Please help I have to turn this in by the end of the night
2 answers:
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Answer:
Step-by-step explanation:
Solution :
Here, 12 - 8 = 4
15 - 11 = 4
18 - 14 = 4
27 - 23 = 4
Thus, every divisor is greater than its remainder by 4. So, the required smallest number is the difference of the L.C.M of the given number and 4
<u>Finding </u><u>the</u><u> </u><u>L.C</u><u>.</u><u>M</u>
First of find the prime factors of each numbers
12 = 2 × 2 × 3
15 = 3 × 5
18 = 3 × 3 × 3
27 = 3 × 3 × 3
Take out the common prime factors : 3 , 3 and 3
Also take out the other remaining prime factors : 2 , 2 and 5
Now, Multiply those all prime factors and obtain L.C.M
L.C.M = Common factors × Remaining factors
= 3 × 3 × 3 × 2 × 2 × 5
= 540
L.C.M of 12 , 15 , 18 and 27 = 540
So, The required smallest number = 540 - 4
= 536
Hope I helped!
Best regards!!