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