Finding the Greatest Common Divisor for Same Remainder
The problem asks for the greatest number that divides 183, 127, and 211, leaving the same remainder each time.
If a number divides three numbers and leaves the same remainder, it must also divide the differences between these pairs of numbers. Let the numbers be $N_1 = 183$, $N_2 = 127$, and $N_3 = 211$. The differences are:
- $N_1 - N_2 = 183 - 127 = 56$
- $N_3 - N_1 = 211 - 183 = 28$
- $N_3 - N_2 = 211 - 127 = 84$
The required greatest number is the Highest Common Factor (HCF) of these differences: 56, 28, and 84.
Calculating the HCF
We find the HCF of 56, 28, and 84.
- Prime factorization of 56: $2 \times 2 \times 2 \times 7 = 2^3 \times 7$
- Prime factorization of 28: $2 \times 2 \times 7 = 2^2 \times 7$
- Prime factorization of 84: $2 \times 2 \times 3 \times 7 = 2^2 \times 3 \times 7$
The common prime factors raised to the lowest power are $2^2$ and $7$.
Therefore, HCF(56, 28, 84) = $2^2 \times 7 = 4 \times 7 = 28$.
The greatest number that divides 183, 127, and 211 leaving the same remainder is 28.
Similar Questions
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Important Questions from LCM and HCF
Six bells begin to toll together and toll, respectively, at intervals of 3, 4, 6, 7, 8 and 12 seconds. After how many seconds, will they toll together again?
A and B are two prime numbers such that A > B and their LCM is 209. The value of A 2 - B is:
Find the least number which when divided by 12, 18, 24 and 30 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.
Calculate the HCF of \(\frac{12}{5}\) , \(\frac{14}{15}\) and \(\frac{16}{17}\) .
Three numbers are in the proportion of 3 : 8 : 15 and their LCM is 8280. What is their HCF?
