The least number which when divided by 24, 28 and 32 leaves zero as it's remainder, is
Answers
Answer:
672
Step-by-step explanation:
find the LCM by factorization method
Step-by-step explanation:
Let us assume the required number as N.
Given conditions are:
When N28
is done we get 8 as remainder.
When N32
is done we get 12 as remainder.
So, first we will find the smallest number let it be K, which when divided by 28 and 32 leaves remainder 0 in both the cases. The LCM of two numbers implies the same description.
K = LCM(28,32).
We find the LCM by using the prime factorisation method.
So by converting 28 and 32 into multiplication of prime numbers we get:
28=22×732=25
So, now we find the least common multiple (LCM), we get:
K=25×7=224
Now, by Euclid’s division lemma,
Dividend = (Divisor ×
Quotient) + Remainder (remainder < divisor)
Now, we will write 224 in terms of 28:
224 = 28 ×
8+0
224 = 224+0
We need the remainder to be 8.
So, making remainder to be 8, we get:
224 = 216 + 8 ….. (1)
Now we need to write 216 in terms of 28,
Again by Euclid’s division lemma, we get:
216 = (28 ×
7) + 20 ….. (2)
By substituting (2) in (1), we get:
224 = ((28 ×
7) + 20) + 8
By simplifying, we get:
224-20 = (28 ×
7) + 8
204 = 28×
7 + 8 ….. (3)
Now we need to write 224 in terms of 32.
Again by Euclid’s division lemma, we get:
224 = (32 ×
7) + 0
We need the remainder to be 12.
By converting remainder to 12, we get:
224 = 212 + 12 ….. (4)
Now we need to write 212 in terms of 32,
Again by Euclid’s division lemma, we get:
212 = (32 ×
6) + 20 ….. (5)
By substituting (5) in (4), we get:
224 = ((32×
6) + 20) + 12
By simplifying, we get:
224-20 = (32 ×
6) + 12
204 = 32 ×
6 + 12 ….. (6)
By equations (3) and (6):
204 = 28 ×
7 + 8 = 32 ×
6 + 12 …..(7)
Now we can see that the number 204 when divided by 28 and 32 leaves remainder 8 and 12.
∴
The smallest number which when divided by 28 and 32 leaves remainder 8 and 12 is 204.
204 divided by 28 leaves 8 as remainder and when 204 divided by 32 leaves 12 as remainder.