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OleMash [197]
3 years ago
12

Is 4320 perfect, abundant, or deficient? also perfect numbers? Explain why all positive multiples of 6 greater than 6 are aburnd

ant numbers.
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
6 0

Answer:

The number 4320 is an abundant number.

Multiples of 6 is a perfect number.

Step-by-step explanation:

Given : Number 4320

To find : Is 4320 is perfect, abundant, or deficient?

Solution :

1) A perfect number is one whose factors are equal to a given number i.e. P(n)=n

2) An abundant number is a composite number whose factors, without the number itself, have a sum greater than the number i.e. P(n) >n

3) A deficient number is a composite number in which the sum of its  factors is less than the given number i.e. P(n) < n

Number - 4320

Factors are

1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,27,30,32,36,40,45,48,54,60,72,80,90,96,108,120,135,144,160,180,216,240,270,288,360,432,480,540,720,864,1080,1440,2160,4320.

If we disregard 4320 as a factor,  

Then the sum is

1+2+3+4+5+6+8+9+10+12+15+16+18+20+24+27+30+32+36+40+45+48+54+60+72+80+90+96+108+120+135+144+160+180+216+240+270+288+360+432+480+540+720+864+1080+1440+2160=10800

Now, 10800 >4320

Therefore, The number 4320 is an abundant number.

If we take positive multiples of 6,

Factors of 6, 1,2,3 and 6

The sum of factors except 6,

1+2+3=6

Which means Multiples of 6 is a perfect number.

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Given $dc = 7$, $cb = 8$, $ab = \frac{1}{4}ad$, and $ed = \frac{4}{5}ad$, find $fc$. Express your answer as a decimal.
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A ↔ B ↔ C ↔ D ↔ E ↔ F

 \frac{1}{4} AD      8      7      \frac{4}{5} AD   ???

AB + BC + CD = AD   <em>segment addition postulate</em>

\frac{1}{4} AD   +   8   +   7   = AD

\frac{1}{4} AD   +       15        = AD

AD         + 60    = 4AD

                  60    = 3AD

                  20    = AD

AB =  \frac{1}{4} AD    =  \frac{1}{4}(20)    = 5

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Answer: 11 + EF

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