\nLet the number be N. The condition states that when N is divided by 12, 15, and 18, the remainder is always 5. This means N – 5 is divisible by 12, 15, and 18. Therefore, N – 5 must be a multiple of the Least Common Multiple (LCM) of 12, 15, and 18.
\nThe prime factorization of 12 is 2\u00b2 \u00d7 3.
\nThe prime factorization of 15 is 3 \u00d7 5.
\nThe prime factorization of 18 is 2 \u00d7 3\u00b2.
\nThe LCM (12, 15, 18) = 2\u00b2 \u00d7 3\u00b2 \u00d7 5 = 4 \u00d7 9 \u00d7 5 = 180.
\nSo, N – 5 = 180k for some integer k.
\nN = 180k + 5.
\nWe are looking for the largest 3-digit number of this form. The largest 3-digit number is 999.
\nWe need 180k + 5 \u2264 999.
\n180k \u2264 994.
\nk \u2264 994 \/ 180 \u2248 5.52.
\nThe largest integer value for k is 5.
\nSubstituting k = 5 into the formula for N:
\nN = 180 \u00d7 5 + 5 = 900 + 5 = 905.
\n905 is a 3-digit number. Let’s check if it gives a remainder of 5 when divided by 12, 15, and 18:
\n905 \u00f7 12 = 75 with remainder 5. (900 = 12 * 75)
\n905 \u00f7 15 = 60 with remainder 5. (900 = 15 * 60)
\n905 \u00f7 18 = 50 with remainder 5. (900 = 18 * 50)
\nSince 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.
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