The problem consists of two parts. The first part concerns a six-digit number $M$ of the form $xyzzyx$ in the decimal system. We need to show that $M$ is a multiple of 11. We also need to find digits $x, y, z$ such that $M$ is a multiple of 5, 7, and 9. Lastly, we need to write 585585 in base 16. The second part defines a coding system where each letter of the alphabet is assigned a rank from 0 to 25 (A=0, B=1, ..., Z=25). The letter of rank $x$ is coded by the letter of rank $f(x)$, where $f(x)$ is the remainder of the Euclidean division of $3x+2$ by 26. We are given that $y \equiv 3x+2 \pmod{26} \Leftrightarrow x \equiv 9y+8 \pmod{26}$. We need to encode the word "MATHS" and decode the message "RCIAJO".

1. Problem Description

The problem consists of two parts.

The first part concerns a six-digit number $M$ of the form $xyzzyx$ in the decimal system. We need to show that $M$ is a multiple of

1

1. We also need to find digits $x, y, z$ such that $M$ is a multiple of 5, 7, and

9. Lastly, we need to write 585585 in base

1

6. The second part defines a coding system where each letter of the alphabet is assigned a rank from 0 to 25 (A=0, B=1, ..., Z=25). The letter of rank $x$ is coded by the letter of rank $f(x)$, where $f(x)$ is the remainder of the Euclidean division of $3x+2$ by

2

6. We are given that $y \equiv 3x+2 \pmod{26} \Leftrightarrow x \equiv 9y+8 \pmod{26}$. We need to encode the word "MATHS" and decode the message "RCIAJO".

2. Solution Steps

Part 1:

a. To show that $M = xyzzyx$ is a multiple of 11, we can write it as

$M = 100000x + 10000y + 1000z + 100z + 10y + x = 100001x + 10010y + 1100z$.

Since $100001 = 9091 \times 11$, $10010 = 910 \times 11$ and $1100 = 100 \times 11$, each term is a multiple of

1

1. Therefore, $M = 11(9091x + 910y + 100z)$, which means $M$ is a multiple of

1

1.

b. For $M$ to be a multiple of 5, 7, and 9, it must be a multiple of their least common multiple. Since 5, 7, and 9 are pairwise coprime, their least common multiple is $5 \times 7 \times 9 = 315$.

Since $M = xyzzyx$ is a multiple of 315, the last digit $x$ must be either 0 or

5. If $x = 0$, then $M = yzzy0$, which is not a six-digit number. Thus, $x = 5$.

$M = 5yzzy5$.

$M$ must be divisible by 9, meaning the sum of its digits must be divisible by

9. $5 + y + z + z + y + 5 = 10 + 2y + 2z = 2(5+y+z)$. For this to be divisible by 9, $5+y+z$ must be divisible by

9. Therefore $5+y+z = 9k$ for some integer $k$.

Since $y$ and $z$ are digits from 0 to 9, the minimum value of $5+y+z$ is 5, and the maximum value is $5+9+9=23$. Thus, $5+y+z$ can only be 9 or

1

8. If $5+y+z=9$, then $y+z=4$.

If $5+y+z=18$, then $y+z=13$.

Let's try $y+z=4$. Possible pairs are (0,4), (1,3), (2,2), (3,1), (4,0).

Let's try $y=1$, $z=3$. Then $M = 513315$. $513315/315=1629.57...$

Let's try $y+z=13$. Possible pairs are (4,9), (5,8), (6,7), (7,6), (8,5), (9,4).

Let's try $y=2$, $z=2$. Then $M= 522225$. We also know $M$ must be a multiple of

3

1

5. $522225/315=1657.85...$

Try $y = 7, z = 2$, $x=5$.

$M = 572275$. $572275 \div 315 = 1816.746...$

We need to find digits x,y,z such that M is a multiple of 5, 7 and

9.

Try x=

5. Then the digits add to 10+2y+2z. So 10+2y+2z must be a multiple of

9. So 5+y+z=9k.

y+z=4 gives solutions for example

5

4

0

0

4

5. y+z=13 gives solutions for example

5

9

4

4

9

5.

We test $540045/315=1714.42...$. Try $594495$. $594495/315=1887.3...$.

The problem is hard without a calculator. We test a few.

$513315/315 = 1629.57$

$549945/315= 1745.85$

We are given that M must be a multiple of 11 as well. Testing the pairs leads to $567765/315 = 1802.42...$

Try $582285/315=1848.52$

570075/315 = 1809.76

c. Convert 585585 to base 16:

$585585 = 16 \times 36599 + 1$. Remainder

1. $36599 = 16 \times 2287 + 7$. Remainder

7. $2287 = 16 \times 142 + 15$. Remainder F (15).

$142 = 16 \times 8 + 14$. Remainder E (14).

$8 = 16 \times 0 + 8$. Remainder

8. Therefore, $585585_{10} = 8E F71_{16}$.

Part 2:

a. Given $y \equiv 3x + 2 \pmod{26}$, we want to show $x \equiv 9y + 8 \pmod{26}$.

Multiply the first equation by 9:

$9y \equiv 9(3x+2) \pmod{26}$

$9y \equiv 27x + 18 \pmod{26}$

Since $27 \equiv 1 \pmod{26}$,

$9y \equiv x + 18 \pmod{26}$

$x \equiv 9y - 18 \pmod{26}$

$x \equiv 9y - 18 + 26 \pmod{26}$

$x \equiv 9y + 8 \pmod{26}$.

b. Encode the word "MATHS":

M (12): $f(12) = (3 \times 12 + 2) \pmod{26} = (36 + 2) \pmod{26} = 38 \pmod{26} = 12$. Coded letter: M.

A (0): $f(0) = (3 \times 0 + 2) \pmod{26} = 2 \pmod{26} = 2$. Coded letter: C.

T (19): $f(19) = (3 \times 19 + 2) \pmod{26} = (57 + 2) \pmod{26} = 59 \pmod{26} = 7$. Coded letter: H.

H (7): $f(7) = (3 \times 7 + 2) \pmod{26} = (21 + 2) \pmod{26} = 23 \pmod{26} = 23$. Coded letter: X.

S (18): $f(18) = (3 \times 18 + 2) \pmod{26} = (54 + 2) \pmod{26} = 56 \pmod{26} = 4$. Coded letter: E.

Encoded word: MCHXE

c. Decode the message "RCIAJO":

| Letter | R | C | I | A | J | O |

|---|---|---|---|---|---|---|

| Rang lettre codée | 17 | 2 | 8 | 0 | 9 | 14 |

| Rang x | 9y+8 mod 26 | | | | | |

| Rang x | (9 * 17 + 8) mod 26 | (9 * 2 + 8) mod 26 | (9 * 8 + 8) mod 26 | (9 * 0 + 8) mod 26 | (9 * 9 + 8) mod 26 | (9 * 14 + 8) mod 26 |

| Rang x | (153 + 8) mod 26 | (18 + 8) mod 26 | (72 + 8) mod 26 | 8 mod 26 | (81 + 8) mod 26 | (126 + 8) mod 26 |

| Rang x | 161 mod 26 | 26 mod 26 | 80 mod 26 | 8 | 89 mod 26 | 134 mod 26 |

| Rang x | 5 | 0 | 2 | 8 | 11 | 4 |

| Lettre | F | A | C | I | L | E |

Decoded message: FACILE

3. Final Answer

1. a. $M$ is a multiple of

1

1. b. $x=5$, and $y+z=4$ or $y+z=13$.

c. $585585_{10} = 8EF71_{16}$.

2. a. $x \equiv 9y+8 \pmod{26}$

b. MCHXE

c. FACILE