what is the smallest no. which leaves remender 8 and 12 when divided by 28 and 32
Answers
it is not possible that any no. has a divisor giving different remainders
Answer:
the required number is 204
Step-by-step explanation:
method: 1
First we calculate the smallest number ,which when divided by 28 & 32 leaves remainder 0.
Which is the LCM of 28, & 32 =
28= 2² x 7
32 = 2^5
So, LCM = 2^5 * 7 = 224
Now, by Euclid's division lemma ,
dividend = divisor*quotient + remainder (r<divisor)
224 = 28 *8 +0
=> 224 = 224 +0
=> 224 = 216 + 8 ( because we want r= 8)
=> 224 = (28*7+20)+8
=> 224–20 = 28*7 +8 ................(1)
Similarly, 224 = 32 *7 +0
=> 224 = 224 +0
=> 224 = 212 +12 ( because we want r= 12)
=> 224 = (32*6+20) +12
=> 224 - 20 = 32*6 +12 ....................(2)
Now, we observe that LHS of both eq (1) & eq(2) are equal. Divisors are also 28 & 32 & remainders are also the required numbers 8 & 12
So the smallest dividend = 224 - 20 = 204
method: 2
Given that the smallest number when divided by 28 and 32 leaves remainder 8 and 12 respectively.
28 - 8 = 20 and 32 - 12 = 20 are divisible by the required numbers.
Therefore the required number will be 20 less than the LCM of 28 and 32.
Prime factorization of 28 = 2 * 2 * 7
Prime factorization of 32 = 2 * 2 * 2 * 2 * 2
LCM(28,32) = 2 * 2 * 2 * 2 * 2 * 7
= 224.
Therefore the required smallest number = 224 - 20
= 204.
Verification:
204/28 = 28 * 7 = 196.
= 204 - 196
= 8
204/32 = 32 * 6 = 192
= 204 - 192
= 12.
Hope this helps