Sure! Let's tackle this problem step by step. We need to find the least number [tex]\( x \)[/tex] that, when divided by 8, 12, 20, and 36, leaves remainders 6, 10, 18, and 34 respectively.
### Step-by-Step Solution:
#### 1. Understanding the problem:
We need to find [tex]\( x \)[/tex] such that:
1. [tex]\( x \equiv 6 \pmod{8} \)[/tex]
2. [tex]\( x \equiv 10 \pmod{12} \)[/tex]
3. [tex]\( x \equiv 18 \pmod{20} \)[/tex]
4. [tex]\( x \equiv 34 \pmod{36} \)[/tex]
#### 2. Recognizing equivalents:
Note that for each congruence:
[tex]\[ x \equiv a \pmod{n} \][/tex]
can be written as:
[tex]\[ x = k \cdot n + a \][/tex]
for some integer [tex]\( k \)[/tex].
##### For the congruences, we get:
1. [tex]\( x = 8k + 6 \)[/tex]
2. [tex]\( x = 12m + 10 \)[/tex]
3. [tex]\( x = 20p + 18 \)[/tex]
4. [tex]\( x = 36q + 34 \)[/tex]
#### 3. Setting expressions equal to each other:
Since [tex]\( x \)[/tex] must satisfy all these conditions simultaneously, we'll use the Chinese Remainder Theorem or manual calculation for a system of linear congruences.
Let's solve two congruences at a time.
First, let's deal with the congruences involving 8 and 12:
[tex]\[ 8k + 6 \equiv 10 \pmod{12} \][/tex]
Let’s adjust the equation:
[tex]\[ 8k + 6 \equiv 10 \pmod{12} \][/tex]
[tex]\[ 8k \equiv 4 \pmod{12} \][/tex]
Divide both terms by 4:
[tex]\[ 2k \equiv 1 \pmod{3} \][/tex]
This simplifies to:
[tex]\[ 2k \equiv 1 \pmod{3} \][/tex]
The multiplicative inverse of 2 modulo 3 is 2 (since [tex]\( 2 \times 2 \equiv 1 \pmod{3} \)[/tex]):
[tex]\[ k \equiv 2 \times 1 \pmod{3} \][/tex]
[tex]\[ k \equiv 2 \pmod{3} \][/tex]
So [tex]\( k = 3m + 2 \)[/tex] for some integer [tex]\( m \)[/tex].
Hence:
[tex]\[ x = 8k + 6 = 8(3m + 2) + 6 = 24m + 16 + 6 = 24m + 22 \][/tex]
So:
[tex]\[ x \equiv 22 \pmod{24} \][/tex]
[tex]\[ x = 24m + 22 \][/tex]
Now let's solve with the next modulo 20 condition:
[tex]\[ 24m + 22 \equiv 18 \pmod{20} \][/tex]
[tex]\[ 24m \equiv -4 \pmod{20} \][/tex]
[tex]\[ 24m \equiv 16 \pmod{20} \][/tex]
[tex]\[ 4m \equiv 4 \pmod{5} \][/tex]
Divide both terms by 4:
[tex]\[ m \equiv 1 \pmod{5} \][/tex]
So [tex]\( m = 5p + 1 \)[/tex]:
Thus:
[tex]\[ x = 24m + 22 = 24(5p + 1) + 22 = 120p + 24 + 22 = 120p + 46 \][/tex]
So:
[tex]\[ x \equiv 46 \pmod{120} \][/tex]
[tex]\[ x = 120p + 46 \][/tex]
Finally, let’s solve with the 36 modulo condition:
[tex]\[ 120p + 46 \equiv 34 \pmod{36} \][/tex]
[tex]\[ 120p \equiv -12 \pmod{36} \][/tex]
[tex]\[ 120p \equiv 24 \pmod{36} \][/tex]
[tex]\[ 12p \equiv 2 \pmod{3} \][/tex]
Divide both terms by 12:
[tex]\[ p \equiv 1 \pmod{3} \][/tex]
[tex]\[ p = 3s + 1 \][/tex]
Thus:
[tex]\[ x = 120(3s + 1) + 46 = 360s + 120 + 46 = 360s + 166 \][/tex]
#### 4. Least Solution:
The smallest solution occurs when [tex]\( s = 0 \)[/tex]:
[tex]\[ x = 360 \cdot 0 + 166 = 166 \][/tex]
So, the least number which, when divided by 8, 12, 20, and 36, leaves remainders 6, 10, 18, and 34 respectively is:
[tex]\[ \boxed{166} \][/tex]
