Question 4: Let x be the least number which when divided by 12,15,18,20,27 and 30, the...
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Let x be the least number which when divided by 12,15,18,20,27 and 30, the remainder in each case is 1 and x is a multiple of 47. The sum of the digits of x is:
Show answer & explanation
The number x must be of the form (LCM of divisors) * k + 1. The LCM of 12, 15, 18, 20, 27, and 30 is 540. We seek the smallest k such that 540k + 1 is divisible by 47, which occurs at k = 2, giving x = 1081. The sum of the digits of 1081 is 1 + 0 + 8 + 1 = 10.
Step-by-step Derivation:
Step 1: Find the Least Common Multiple (LCM) of 12, 15, 18, 20, 27, and 30.
- Prime factorizations: 12=2²·3, 15=3·5, 18=2·3², 20=2²·5, 27=3³, 30=2·3·5.
- LCM = 2² · 3³ · 5 = 4 · 27 · 5 = 540.
Step 2: Set up the general form for x.
Since x leaves a remainder of 1 when divided by these numbers, x = 540k + 1 for some integer k ≥ 1.
Step 3: Find the smallest k such that x is a multiple of 47.
- 540k + 1 ≡ 0 (mod 47)
- Divide 540 by 47: 540 = 47 * 11 + 23.
- So, 23k + 1 ≡ 0 (mod 47).
- Testing small values of k:
- k = 1: 23(1) + 1 = 24 (Not divisible by 47)
- k = 2: 23(2) + 1 = 46 + 1 = 47 (Divisible by 47)
Step 4: Calculate x.
- x = 540(2) + 1 = 1080 + 1 = 1081.
Step 5: Calculate the sum of the digits of x.
- Sum = 1 + 0 + 8 + 1 = 10.