find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively
Anonymous:
___k off
Answers
Answered by
2
Answer:
Step-by-step explanation:
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
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
Similar questions