Find the smallest number which when decreased by 10 is exactly divisible by 16,21 ,24 and 42 ?

Answer

To find the smallest number which when decreased by 10 is exactly divisible by 16, 21, 24, and 42, we can follow these steps:

### Step 1: Find the LCM of the given numbers
We need to find the least common multiple (LCM) of 16, 21, 24, and 42.

**Prime Factorization:**
- 16 = 2^4
- 21 = 3^1 × 7^1
- 24 = 2^3 × 3^1
- 42 = 2^1 × 3^1 × 7^1

**Finding LCM:**
To find the LCM, we take the highest power of each prime number that appears in the factorizations:
- For 2: the highest power is 2^4 (from 16)
- For 3: the highest power is 3^1 (from 21, 24, and 42)
- For 7: the highest power is 7^1 (from 21 and 42)

Thus, the LCM is:
\[ LCM = 2^4 × 3^1 × 7^1 = 16 × 3 × 7 \]

### Step 2: Calculate the LCM
Now, we calculate:
\[ 16 × 3 = 48 \]
\[ 48 × 7 = 336 \]

So, the LCM of 16, 21, 24, and 42 is 336.

### Step 3: Find the required number
The problem states that we need a number which when decreased by 10 gives us a number that is divisible by the LCM we found. Therefore, we need to add 10 to the LCM:
\[ \text{Required number} = LCM + 10 = 336 + 10 = 346 \]

### Step 4: Verification
To verify, we check if 336 (which is 346 - 10) is divisible by 16, 21, 24, and 42:
- \(336 ÷ 16 = 21\) (exactly divisible)
- \(336 ÷ 21 = 16\) (exactly divisible)
- \(336 ÷ 24 = 14\) (exactly divisible)
- \(336 ÷ 42 = 8\) (exactly divisible)

Since 336 is divisible by all four numbers, our solution is confirmed.

### Final Answer
The smallest number which when decreased by 10 is exactly divisible by 16, 21, 24, and 42 is **346**.

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