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What is the smallest number which when divided by 10 leaves
a remainder of 9, when divided by 9 leaves a remainder of 8,
when divided by 8 leaves a remainder of 7, when divided by 7
leaves a remainder of 6 and so on until when divided by 2
leaves a remainder of 1?
- 3 Answers
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Answer / guest
The smallest such number is 2519.
The easiest way is to find the Least Common Multiple (LCM)
of 2, 3, 4, 5, 6, 7, 8 and 9. And subtract 1 from it.
The LCM of 2, 3, 4, 5, 6, 7, 8 and 9 is given by 2520.
Hence, the required number is 2519
| Is This Answer Correct ? | 90 Yes | 8 No |
The answer is LCM of 2,3,....9 and subtract 1 from the
result. Please not that every no divisible by 6 is
divisible by 2 & 3, also every number divisible by 8 is
divisible by 4, and any number divisible by 8 & 9 will be
divisible by 6. These are the fundas. hence the answer is
9*8*7*5 - 1 = 2519.
ofcourse this may not as easy to figure out as i siad
unless one is good at the math fundas.
| Is This Answer Correct ? | 34 Yes | 12 No |
Answer / jason
Or write a simple computer program...
while(true){
if(x%10 = 9 && x%9 = 8 && x%8 = 7 && x%7 = 6 && x%6 = 5
&& x%5 = 4 && x%4 = 3 && x%3 = 2 && x%2 = 1){
print(x);
break;
}
else{
x++;
}
}
| Is This Answer Correct ? | 16 Yes | 4 No |
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