the fourth column there will be a 1 carrying into the third column.

So from the third column, $1 + N + 8 = 12 .$ This means $N = 3 .$

$\begin{array} {ccccc} \large & & \color{#D61F06}2 & \color{#3D99F6}8 & \color{#20A900}3 & \color{#69047E}G\\ \large + & & & \color{#D61F06}2& \color{#3D99F6}8 & \color{#20A900}3 \\ \hline \large & & 3 & 1 & 2 & 2 \end{array}$

The fourth column is then $G + 3 = 12 ,$ so $G = 9 .$

$\overline{FANG}+\overline{FAN}=G+11\overline{FAN}$ ,so $G=3122\pmod{11}=9,\overline{FAN}=283$

The given condition is :

$({\color{#D61F06}F} \quad {\color{#3D99F6}A} \quad {\color{#20A900}N} \quad {\color{#69047E}G}) + ({\color{#D61F06}F} \quad {\color{#3D99F6}A} \quad N) = 3122$

$\implies (1000{\color{#D61F06}F} + 100{\color{#3D99F6}A} + 10{\color{#20A900}N} + {\color{#69047E}G}) + (100{\color{#D61F06}F} + 10{\color{#3D99F6}A} + {\color{#20A900}N}) = 3122$

$\implies 1100{\color{#D61F06}F} + 110{\color{#3D99F6}A} + 11{\color{#20A900}N} + {\color{#69047E}G} = 3122$

$\implies 11(100{\color{#D61F06}F} + 10{\color{#3D99F6}A} + {\color{#20A900}N}) + {\color{#69047E}G} = 3122$

Let $(100{\color{#D61F06}F} + 10{\color{#3D99F6}A} + {\color{#20A900}N})$ be $x$

$\implies {\color{#69047E}G} = 3122 - 11x$

Now, we have to find the greatest multiple of 11 which is less than or equal to 3122. Now, $11 \times 283 = 3113$ is the greatest multiple of 11 less than or equal to 3122.

So, ${\color{#69047E}G} = 3122 - 3113 = 9$

From this we can also find the value of ${\color{#D61F06}F},{\color{#3D99F6}A} ,{\color{#20A900}N}$ . We got that $11 \times 283 = 3113 \implies x = 283$

Now, $100{\color{#D61F06}F} + 10{\color{#3D99F6}A} + {\color{#20A900}N} = 2(100) + 8(10) + 3$

$\implies {\color{#D61F06}F} = 2, {\color{#3D99F6}A} = 8, {\color{#20A900}N} = 3$

Note: FAN is multiple of 11. Hence G is the remainder while 3122 divided by 11 which is equal to 9 and FAN is the quotient of 3122 divided by 11 winch is equal to 283.

The sum is $11\ \overline{FAN} + G = 3122.$ Thus $G$ is the remainder of the division of 3122 by 11. This is equal to $(-3) + 1 - 2 + 2 = -2 \equiv \boxed{9}\ \ \text{modulo}\ 11.$

FANG+FAN=G+11(FAN) 11(FAN) should be 3113 so G=9

If you consider the sum: FANG + FAN = 3122 as a sum of its parts you see that the sum is equivalent to 1100F + 110A + 11N + G = 3122. The first 3 portions of the sum are all divisible by 11. So, they can be written as 11(100F + 10A + N) + G = 3122. Which means that G must be a single digit number that when subtracted from 3122 produces a multiple of 11. The only number that produces this quality is 9. As 3113 is 11 x 283.

The official answer to the dolphin trying to escape is wrong (July 16, Intermediate problem 1). For some reason, I am not offered the option to add my own solution. Maybe because I supposedly got it wrong. Anyway I am adding it here.

The official answer is that the hole seems higher (That is correct) and that it seems harder to escape because it is smaller (That is incorrect, the apparent hole is actually larger)

This is what happens:

Now lets look more closely at the mathematics of one of the rays.

