Which one of the following is the largest 3-digit number which when divided by 12, 15 and 18 respectively, gives a remainder 5 in each case ?
955
905
995
755
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CISF-AC-EXE – 2022
The prime factorization of 12 is 2² × 3.
The prime factorization of 15 is 3 × 5.
The prime factorization of 18 is 2 × 3².
The LCM (12, 15, 18) = 2² × 3² × 5 = 4 × 9 × 5 = 180.
So, N – 5 = 180k for some integer k.
N = 180k + 5.
We are looking for the largest 3-digit number of this form. The largest 3-digit number is 999.
We need 180k + 5 ≤ 999.
180k ≤ 994.
k ≤ 994 / 180 ≈ 5.52.
The largest integer value for k is 5.
Substituting k = 5 into the formula for N:
N = 180 × 5 + 5 = 900 + 5 = 905.
905 is a 3-digit number. Let’s check if it gives a remainder of 5 when divided by 12, 15, and 18:
905 ÷ 12 = 75 with remainder 5. (900 = 12 * 75)
905 ÷ 15 = 60 with remainder 5. (900 = 15 * 60)
905 ÷ 18 = 50 with remainder 5. (900 = 18 * 50)
Since k=5 gives 905, and any larger integer k would result in a number greater than 999 (e.g., k=6 gives 180*6+5 = 1085), 905 is the largest 3-digit number satisfying the condition.