# Aptitude - Problems on H.C.F and L.C.M - Discussion

Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 5)

5.

The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

Answer: Option

Explanation:

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

Required number (9999 - 399) = 9600.

Discussion:

118 comments Page 1 of 12.
Arun said:
8 years ago

Guys, In question, they ask 15, 25, 40 & 75 is divisible by the greatest number of four digits.

So they give 4 digits means.

9999 is the big digit, okey leave this one .

Given options a)9000 b) 9400 c)9600 d)9800.

First we take L.C.M of 15, 25, 40 & 75 = 600.

In options first big one is 9800 so divide with 600 (9800/600=is not divisible).

So take second big one is 9600 so divide with 600(9600/600=is divisible with 16).

Third big one is 9400 so divide with 600(9400/600 = is not divisible).

Fourth big one is 9000 so divide with 600(9000/600 = is divisible with 15).

Then what's the question is the biggest 4 digit number is divisible with 15,25,40&75.

Here given options two 4 digit's number is divisible (i.e.9000, 9600).

In question ask the greatest 4 digit number (9000 & 9600 which is big one) And obviously answer is 9600.

So they give 4 digits means.

9999 is the big digit, okey leave this one .

Given options a)9000 b) 9400 c)9600 d)9800.

First we take L.C.M of 15, 25, 40 & 75 = 600.

In options first big one is 9800 so divide with 600 (9800/600=is not divisible).

So take second big one is 9600 so divide with 600(9600/600=is divisible with 16).

Third big one is 9400 so divide with 600(9400/600 = is not divisible).

Fourth big one is 9000 so divide with 600(9000/600 = is divisible with 15).

Then what's the question is the biggest 4 digit number is divisible with 15,25,40&75.

Here given options two 4 digit's number is divisible (i.e.9000, 9600).

In question ask the greatest 4 digit number (9000 & 9600 which is big one) And obviously answer is 9600.

Sravanreddypailla said:
1 decade ago

We have to find greatest 4 digit number so it may be between 9000 to 9999 let the number be x

this x should be divisible by 15, 25, 40 and 75 so take common least nnumber from all factors i.e l.c.m=600

therefore x is also divisible by 600 that means when x is divided by 600 remainder should be '0' but how can we find that number??

that number x lies between 9000 to 9999

take any number between them i.e take 9000,..9010,...9090...9900...9997,9998,9999 as your wish

so insted of 9999 here iam taking 9998 which is divided by 600

so remainder is 398 so substract 398 from 9998 i.e 9998-398=9600

why we wre substracting is because x should be divisible 600 and

remaider should be '0' so x is 9600 thats it

this x should be divisible by 15, 25, 40 and 75 so take common least nnumber from all factors i.e l.c.m=600

therefore x is also divisible by 600 that means when x is divided by 600 remainder should be '0' but how can we find that number??

that number x lies between 9000 to 9999

take any number between them i.e take 9000,..9010,...9090...9900...9997,9998,9999 as your wish

so insted of 9999 here iam taking 9998 which is divided by 600

so remainder is 398 so substract 398 from 9998 i.e 9998-398=9600

why we wre substracting is because x should be divisible 600 and

remaider should be '0' so x is 9600 thats it

Subir shahu said:
5 years ago

Sure the greatest 4 digit no. is 9999.

Here we have to find the greatest 4 digit number divisible by 15,25,40 and 75( that is the number divisible by 600 as the LCM of 15,25,40 and 75 is 600).it means the greatest 4 digit number divisible by 600 is only the greatest number divisible by 15,25,40 and 75.

To find 4 digit greatest number here we must find the nearest 4 digit number less 9999 that is possible by dividing the 9999 by the LCM and subtracting LCM.

=> 9999Ã·600

=> 9999-(600*16)

=>9999-(9600)

=>399 is remainder.

We can calculate the required number by subtracting 399= 9999-399 or,

we can directly find the nearest number divisible by 600= 600*16=9600.

Here we have to find the greatest 4 digit number divisible by 15,25,40 and 75( that is the number divisible by 600 as the LCM of 15,25,40 and 75 is 600).it means the greatest 4 digit number divisible by 600 is only the greatest number divisible by 15,25,40 and 75.

To find 4 digit greatest number here we must find the nearest 4 digit number less 9999 that is possible by dividing the 9999 by the LCM and subtracting LCM.

=> 9999Ã·600

=> 9999-(600*16)

=>9999-(9600)

=>399 is remainder.

We can calculate the required number by subtracting 399= 9999-399 or,

we can directly find the nearest number divisible by 600= 600*16=9600.

(1)

Ankit said:
1 decade ago

When greatest FOUR Digit number is to be found it means the number wont exceed 9999 so when we take 600 modulus 9999 we get remainder 399 which means that we are having 399 left after dividing 9999 with 6000 so we subtract 399 from 9999 so that remainder becomes zero and we get the greatest 4 digit number.

As the LCM of four number is 600 the number divisible by them are multiple of 600 i.e.

600

1200

1800

2400

3000

3600

4200

4800

5400

6000

6600

7200

7800

8400

9000

9600 <----- this is the greatest 4 digit no. divisible

10200

As the LCM of four number is 600 the number divisible by them are multiple of 600 i.e.

600

1200

1800

2400

3000

3600

4200

4800

5400

6000

6600

7200

7800

8400

9000

9600 <----- this is the greatest 4 digit no. divisible

10200

Yash said:
5 years ago

When greatest FOUR Digit number is to be found it means the number won't exceed 9999 so when we take 600 modulus 9999 we get remainder 399 which means that we are having 399 left after dividing 9999 with 6000 so we subtract 399 from 9999 so that remainder becomes zero and we get the greatest 4 digit number.

