Pandora's box

Let $n$ be the number we are trying to find. We know that

$5|n+4$ $\Rightarrow$ $5|2n+8$ $\Rightarrow$ $5|2n+3$

$7|n+5$ $\Rightarrow$ $7|2n+10$ $\Rightarrow$ $7|2n+3$

$9|n+6$ $\Rightarrow$ $9|2n+12$ $\Rightarrow$ $9|2n+3$

$11|n+7$ $\Rightarrow$ $11|2n+14$ $\Rightarrow$ $11|2n+3$

Since 5, 7, 9 and 11 are co-primes, the smallest number that is divisible by all of them is $5 \times 7 \times 9 \times 11 = 3465$

$2n+3 = 3465$ $\Rightarrow$ $n = 1731$

We need to find the smallest number $x$ such that $x\equiv 1\pmod 5,x\equiv 2\pmod 7, x\equiv 3\pmod 9,$ and $x\equiv 4\pmod {11}.$ Let $y_1= 7*9*11 = 693\equiv 3 \pmod 5,$ $y_2=5*9*11 =495 \equiv 5 \pmod 7,$ $y_3= 5*7*11 = 385 \equiv 7 \pmod 9,$ and $y_4= 5*7*9 = 315 \equiv 7 \pmod 5.$

Then, according to the proof of Chinese Remainder Theorem a solution $x$ of the simultaneous congruences above can be obtained in the form $x=1*y_1*x_1+2*y_2*x_2+3*y_3*x_3+4*y_4*x_4,$ where $y_1*x_1, y_2*x_2, y_3*x_3,$ and $y_4*x_4$ are congruent to 1 modulus 5, 7, 9, and 11, respectively. There is a systematic way of finding $x_1, x_2, x_3,$ and $x_4$ by using the Euclid extended algorithm, but in this case we can find them just by trial and errors. For example, since $y_1\equiv 3 \pmod 5,$ then $3*x_1\equiv \pmod 5$ , replacing $x_1$ by 1, 2, 3, and 4, we get that the value that solves the latter congruence is 2. Similarly, we obtain that $x_2=3, x_3=4, x_4=8.$

So, according to the expression of $x$ that we mention above, we obtain that $x=1*693*2+2*495*3+3*385*4+4*315*8=19056.$ Then according to the Chinese Remainder Theorem, any solution of the given congruences must be congruent to 19056 modulus $5*7*9*11=3465.$ The smallest number satisfying this condition is the remainder of the division of 19056 by the 3465,which is 1731.

Start by searching for numbers that give you a remainder of 1 when divided by 5. These are numbers that end in either 1 or 6. Then, try to find numbers that end in 1 or 6 that give you a remainder of 2 when you divide by 7. The first number of this kind turns out to be 16. In order to find other numbers that satisfy this property, notice that if $X=16+(5\times7)\times n$ , dividing by 5 still leaves remainder 1 and dividing by 7 still leaves remainder 2. This means if we keep adding by $5\times7 = 35$ , our first two properties are maintained. Now, search for a number that satisfies the first 3 properties using this method. The first number that works is $16+35\times4 = 156$ . Again we find the pattern, AKA then number we can add by to maintain our properties. Let $X = 156+(5\times7\times9)\times n$ and as you can see the properties hold. Now add $5\times7\times9=315$ to 156 continuously to find the first instance of a number that works with your fourth and final property. The first value that comes up is $156+315\times 5=1731$ , giving us our final answer.

Let number is X. X=11k+4=3(mod9) So k=9a+4 X=11(9a+4)+4=99a+48 Now X=99a+48=2(mod7) a=7b+3 Put value of a we will get X=693b+345=1(mod5) So here b=2 for smallest number X=693×2+345=1731

X = 1 mod 5

X = 2 mod 7

X = 3 mod 9

X = 4 mod 11

The Chinese remainder theorem, we will get this 19056=1731 mod 3465 the smallest number is 1731.