L.C.M method to find the Smallest Numbers:
We use L.C.M method in finding the Smallest numbers in below cases.
A) To find the Smallest or least
number:
(i) The smallest number which is
exactly divisible by the numbers.
(ii) The smallest number when divided
by the numbers: a, b, c leaves remainder
a) Remainder ‘K’ in each case.
b) Remainders x, y, z respectively.
A) To find the smallest number which is exactly divisible:
(i) We need to find the L.C.M of the given numbers, which gives the
least number of the given numbers.
1) Find the smallest number which is exactly divisible by 72, 90 and
120.
Solution:
We need to find the L.C.M of the given numbers :
L.C.M( 72, 90, 120) = 360.
Therefore, L.C.M is the smallest number which
is divisible exactly by the given numbers = 360.
2) Find the smallest number which is divisible exactly by the numbers 7,
18, 56, 36 is?
Solution:
We need to find the L.C.M of the given numbers: L.C.M ( 7, 18, 56, 36 )
=
Therefore, L.C.M is the smallest number which is divisible exactly by the numbers 7, 18, 56, 36 = 504.
A) (ii) (a) Steps to find the smallest number when divided by the
numbers leaves remainder ‘K’ in each case:
1. We need to find the L.C.M:‘L’ of the given numbers.
2. The required Smallest number is obtained by adding the L.C.M obtained
in the step-1 and the Remainder:‘K’ given.
Smallest number = L + K.
3) Find the least number when divided by 36, 24, 16 leaves 11 as
remainder in each case?
Solution:
Given : Remainder ‘K’ = 11 ;
Numbers = 36, 24, 16.
(i) We need to find the L.C.M : L of the given numbers.
L.C.M(36, 24 ,16) = 144.
The least number = L.C.M of numbers’L’ + Remainder’K’.
=
144 + 11 = 155.
4) What is the least
number, which when divided by 98 and 105 has in each case 10 as remainder?
Solution:
Given : Numbers = 98 and 105; Remainder ’K’ = 10.
1. We need to find the L.C.M ‘L’ of
the given numbers:
L.C.M(98, 105) ‘L’= 1470.
Therefore, Smallest
number = L.C.M ‘L’ + Remainder ‘K’
Smallest Number = 1470 + 10 = 1480.
5) Find the least number which being
divided by 2, 3, 4, 5, 6, leaves in each case a remainder 1, but when divided
by 7 leaves no remainder.
Solution :
Given : Numbers = 2, 3, 4, 5, 6 ;
Reaminder ‘K’ = 1.
We need to find the L.C.M of given
numbers.
L.C.M( 2, 3, 4, 5, 6) ,’L’ = 60.
Therefore, the smallest
number = L + K = 60 + 1 = 61.
Here, we have to check the second
condition; the smallest number when divided by 7 leaves no remainder.
The smallest number we
got is 61, when divided by 7 leaves remainder; which doesn’t satisfy the given
condition. So, we have to find the multiples of Smallest number.
Multiples of 60 = 60, 120, 180, 240,
300, 360, 420, 480, 540, 600.
We have to check : The
multiples of 60 + Remainder ‘K’ is to be divisible by 7 exactly.
Therefore, the smallest number is 301, which is divisible by 7 exactly.
(A)(ii)(b) Steps to find the Smallest
number when divided by the numbers a, b, c leaves remainders x, y, z
respectively:
1.Find out the L.C.M of the given
numbers; L.C.M(a, b c) = L.
2. Check the difference between the
numbers and the remainders, which is equal to ‘K’.
3. The difference between the L.C.M ‘L’
and the Remainder ’K’ gives the required
Smallest number.
Smallest number = L – K .
6) What is the smallest number when
divided by the numbers 20, 48, 36 leaves remainders 13, 41, 29 respectivey.
Solution:
Given : Numbers: a, b, c = 20, 48, 36
; Remainders: x, y, z = 13, 41, 29.
1. L.C.M( 20, 48, 36 ) = 720.
2. The difference between the numbers and the remainders = a-x = b-y = c-z = K, say.
i.e; 20 – 13 = 48 – 41 = 36 – 29 = 7
= K.
3. The Smallest number = L – K = 720 –
7 = 713.
8) Find the smallest number which on
dividing by 6 8 12 leaves the remainder 4 6 10 respectively?
Solution:
Given:
Numbers: a, b, c = 6, 8, 12 ; Remainders: x, y, z = 4, 6, 10.
1.
L.C.M( a, b, c ) = L.C.M(6, 8, 12) = 24.
2.
The difference between the numbers and remainders: a – x = b – y = c – z = K ;
6 – 4 = 8 – 6 = 12 – 10 = 2 = K.
3.
The Smallest number = L – K = 24 – 2 = 20.
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