Question
Download Solution PDFWhat is the smallest number which when divided by 3, 5, 6, 8, 10 and 12 leaves a remainder of 2 in each case but which when divided by 13 leaves no remainder?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation
LCM of 3, 5, 6, 8, 10 and 12 is 120.
If the remainder in each case is 2 and the number is divisible by 13, then
The number is (120x + 2) divisible by 13, so
(120x + 2)/13
⇒ 9x + (3x + 2)/13
(3x + 2) is divisible by 13 if we put x = 1, 2, 3, 4, ……..
Put x = 8, then
⇒ (3 × 8 + 2)
⇒ 26
As we know, 13 is divisible by 13 so put x = 8
120x + 2
⇒ 120 × 8 + 2
⇒ 962
The answer is 962.
Last updated on Jan 29, 2025
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