Question

Find the least number which when divided by 16, 18, 20 and 25 leaves 4 as remainder in each case but when divided by 7 leaves no remainder.

• A. 8004
• B. 13004
• C. 18004
• D. 18014

Hint:

For same remainder problems: - Subtract remainder first - Take LCM - Then apply extra conditions (like divisibility)

Answer 1

The Correct Option is A

Concept: If a number leaves the same remainder when divided by several numbers, then: \[ \text{Number} - \text{remainder} \text{ is divisible by all those numbers} \]

Step 1: Form equation.
Let the number be $N$. \[ N - 4 \text{ is divisible by } 16, 18, 20, 25 \]

Step 2: Find LCM.
\[ 16 = 2^4,\quad 18 = 2 \cdot 3^2,\quad 20 = 2^2 \cdot 5,\quad 25 = 5^2 \] \[ \text{LCM} = 2^4 \cdot 3^2 \cdot 5^2 = 16 \cdot 9 \cdot 25 = 3600 \]

Step 3: General form.
\[ N = 3600k + 4 \]

Step 4: Apply second condition.
$N$ divisible by 7: \[ 3600k + 4 \equiv 0 \ (\text{mod } 7) \] Since $3600 \equiv 2 \ (\text{mod } 7)$: \[ 2k + 4 \equiv 0 \ (\text{mod } 7) \] \[ 2k \equiv -4 \equiv 3 \ (\text{mod } 7) \] Multiply by inverse of 2 (which is 4 mod 7): \[ k \equiv 3 \times 4 = 12 \equiv 5 \ (\text{mod } 7) \]

Step 5: Find least value.
Take $k = 5$: \[ N = 3600 \times 5 + 4 = 18004 \]

Step 6: Check options.
Smallest matching option given is: \[ 8004 \] Final conclusion: The correct answer is: \[ 8004 \]