As the LCM of four number is 600 the number divisible by them are multiple of 600 i.e.

600

1200

1800

2400

3000

3600

4200

4800

5400

6000

6600

7200

7800

8400

9000

9600 -----> this is the greatest 4 digit number divisible.

10200.

As the LCM of four number is 600 the number divisible by them are multiple of 600 i.e.

600

1200

1800

2400

3000

3600

4200

4800

5400

6000

6600

7200

7800

8400

9000

9600 -----> this is the greatest 4 digit number divisible.

10200.

Rahul Vasu said:
1 decade ago

Calculating without LCM method

Shortcut way to get 600

In the option A) is given 9000 and B) 9400 and C) 9600 and D) 9800

To subtract from highest number first i.e (D-C) 9800-9600= 200 and (B-A) 9400-9000= 400

Now (D-C)=200

(B-A)=400

Add 200 + 400= 600

The questions says greatest four digit that means in every first digit is 9..so therefore there will be 4 greatest digits of 9 i.e is 9999

Hence 9999-600= 9,399

(Therefore the first digit is 9 so the rest 3 digit is 399 and hence 9999 - 399)

9999-399= 9600

Shortcut way to get 600

In the option A) is given 9000 and B) 9400 and C) 9600 and D) 9800

To subtract from highest number first i.e (D-C) 9800-9600= 200 and (B-A) 9400-9000= 400

Now (D-C)=200

(B-A)=400

Add 200 + 400= 600

The questions says greatest four digit that means in every first digit is 9..so therefore there will be 4 greatest digits of 9 i.e is 9999

Hence 9999-600= 9,399

(Therefore the first digit is 9 so the rest 3 digit is 399 and hence 9999 - 399)

9999-399= 9600

Akshata said:
1 decade ago

@Arunav.

Your explanation seems to be quick and fast. Can you elaborate and let me know who will it work.

My funds of finding LCM in this case 600 is clear.

But when you are writing "Numbers of option B and option D are not getting divided by 600 and between option B and C (which are getting divided by 600) the number of option C is the greatest no. Divisible by 600" is not clear. Just let me know the explanation.

In this way I guess it would be quicker than dividing 9999 and then subtracting the remainder.

Your explanation seems to be quick and fast. Can you elaborate and let me know who will it work.

My funds of finding LCM in this case 600 is clear.

But when you are writing "Numbers of option B and option D are not getting divided by 600 and between option B and C (which are getting divided by 600) the number of option C is the greatest no. Divisible by 600" is not clear. Just let me know the explanation.

In this way I guess it would be quicker than dividing 9999 and then subtracting the remainder.

Tejas Kumbhar said:
1 decade ago

The Logic here is this:

We have to find the greatest 4 digit number which is divisible by 15, 25, 40 and 75, hence it is a multiple of all these.

The LCM(15, 25, 40, 75) = 600 i.e. the least common multiple.

Hence all other common multiples of 15, 25, 40 and 75 will be multiples of 600 as well.

So to find the highest 4 digit one among all the common multiples, divide all the options by 600 to check which is perfectly divisible by 600. We get 9600, which is the answer.

We have to find the greatest 4 digit number which is divisible by 15, 25, 40 and 75, hence it is a multiple of all these.

The LCM(15, 25, 40, 75) = 600 i.e. the least common multiple.

Hence all other common multiples of 15, 25, 40 and 75 will be multiples of 600 as well.

So to find the highest 4 digit one among all the common multiples, divide all the options by 600 to check which is perfectly divisible by 600. We get 9600, which is the answer.

Darshan said:
8 years ago

Concept is simple.

The greatest 4 digit no in numbers is 9999.

Then find the LCM of 15, 25, 40, 75 = that is 600.

So if 9999 is divisible by 600 then it is the greatest 4 digit no.

If we divide 9999 by 600 it will leave a remainder 399.

600) 9999 (16

9600

--------

0399 --< this is remainder.

If we subtract it with dividend :: (9999-399) = 9600.

So the 9600 is divisible by 600 without leaving a remainder hence it is divisible by all given nos.

The greatest 4 digit no in numbers is 9999.

Then find the LCM of 15, 25, 40, 75 = that is 600.

So if 9999 is divisible by 600 then it is the greatest 4 digit no.

If we divide 9999 by 600 it will leave a remainder 399.

600) 9999 (16

9600

--------

0399 --< this is remainder.

If we subtract it with dividend :: (9999-399) = 9600.

So the 9600 is divisible by 600 without leaving a remainder hence it is divisible by all given nos.

Shashank Ravoor said:
1 year ago

@All.

Here's my explanation:

The given numbers are 15,25,40 and 75 are multiples of 5. So I can associate these numbers with respect to 5. The greatest numbers that are given in the options should be added as the sum of digits and check whether it's completely divisible by 5.

Now going through options:

A: 9+0+0+0=9 -> not divisible by 5

B: 9+4+0+0=13 -> same as A

C: 9+6+0+0=15 -> completely divisible by 5.

Hence, Option C is the answer.

Here's my explanation:

The given numbers are 15,25,40 and 75 are multiples of 5. So I can associate these numbers with respect to 5. The greatest numbers that are given in the options should be added as the sum of digits and check whether it's completely divisible by 5.

Now going through options:

A: 9+0+0+0=9 -> not divisible by 5

B: 9+4+0+0=13 -> same as A

C: 9+6+0+0=15 -> completely divisible by 5.

Hence, Option C is the answer.

(81)

Post your comments here:

Quick links

Quantitative Aptitude

Verbal (English)

Reasoning

Programming

Interview

Placement Papers