Lets do some maths starting with Snells law of refraction:

$\[\begin{array}{l} \frac{{\sin {\theta _i}}}{{\sin {\theta _r}}} = \frac{{{v_i}}}{{{v_r}}} = \frac{{{n_r}}}{{{n_i}}} = n\\ {\theta _r} = {\sin ^{ - 1}}(\frac{{\sin {\theta _i}}}{n})\\ x = {y_i}\tan ({\theta _i})\\ {y_r} = {y_i}\frac{{\tan ({\theta _i})}}{{\tan ({\theta _r})}}\\ {y_r} = {y_i}\frac{{\sqrt {{n^2} - {{\sin }^2}({\theta _i})} }}{{\cos ({\theta _i})}} \end{array}$ \)

So how does yr/yi change with the angle of incidence? First with a refractive index of n=1.33: Then with n=2.0 This show that both edges of the hole get magnified, so the difference will also get magnified. Here are two examples: First with n=1.33. The hole edges are at heights of 1 and 2 and angles of incidence of 45 degrees and 60 degrees. Now the same situation with n=2.0

The hole will always appear larger to the dolphin.

I was too lazy to do this on my own, so I wrote some code to do it for me. (Note, it’s not the cleanest code ever, but it works. Written in Swift 4.) ———————————————

import UIKit

let answer : Int = 3122 // the answer

var fang : Int = 2123 // 3122 - 999 var fan : Int = 999 // The highest 3 digit number possible

var fangArray = Array(String(fang)).map{Int(strtoul((String($0)),nil,16))} //converting the FANG integer into an array of digits [2,1,2,3] var fanArray = Array(String(fan)).map{Int(strtoul((String($0)),nil,16))} //converting the FAN integer into an array of digits [9,9,9]

var fangString : String = ("(fangArray[0])(fangArray[1])(fangArray[2])") // pulling the values of the F, A and N digits out of FANG and setting it as a String var fanString : String = String(fan) // setting value of FAN as a string

for _ in stride(from: 101, through: 999, by: 1) {

fang = fang + 1 // Add one to FANG
fan = fan - 1 // Subtract one from FAN

fangArray = Array(String(fang)).map{Int(strtoul((String($0)),nil,16))} // Update FANG array numbers
fanArray = Array(String(fan)).map{Int(strtoul((String($0)),nil,16))} // Update FAN array numbers

fangString = ("\(fangArray[0])\(fangArray[1])\(fangArray[2])") // update FANG string
fanString = String(fan) // update FAN string

if fangString == fanString { // if the digits F-A-N from FANG and F-A-N from FAN match, then print the results.

    print("Solution Found!\nThe value of FANG is \(fang),\nthe value of FAN is \(fan).\nThe correct answer is \(fangArray[3])")

break

}

}

————————————————

Copy and paste it into an Xcode playground to see it in action.

Wolfram alpha system: G+N=a10+2 , N+A = b10+2-a , A+F=c10+1-b , F = 3-c, (a = 0 || a = 1) && (b = 0 || b = 1) && (c = 0 || c =1)

generic solution by https://norsigma.com

FANG + FAN = 10 FAN + G + FAN = 11 FAN + G. (FANG + FAN) mod 11 = 3122 mod 11 = 9, therefore (11*FAN + G) mod 11 = G mod 11 is also 9. Since G is a positive digit, G = 9.

$\begin{array} {ccccc} \large & & \color{#D61F06}F & \color{#3D99F6}A & \color{#20A900}N & \color{#69047E}G\\ \large + & & & \color{#D61F06}F& \color{#3D99F6}A & \color{#20A900}N \\ \hline \large & & 3 & 1 & 2 & 2 \end{array}$ Note that the thousands digit is greater than the hundreds digit. Because $F$ appears in both columns, the hundreds sum must have carried over a 1, by the fact that all the variables are all digits 1 to 9. Therefore, $F=3-1=2$ . $\begin{array} {ccccc} \large & & 2 & \color{#3D99F6}A & \color{#20A900}N & \color{#69047E}G\\ \large + & & & 2& \color{#3D99F6}A & \color{#20A900}N \\ \hline \large & & 3 & 1 & 2 & 2 \end{array}$ This next part is a little tricky. It could be that $N+A \gt 10$ , and we could easily assume $A=9$ without thinking about it. To err on the side of caution, let's not touch any of the variables and instead simplify this formula, using the fact that the hundreds sum resolves as greater than 10: $\begin{array} {ccccc} \large & & & \color{#3D99F6}A & \color{#20A900}N & \color{#69047E}G\\ \large + & & & & \color{#3D99F6}A & \color{#20A900}N \\ \hline \large & & & 9 & 2 & 2 \end{array}$ The tens column is now less than the hundreds column. Therefore, $A=9-1=8$ . $\begin{array} {ccccc} \large & & & 8 & \color{#20A900}N & \color{#69047E}G\\ \large + & & & & 8 & \color{#20A900}N \\ \hline \large & & & 9 & 2 & 2 \end{array}$ And again: $\begin{array} {ccccc} \large & & & & \color{#20A900}N & \color{#69047E}G\\ \large + & & & & & \color{#20A900}N \\ \hline \large & & & & 4 & 2 \end{array}$ The ones column is now less than the tens column. Therefore, $N=4-1=3$ . $\begin{array} {ccccc} \large & & & & 3 & \color{#69047E}G\\ \large + & & & & & 3 \\ \hline \large & & & & 4 & 2 \end{array}$ Now we just simplify again to get our answer: $G=(42-33)=(12-3)=9$

