FIND GREATEST FOUR DIGIT NUMBER WHICH IS DIVISIBLE BY GIVEN NUMBERS

Problem 1 :

Find the largest four digit number exactly divisible by 12, 15, 18 and 27.

Solution :

The largest four digit number is 9999.

Since the required four digit number exactly divisible by 12, 15, 18 and 27, the required number is divisible by the given numbers.

LCM (12, 15, 18 and 27) = 54 x 10 ==> 540

9999 - 279 = 9720

9720 is the largest four digit number which is divisible by 12, 15, 18 and 27.

Problem 2 :

The largest four digit number which when divided by 4, 7 or 13 leaves the remainder of 3 in each case.

Solution :

LCM of 4, 7 and 13 = 364

9999 - 171 ==> 9828

When we divide 9828 by 4, 7 and 13, we will get 0 as remainder. But accordingly the question, we should get 3 as remainder.

So, the required number 3 greater than 9828.

required number = 9828 + 3

= 9831

Problem 3 :

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

Solution :

LCM (15, 25, 40, 75)

LCM (15, 25, 40, 75) = 600

The greater four digit number = 9999

Divide 9999 by 600

9999 - 399 ==> 9600

So, the required greatest four digit number which is divisible by 15, 25, 40 and 75 is 9600.

Problem 4 :

Find the least number when divided by 6, 7, 8, 9 and 12 leaves the same remainder 1 in each case.

Solution :

To find the least number which is common for 6, 7, 8, 9 and 12, we find the least common multiple of these numbers.

Least common multiple = 3 x 2 x 2 x 7 x 2 x 3

= 504

Since 504 is the least common multiple of 6, 7, 8, 9 and 12, while dividing 504 by each number we will get 0 as remainder. But accordingly the question given we should get the remainder 1. Then 504 should be 1 more. Then the required number while dividing by 6, 7, 8, 9 and 12 and getting the remainder 1 is 505.

Problem 5 :

Find the largest number of four digits exactly divisible by 12, 15, 18 and 27.

Solution :

To find the least number which is common for 12, 15, 18 and 27 we find the least common multiple of these numbers.

Least common multiple = 3 x 3 x 2 x 2 x 5 x 1 x 3

= 540

While dividing 540 by each of the given numbers 12, 15, 18 and 27 we will get the remainder as 0.

Since we try to find the largest four digit number, let us divide the largest four digits number 9999 by 540, we will get 279 as remainder. So, the required number will be 279 lesser than the greatest four digit number.

= 9999 - 279

= 9720

So, the required number is 9720.

Problem 6 :

Find the smallest number of five digits exactly divisible by 16, 24, 36 and 54.

Solution :

To find the least number which is common for 16, 24, 36 and 54 we find the least common multiple of these numbers.

Least common multiple = 2 x 2 x 3 x 2 x 3 x 2 x 3

= 432

The smallest five digit number is 10000 . While dividing 100000 by 432 we will get the remainder 64. So, the required number should be 64 more than the smallest five digit number.

= 100000 + (432 - 64)

= 100000 + 368

= 100368

So, 100368 is the smallest five digit number which is divisible by 16, 24, 36 and 54.

Problem 7 :

Find the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3 but when divided by 9 leaves no remainder.

Least common multiple of 5, 6, 7 and 8 is

= 2 x 5 x 3 x 7 x 4

= 840

Since the required number while dividing it by 5, 6, 7 and 8 it should get the remainder 3, so the required number must be multiple of 840 and added by 3,

843 is not divisible by 9

= 840 x 2 + 3

= 1680 + 3

= 1683

Here 1683 is divisible by 9, then the required least number  when divided by 5, 6, 7 and 8 leaves a remainder 3 but when divided by 9 leaves no remainder is 1683.

Problem 8 :

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

Solution :

Difference between the first number and the second number = 91 - 43

= 48

Difference between the second number and third number = 183 - 91

= 92

Difference between the third number and the first number = 183 - 43

= 140

Highest common factor of 48, 92 and 140 is 4.

So, the required number is 4..