Answer :
Let's tackle the problem step-by-step.
### Part 1: Finding the greatest number that can divide 166 and 237 leaving remainders 5 and 7 respectively.
First, we need to understand the condition that the remainders provide us.
- When 166 is divided by an integer [tex]\( d \)[/tex], it leaves a remainder of 5. Mathematically, we can express this as:
[tex]\[ 166 = dq + 5 \][/tex]
where [tex]\( q \)[/tex] is the quotient.
- Similarly, when 237 is divided by the same integer [tex]\( d \)[/tex], it leaves a remainder of 7. Mathematically, we can express this as:
[tex]\[ 237 = dp + 7 \][/tex]
where [tex]\( p \)[/tex] is the quotient.
To eliminate the remainders and find [tex]\( d \)[/tex], we subtract these two equations:
[tex]\[ 237 - 166 = dp + 7 - (dq + 5) \][/tex]
[tex]\[ 71 = d(p - q) + 2 \][/tex]
Rearranging the equation, we get:
[tex]\[ 69 = d(p - q) \][/tex]
Therefore, [tex]\( d \)[/tex] must be a divisor of 69. Let's find the divisors of 69:
[tex]\[ 69 = 1 \times 69,\quad 3 \times 23 \][/tex]
Thus, the possible values for [tex]\( d \)[/tex] are 1, 3, 23, and 69. Among these, the greatest number is:
[tex]\[ \boxed{69} \][/tex]
### Part 2: Finding the smallest number which, when 5 is added to it, is exactly divisible by the greatest number found in Part 1.
From Part 1, the greatest number that can divide both 166 and 237, leaving remainders 5 and 7 respectively, is 69.
To fulfill the requirement:
[tex]\[ \text{Find the smallest number } n \text{ such that } (n + 5) \text{ is exactly divisible by 69}. \][/tex]
This can be expressed as:
[tex]\[ n + 5 \equiv 0 \pmod{69} \][/tex]
or equivalently:
[tex]\[ n + 5 = 69k \quad \text{for some integer } k \][/tex]
Rearranging for [tex]\( n \)[/tex]:
[tex]\[ n = 69k - 5 \][/tex]
We need the smallest positive [tex]\( n \)[/tex], so let's consider the smallest integer value for [tex]\( k \)[/tex]:
[tex]\[ k = 1 \implies n = 69 \times 1 - 5 = 64 \][/tex]
Therefore, the smallest number [tex]\( n \)[/tex] is:
[tex]\[ \boxed{64} \][/tex]
### Part 1: Finding the greatest number that can divide 166 and 237 leaving remainders 5 and 7 respectively.
First, we need to understand the condition that the remainders provide us.
- When 166 is divided by an integer [tex]\( d \)[/tex], it leaves a remainder of 5. Mathematically, we can express this as:
[tex]\[ 166 = dq + 5 \][/tex]
where [tex]\( q \)[/tex] is the quotient.
- Similarly, when 237 is divided by the same integer [tex]\( d \)[/tex], it leaves a remainder of 7. Mathematically, we can express this as:
[tex]\[ 237 = dp + 7 \][/tex]
where [tex]\( p \)[/tex] is the quotient.
To eliminate the remainders and find [tex]\( d \)[/tex], we subtract these two equations:
[tex]\[ 237 - 166 = dp + 7 - (dq + 5) \][/tex]
[tex]\[ 71 = d(p - q) + 2 \][/tex]
Rearranging the equation, we get:
[tex]\[ 69 = d(p - q) \][/tex]
Therefore, [tex]\( d \)[/tex] must be a divisor of 69. Let's find the divisors of 69:
[tex]\[ 69 = 1 \times 69,\quad 3 \times 23 \][/tex]
Thus, the possible values for [tex]\( d \)[/tex] are 1, 3, 23, and 69. Among these, the greatest number is:
[tex]\[ \boxed{69} \][/tex]
### Part 2: Finding the smallest number which, when 5 is added to it, is exactly divisible by the greatest number found in Part 1.
From Part 1, the greatest number that can divide both 166 and 237, leaving remainders 5 and 7 respectively, is 69.
To fulfill the requirement:
[tex]\[ \text{Find the smallest number } n \text{ such that } (n + 5) \text{ is exactly divisible by 69}. \][/tex]
This can be expressed as:
[tex]\[ n + 5 \equiv 0 \pmod{69} \][/tex]
or equivalently:
[tex]\[ n + 5 = 69k \quad \text{for some integer } k \][/tex]
Rearranging for [tex]\( n \)[/tex]:
[tex]\[ n = 69k - 5 \][/tex]
We need the smallest positive [tex]\( n \)[/tex], so let's consider the smallest integer value for [tex]\( k \)[/tex]:
[tex]\[ k = 1 \implies n = 69 \times 1 - 5 = 64 \][/tex]
Therefore, the smallest number [tex]\( n \)[/tex] is:
[tex]\[ \boxed{64} \][/tex]
