Greatest number, which is to be divided by 280 and 1245 leaves the remainder 4 and 3 respectively, is

Options

A. 138
B. 148
C. 145
D. 178

Answer

To find the greatest number that, when divided by 280 and 1245, leaves remainders of 4 and 3 respectively, we can follow these steps:

### Step 1: Adjust the Divisors
Since the number leaves a remainder when divided, we can adjust the divisors:
- For 280, if the remainder is 4, the number we are looking for must be 4 less than a multiple of 280. Therefore, we consider \(280 - 4 = 276\).
- For 1245, if the remainder is 3, the number must be 3 less than a multiple of 1245. Therefore, we consider \(1245 - 3 = 1242\).

### Step 2: Find the HCF of the Adjusted Numbers
Now we need to find the highest common factor (HCF) of 276 and 1242.

### Step 3: Prime Factorization of 276
1. Divide 276 by 2:
\[ 276 \div 2 = 138 \]
2. Divide 138 by 2:
\[ 138 \div 2 = 69 \]
3. Divide 69 by 3:
\[ 69 \div 3 = 23 \]
4. 23 is a prime number.

Thus, the prime factorization of 276 is:
\[ 276 = 2^2 \times 3^1 \times 23^1 \]

### Step 4: Prime Factorization of 1242
1. Divide 1242 by 2:
\[ 1242 \div 2 = 621 \]
2. Divide 621 by 3:
\[ 621 \div 3 = 207 \]
3. Divide 207 by 3:
\[ 207 \div 3 = 69 \]
4. Divide 69 by 3:
\[ 69 \div 3 = 23 \]
5. 23 is a prime number.

Thus, the prime factorization of 1242 is:
\[ 1242 = 2^1 \times 3^3 \times 23^1 \]

### Step 5: Determine the HCF
Now, we find the HCF by taking the lowest power of each common prime factor:
- For 2: minimum power is \(1\) (from 1242)
- For 3: minimum power is \(1\) (from 276)
- For 23: minimum power is \(1\) (common in both)

Thus, the HCF is:
\[ HCF = 2^1 \times 3^1 \times 23^1 = 2 \times 3 \times 23 = 138 \]

### Conclusion
The greatest number that can be divided by 280 and 1245 leaving remainders of 4 and 3 respectively is:
\[ \text{Greatest Number} = 138 \]

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