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Post by Ruthy on Feb 14, 2006 20:28:54 GMT -5
Lets try this and post the answer! What is the smallest number that is divisible by 2, 3, 4, 5 and 6 with one digit left over, yet is evenly divided by 7?  |
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Post by Bala on Dec 7, 2006 1:10:11 GMT -5
The answer is 301.
First you must find the last digit. Since 5 can only divide into itself or 0 and 4 cannot divide into 5 the last digit must be 1 (as there is a remainder of 1 when it is divided by 2,3,4,5,6).
Since the number must be divisible by 7, you must find the multiples of 7 that end in 1 (ie. 21, 91, etc). You can find these multiples by finding how many times 7 takes to cycle through the digits (1
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Post by Bala on Dec 7, 2006 1:12:47 GMT -5
cont'd...
(10). Therefore you know that after 21 you keep adding 70 to attain the next number in the sequence that ends in 1. After some trial and error, the number 301 fits the criteria.
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KanagasingamMohanalingamமனிதன்
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Post by KanagasingamMohanalingamமனிதன் on Feb 29, 2020 23:42:59 GMT -5
👀
2/29/2020 February 29, also known as leap day or leap year day, is a date added to most years that are divisible by 4, such as 2016, 2020, and 2024.
420,840,1680,⋯are the numbers perfectly divisible by 2, 3, 4, 5, 6 and 7.
LCM: 2x2x3x5x7 = 420 👍
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