Find the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively plzzz give answer with full explanation....
Answers
Answer:
Step-by-step explanation: see in explanation i can tell u in an elaborate manner
1.Lets assume n to be smallest number
2. Such that n/28 ,remainder =8 & when n/32 , remainder=12
3. So as per the question we need to calculate the smallest number when divided by 28 and 32 leaves remainder 0.
4. L.c.m of 28 & 32
5. 28= 2^2×7
6. 32= 2^5
7. So LCM = 2^5×7 = 224
8. I think u know this ( dividend = divisor ×quotient + remainder ) .
9. Remainder < divisor
10. 224= 28×8+0
11. => 224 = 224 +0
12.=> 224 = 216 + 8 ( because we want r= 8)
13. => 224 = (28*7+20)+8 ( since divisor given is 28)
14. => 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 ( since divisor given is 32)
=> 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
●ANS● 204
Justification: 204 divided by 28, remainder = 8
& 204 divided by 32, remainder = 12
9.