Answer :
To find the greatest four-digit number that, when divided by 6, 20, 33, and 66, leaves remainders of 2, 16, 29, and 62 respectively, you can follow these steps:
Understand the Problem:
- You need a four-digit number that, when divided by a set of divisors, leaves specific remainders.
Set Up Equations:
Let the number be represented by [tex]N[/tex].
Based on the problem's conditions:
[tex]N \equiv 2 \pmod{6}[/tex]
[tex]N \equiv 16 \pmod{20}[/tex]
[tex]N \equiv 29 \pmod{33}[/tex]
[tex]N \equiv 62 \pmod{66}[/tex]
Conceptual Approach:
- Note that these relations set up a simultaneous system of congruences.
- Specifically, each congruence is of the form [tex]N \equiv a_k \pmod{b_k}[/tex], where [tex]a_k[/tex] are the remainders, and [tex]b_k[/tex] are the divisors.
Solve the Congruences Jointly:
- Here you recognize from the problem statement that these conditions are often handled systematically using a method like the Chinese Remainder Theorem, but simplified since divisions correlate:
- Each remainder is such that the remainder + 4 equals the divisor (i.e., 2+4=6, 16+4=20, ...)
- Thus, we focus on the simplest case where [tex]N = x \cdot \text{lcm}(6, 20, 33, 66) - 4[/tex], where x is an integer.
Finding the LCM:
Compute for LCM:
[tex]\text{LCM}(6, 20, 33, 66)[/tex]
Express these as products of prime factors:
- 6 = 2 × 3
- 20 = 2^2 × 5
- 33 = 3 × 11
- 66 = 2 × 3 × 11
Thus [tex]\text{LCM} = 2^2 \times 3 \times 5 \times 11 = 660[/tex].
Determine the Largest Value for N:
Your number [tex]N[/tex] must be a four-digit number.
Set [tex]N = 660k - 4[/tex] where [tex]k[/tex] is kept maximum such that [tex]N < 10000[/tex].
Find [tex]k[/tex]:
[tex]660k - 4 < 10000 \implies 660k < 10004 \implies k < \frac{10004}{660} \approx 15.15[/tex]
The largest integer [tex]k[/tex] is 15.
Therefore, substituting back, [tex]N = 660 \times 15 - 4 = 9896[/tex].
Therefore, the greatest four-digit number is [tex]\boxed{9896}[/tex].
