What is the largest number that divides 1206 and 321 leaving remainder
let the largest no. be y
1206/y=6
1206=6y
y=1206/6
y=201
similarly
321/y=6
6y=321
y=321/6
y=53.5
To find the largest number that divides both 1206 and 321 and leaves a remainder, we can use the Euclidean algorithm.
Step 1: Divide 1206 by 321:
1206 ÷ 321 = 3 remainder 243
Step 2: Divide 321 by 243:
321 ÷ 243 = 1 remainder 78
Step 3: Divide 243 by 78:
243 ÷ 78 = 3 remainder 9
Step 4: Divide 78 by 9:
78 ÷ 9 = 8 remainder 6
Step 5: Divide 9 by 6:
9 ÷ 6 = 1 remainder 3
Step 6: Divide 6 by 3:
6 ÷ 3 = 2 remainder 0
The largest number that divides 1206 and 321 leaving a remainder is the remainder obtained in the last step, which is 3. So, the answer is 3.
To find the largest number that divides both 1206 and 321, leaving the same remainder for both divisions, you can use the concept of the greatest common divisor (GCD) or highest common factor (HCF). Here's how you can find it step-by-step:
1. Find the remainders when 1206 and 321 are divided by any divisor.
Let's say the divisor is "x".
1206 divided by "x" leaves a remainder of "r1".
321 divided by "x" leaves a remainder of "r2".
2. Set up the equations using the given information:
1206 = qx + r1 (where q is the quotient)
321 = px + r2
3. Since the remainders (r1 and r2) are the same:
qx + r1 = px + r2
4. Rearrange the equation:
qx - px = r2 - r1
(q - p)x = r2 - r1
5. Since "x" divides both sides of the equation, "x" must also divide the difference (r2 - r1).
6. Determine the difference between the remainders:
Let's say r2 - r1 = k
7. Find the divisors of "k" that divide it evenly:
Determine all the divisors of "k" and choose the largest one.
Therefore, the largest number that divides 1206 and 321, leaving the same remainder, is the largest divisor of the difference between their remainders.
For example, if the remainder for 1206 is 5 and the remainder for 321 is 2, then the difference (k) between the remainders would be 5 - 2 = 3. The divisors of 3 are 1 and 3, so the largest number that divides them evenly is 3.