Observing the leftmost place ; F can be either 3 or 2 ( F = 2 when there is a carry from previous stage) . A + F = 11 (as F is 2 or 3) ; so its evident that a carry has been sent to leftmost place . Hence F= 2. A + F =11 ; so A can be 9 or 8 ( A = 8 when there is carry from previous stage; which holds as N + A should be 12 as A = 8 or 9. Therefore A = 8; N should be 4 or 3 as A + N = 12; G + N should be 12 as N can take vales either 4 or 3 only ; there is a carry from previous place so N =3 ; hence G =9.

I answered this question yesterday 7/16/18 to continue my streak..this morning it shows as unanswered again...so I did it again..and did third in easy set..still did not take it..streak should be 112 days not one...what is the matter with app

Since G can equal only one of the following: 5,6,7,8, or 9, you can work it out by the exhaustion method, and you finally get the following: G=9, N=3, A=8, and F=2.

G + N = 2 or 12

N + A = 2 or 11 (because if G + N = 12, we would carry over a 1 so 11 + 1 would give us the "2" in the tens spot)

A + F = 1 or 10 (same reasoning as above)

So F must equal 2 or 3 - let's see which one it is.

>

Let F = 2

A = 8 since A + F = 10 (because A + F cannot equal 1)

N = 3 since N + A = 11(because N + A cannot equal 2)

G = 9 since G + N = 12 (because G + N cannot equal 2)

Test: 2839 + 283 = 3122, it works!

>

Let F = 3

A = 7 (same reasoning as above)

N = 4 (same reasoning as above)

G = 8 (same reasoning as above)

Test 3748 + 374 = 4122, which is too much so this is not correct.

>

Thus, G = 9 is the answer.

Though this is an algebra question, humour me We can first identify that for this particular problem we are working with non-zero digits, as such let us consider how we can arrive at 2 from the first column. We can have either 1+1 or some sum to 12, which cannot be the same digit twice; not 6+6. So the combinations we can use are: 7+5, 8+4, 9+3. Intuitively, we realise for the first two columns, the solution will have the same value with different digits because of the extra 1 from the carry, To arrive at 3 for the result from the final column we must realise that since one of the digits reduces by one, and the other is conserved, the puzzle must begin with the largest combination, (9+3). Please poke as many holes and point out as many things as possible in this solution :)

divide 3122 by 11. G is remainder

Since each letter represents a unique digit

                                                                  FANG

                                                                    FAN
                                                             ---------------
                                                                  3122

G+N = 12 A+N = 11 A+F = 10 F = 2

Hence A = 8; N = 3 and so G = 9

Verifying the addition

2839 + 283 = 3122

I did this using the guess and check method or the trial and error method. Since the sum of $FANG$ and $FAN$ is a four digit number, $G+N=12$ . So $G+N$ can be

$3+9, 4+8, 5+7, 7+5, 8+4,$ or $9+3$

Now if you start from $3+9=12$ and observed the result, you can try $9+3$ immediately.

I did it by trying all the numbers: solve: "Each letter represents a different nonzero digit"

G+N= 12 if

G = 9,8,7,5,4,3 according to above: G ≠ 0,1,2,6

N = 3,4,5,7,8,9 according to above: N ≠ 0,1,2,6

N+A= 11 if

N=9,8,7,5,4,3 according to above: N≠0,1,2,6,

A= 2,3,4,6,7,8 according to above: A≠0,1,5,9

A+F= 10 if

A=2,3,4,6,7,8 according to above: A≠0,1,5,9

F=8,7,6,4,3,2 according to above: F≠9,5,1

F+1= 3 if

F= 2 we must now use all the information we have learned!!

f=2 A=8

A=8 N=3

N=3 G=9

Assume G to be 9:

Then must N = 3 Then must A = 8 Then must F = 2

Just one answer can be true. Therefore 9 is the correct solution.