Question

Find the lowest of all the given numbers 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. 19000
• B. 18004
• C. 17004
• D. 18000
• E. 18002

Hint:

For "leaves same remainder" problems, first find the LCM of the divisors. The number is of the form LCM $\times k$ + remainder.

Answer 1

The Correct Option is B

Step 1: Find the LCM:
Let the number be $N$. $N$ leaves a remainder of 4 when divided by 16, 18, 20, and 25. So, $N-4$ is divisible by each of these numbers. Thus, $N-4$ is a common multiple of 16, 18, 20, and 25. Find the LCM(16, 18, 20, 25). $16 = 2^4$, $18 = 2 \times 3^2$, $20 = 2^2 \times 5$, $25 = 5^2$. LCM = $2^4 \times 3^2 \times 5^2 = 16 \times 9 \times 25 = 16 \times 225 = 3600$. So, $N-4 = 3600k$, where $k$ is a positive integer. Thus, $N = 3600k + 4$.
Step 2: Apply the Second Condition:
$N$ is divisible by 7, i.e., $3600k + 4 \equiv 0 \pmod{7}$. Find $3600 \mod 7$: $7 \times 514 = 3598$, remainder $2$. So $3600 \equiv 2 \pmod{7}$. The condition becomes $2k + 4 \equiv 0 \pmod{7}$. $2k \equiv -4 \equiv 3 \pmod{7}$. Multiply both sides by the modular inverse of 2 modulo 7 (which is 4, since $2 \times 4 = 8 \equiv 1 \pmod{7}$): $k \equiv 3 \times 4 = 12 \equiv 5 \pmod{7}$. So, $k = 7m + 5$ for some integer $m \ge 0$. Then $N = 3600(7m+5) + 4 = 25200m + 18000 + 4 = 25200m + 18004$.
Step 3: Find the Lowest:
For the lowest positive $N$, set $m=0$. Then $N = 18004$. Check optionss: 18004 is present.
Step 4: Final Answer:
The lowest number is 18004.