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Find the least number of five digits which when divided by 40, 60, and 75, leave remainders 31, 51 and 66 respectively.
Answer & Solution
Correct Answer:
Option
C
Difference, 40 - 31 = 9
60 - 51 = 9
75 - 66 = 9
Difference between numbers and remainder is same in each case.
Then,
The answer = {(LCM of 40, 60, 75) - 9}
40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
75 = 3 × 5 × 5
LCM = 2 × 2 × 2 × 5 × 5 × 3 = 600
But, the least number of 5 digits = 10000
$$\frac{{10000}}{{600}},$$ we get remainder as 400
Then, the answer = 1000 - (600 - 400) - 9 = 10191
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Login60 - 51 = 9
75 - 66 = 9
Difference between numbers and remainder is same in each case.
Then,
The answer = (Least number of 5 digit) + {(LCM of 40, 60, 75)- (Least number of 5 digit)/ LCM} - difference between divider & remainder
40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
75 = 3 × 5 × 5
LCM = 2 × 2 × 2 × 5 × 5 × 3 = 600
But, the least number of 5 digits = 10000
10000/600, we get remainder as 400
Then, the answer = 10000 + (600 - 400) - 9 = 10191