Find the smallest no . Which is divided by 25,15,30 and leaves remainder 21,11,26 resp.

Answer

To find the smallest number that, when divided by 25, 15, and 30, leaves remainders of 21, 11, and 26 respectively, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( x \) such that:
- \( x \equiv 21 \mod 25 \)
- \( x \equiv 11 \mod 15 \)
- \( x \equiv 26 \mod 30 \)

### Step 2: Rewrite the Congruences
From the congruences, we can express \( x \) in terms of each modulus:
1. \( x = 25k + 21 \) for some integer \( k \)
2. \( x = 15m + 11 \) for some integer \( m \)
3. \( x = 30n + 26 \) for some integer \( n \)

### Step 3: Find the Least Common Multiple (LCM)
To solve these congruences, we first find the least common multiple (LCM) of the divisors:
- The prime factorization of 25 is \( 5^2 \).
- The prime factorization of 15 is \( 3^1 \times 5^1 \).
- The prime factorization of 30 is \( 2^1 \times 3^1 \times 5^1 \).

The LCM is calculated as follows:
- The highest power of 2 is \( 2^1 \).
- The highest power of 3 is \( 3^1 \).
- The highest power of 5 is \( 5^2 \).

Thus,
\[ \text{LCM}(25, 15, 30) = 2^1 \times 3^1 \times 5^2 = 2 \times 3 \times 25 = 150. \]

### Step 4: Adjust for the Remainders
Next, we need to adjust the LCM to account for the remainders. We notice that:
- For 25, the difference between 25 and 21 is \( 4 \).
- For 15, the difference between 15 and 11 is \( 4 \).
- For 30, the difference between 30 and 26 is \( 4 \).

This shows that we can subtract 4 from the LCM to find the smallest number that satisfies all conditions.

### Step 5: Calculate the Final Answer
Now we can calculate:
\[ x = \text{LCM} - 4 = 150 - 4 = 146. \]

Thus, the smallest number that satisfies the given conditions is **146**.