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Let xbe the least number, which when divided by 5,6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves no remainder. The sum of digits of x is

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Let xbe the least number, which when divided by 5,6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves no remainder. The sum of digits of x is
1). 21
2). 22
3). 18
4). 24

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2 answers

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Answered by on | Votes 2 |
option 3 is the right answer

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Answered by on | Votes 0 |
LCM of 5, 6, 7 and 8 = 840 LCM = 2 × 5 × 3 × 7 × 4 = 840 \ Required number = 840x + 3 which is divisible by 9 for a certain least value of x. Now, 840x + 3 = 93x × 9 + 3x + 3 3x + 3, is divisible by 9 for x = 2 Required number = 840 × 2 + 3 = 1680 + 3 = 1683 Sum of digits = 1 + 6 + 8 + 3= 18




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