What is the smallest number of five digits which is divisible by 15, 27 and 35?

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28 Mar 2017 Divisibility

What is the smallest number of five digits which is divisible by 15, 27 and 35?

OPtion

1) 10230
2) 10240
3) 10800
4) 10085
5) 10090
6) 10220
7) 10385
8) 10235
9) 10250
10) None of these

Solution

LCM of 15, 27, 35 = 945

9999 ÷ 945 = 10, remainder = 549
Hence, greatest number of four digits which is divisible by 15, 27 and 35
= 9999 - 549 = 9450
So Smallest 5 digit number divisible by 15, 27 and 35 = 9450+945 = 10395

Option 10)

Correct Option: 10 Best Solution (16)

G.S.Kantharao   7 years AGO

L. C. M of 15,27,35=945
Least 5 digited No. 10000
10000/945= 10.58
So the 11 th multiple should be least 5 digited number which should be divisible by 15,27,35. That is 945*11= 10395
Which is not given Hence my option is 10.

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Balakumaresan Pappuraj   7 years AGO

LCM of 15, 27, 35 is 945.
Remainder for 10000÷945 is 550
when we subtract 550 from 945 the ans is 395.
So the smallest five digit number which is divisible by 15, 27, 35 is (=10000+395) 10,395

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Mukund Manohar   7 years AGO

LCM of 15,27 and 35 = 945
Multiple of 945 close to 10000 is 10395
Smallest 5 digit number divisible by 15,27 and 35 is 10395

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sirisha   7 years AGO

first take the LCM of 15,27,35.
then we get the factors as 9,7,3,5.
Then you verify each and every option whether the number is divisible by all the factors.
If the number is divisible by all the factors then that number is divisible by the given three numbers also.
By option verification it is clear that no number is divisible by all the given numbers
so the right option is 10(none of these).

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Azim Uddin Siddiqui   7 years AGO

none of the given nos are divisible by all 15,27 & 35. so option 10

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Payal   7 years AGO

The LCM of 15, 27, 35 is 945.
the smallest 5 digit number that is multiple of 945 is 945*11=10395

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Jayesh   7 years AGO

Numbers ending with zeros can not be divided by 27 and rest 3 tried pracriacally...they are not divisible by 27 so answer is 10

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chandan rawat   7 years AGO

LCM of 15,27,35 is 945
if we devide 10000 by 945 remainder is 550
10000+(945-550)
10000+395=10395

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abhishek   7 years AGO

L.C.M OF(15,27,35) is 945

Therefore number is multiple of 945. Then smallest 5-digit number which is divisible by 945 is 10395. Which is not available in given option.
Therefore option 10).

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Sahil Mehta   7 years AGO

since the numbers are 15 , 27 and 35 therefore we can conclude that
15 contains number of threes equal to 1 and 27 contains number of threes equal to 3 , and 35 contains no threes. 15 contains one 5 and 35 contains one 5 and 7 .
therefore the number can be:
3*3*3*3*5*5*7=14155

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SORNA KAMALAM S   7 years AGO

Ans:10395
L.C.M of 15,27,35=945
945*10=9450
945*11=10395

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aparna padakanti   7 years AGO

the. LCM of 15 27 35 is 945
five digits smallest number is 10000 divisible by 945 lives reminder 540
10000+(945-540)
10405

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santosh kumar   7 years AGO

Lcm of(15 27 35)=945 when divided by 945 leaves remainder 550
So ans is 10000+945-550=10395

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Amit   7 years AGO

LCM of 15,27,35=945
to get smallest 5 digit number= 945*10=9450+945=10395

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Harsh Poddar   7 years AGO

L.C.M of 15,27,35=945
Smallest five digit number which is divisible by 945 is 945×11=10395
Option 10

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Amit   7 years AGO

LCM of 15,27,35 is 4725
so smallest 5 digit no. would be 4725*3=14175

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