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Let x be the smallest 4-digit number such that when it is divided by 5, 6 and 7, it leaves the remainders 2, 3 and 4, respectively.
The sum of the digits of x is:
8
11
12
14
- We need the smallest 4-digit number \( x \) such that:
- When divided by 5, remainder is 2
- When divided by 6, remainder is 3
- When divided by 7, remainder is 4
- This can be expressed as:
- \( x \equiv 2 \pmod{5} \)
- \( x \equiv 3 \pmod{6} \)
- \( x \equiv 4 \pmod{7} \)
- Notice the remainders are one less than the divisors.
- So, \( x+3 \) is divisible by 5, 6, and 7.
- LCM of 5, 6, 7 = 210.
- Smallest 4-digit number: 1000.
- \( 1000 + 3 = 1003 \)
- \( 1003/210 \approx 4.776 \), next multiple: \( 210 \times 5 = 1050 \)
- \( x + 3 = 1050 \Rightarrow x = 1047 \)
- Sum of digits: \( 1 + 0 + 4 + 7 = 12 \)
- Option 3: 12
- Option 1 (8), Option 2 (11), Option 4 (14) are not correct.
By: santosh ProfileResourcesReport error
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