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What is the smallest whole number that, when divided by 2,
leaves a remainder of 1; when divided by 3, leaves a
remainder of 2; and so on, up to leaving a remainder of 9
when divided by 10?
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Answer / anil
Ans. is 2519.
we have to multiply all numbers from 2-10 keeping aside
common number.
i.e.
2*3*4*5*7*3 = 2520
6 is left b'coz 2*3 is 6
9 = 3*3 hence, 3 is taken twice.
At last 2520-1.
| Is This Answer Correct ? | 29 Yes | 10 No |
Answer / sudhanshu_kmr
According to the question when the number is divided by 2,3,..10 leaves remainder 1,2,...9 respectively. It means that if we increase the that number by 1 then it is divisible by each of the numbers 2,3,..10.
So, find a number that is common divisor of 2,3,4,...and 10.
And subtract 1 from it.
LCM of 2,3,..10 is 2520.
2520-1 = 2519
thus, 2519 is the required number.......
| Is This Answer Correct ? | 11 Yes | 2 No |
Answer / anil
2519
The ans is correct but still searching how to arrive at ans.
| Is This Answer Correct ? | 12 Yes | 7 No |
Answer / ramesh kumar
2519/10=251 @ 9
2519/9=279 @ 8
/8=312 @ 7
/7=358 @ 6
/6=419 @ 5
/5=503 @ 4
/4=629 @ 3
/3=839 @ 2
/2=1259 @ 1
so that the answer should be 2519
| Is This Answer Correct ? | 9 Yes | 4 No |
Answer / nazeera jaffar
First take LCM of 2,3,4,...upto 10
The answer is 2520.
Now when the number is divided and the results are compared with each other,it leaves 1 in each case.
So at last 2520-1=2519 is the answer.
| Is This Answer Correct ? | 3 Yes | 2 No |
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