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 / martin zhu
1.devided by 10 remaining 9, the last-right digit is 9
(also for 2 remaining 1, for 5 remaining 4)
so ok now for 10,2,5
devided 9 remaining 8, 89+10x9n ( n is whole number)(also
for 3 remaining 2, for 6 remaing 5)
so ok now for 10,2,5,9,3,6
devided 8 remaining 7, 359+10x9X4n ( n is whole number)
(also for 4 remaing 3)
so ok now for 10,2,5,9,3,6,8,4
left: devided 7 remaining 6, find the smallest n in
359+10X9X4n to meet 7 remaining 6
then you can get 2519 is the final answer to the question.
| Is This Answer Correct ? | 3 Yes | 6 No |
Answer / duke
23....if you divide the # by ur reminder is 1.....then when
u divided 23 by 3.....ur reminder is 2.......
| Is This Answer Correct ? | 1 Yes | 6 No |
Answer / jaswinder
when 3/2 leaves reminder 1..
when 5/3 leaves reminder 2..
when 19/10 leaves reminder 9..
so answer is 3,5,19
| Is This Answer Correct ? | 9 Yes | 19 No |
Answer / abhijeet
29/2 => qoutient = 14 (2*14=28), REMAINDER = 1 (29-28 = 1)
29/3 => qoutient = 9 (3*9=27) , REMAINDER = 2 (29-27 = 2)
29/10 => qoutient = 2 (10*2=20), REMAINDER = 2 (29-20 = 9)
SO, WE GET REMAINDERS 1, 2 and 9 respectively.
| Is This Answer Correct ? | 6 Yes | 20 No |
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