determine two no. nearest to 10,000 are exactly divisible by each of 2,3,4,5,6and 7
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Answer:
How do you find the two numbers which are nearest to 10000 and are exactly divisible by each of 7,6,5,4,3,2?
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The first step is identifying the least common multiple (LCM) of 7,6,5,4,3,2 which happens to be
2*3*5*7*2 = 420.
(Shortcut to LCM is directly multiply all prime numbers in the list which 2,3,5 and 7. That leaves us with 4 and 6. 4 is 2*2, there is already one 2 factored in, so factor in another 2. And 6 can be ignored as 6 = 2 *3 and both are factored in already. In short, if a number is divisible by both 2 and 3 then the number is also divisible by 6)
Now let's divide 10000/420
The quotient is 23 and the remainder is 340.
So the two numbers nearest - attention to the word nearest - not smaller, not larger - it's nearer - so the two numbers would obviously be
The highest number that is divisible by 2,3,4,5,6 and 7 and is less than 10000 which is 23*420 = 9660 and
The lowest number that is divisible by 2,3,4,5,6 and 7 and is greater than 10000 which is 24*420 = 10080
The other way to visualise this is - the nearest two numbers would definitely be the multiples of the six numbers (2,3,4,5,6,7) on either side of the given number (10000) in the number line.
So the numbers you are looking for are 9660 and 10080.
Hope this helps. Thanks.
∴ The two numbers nearest to 10000 which are exactly divisible by each of 2, 3, 4, 5, 6 and 7 are 9660 and 10080.
The numbers which are exactly divisible by 2, 3, 4, 5, 6 and 7 are the multiples of the LCM of the given numbers.
∴ LCM = 2 x 2 x 3 x 5 x 7 = 420
Now, dividing 10000 by 420, we get remainder = 340
∴ Number just less than 10000 and exactly divisible by the given numbers = 10000 - 340 = 9660
Number just greater than 10000 and exactly divisible by the given numbers = 10000 + (420 - 340) = 10080