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A 6 digit number 123ABC is exactly divisible by $5,7$ and $9$. How many such possible numbers are there ?

1. 2
2. 3
3. 4
4. 5

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what is easy to do this?

For 7 divisibility is causing problem.

Total 6 digit numbers possible with start digits as 123 are 123000 to 123999.

Hence, 1000 such numbers are there .

Now, the number should be exactly divisible by 5,7,9 . Take their LCM which is 315 .

Now, possible numbers which are divisible are =>

Floor (1000 / 315) = Floor ( 3.17) = 3

Hence, 3 numbers are possible between 123000 and 123999 which are divisible by 5,7 and 9.

by

digvijay Only 123480(315*392) and 123795(315*393) are the numbers which are divisible by 5,7 and 9.I can't find any number other than this.After this 393 numbers are of form 124xxx .please check the answer.

@suniljha

123480-315= 123165 is also diviible.

Yeah, thanks @Apoorva Jain

For any k random numbers, the minimum number that those k numbers exactly divide is called the LCM.

And the greatest number that exactly divides those k random numbers is called the GCD/HCF.

LCM of 5, 7 and 9 is 315. It means that in the (Natural) Number Line every 315th number can be exactly divided by 5, 7 and 9.

123_ _ _. There can be $10^3=1000$ such numbers.

Since every 315th number is divisible by 5,7 and 9; for any range of 1000 numbers, there'll be $\frac{1000}{315}=3.17=3$ such numbers.

So, Option B

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