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zhannawk [14.2K]
3 years ago
6

How many 4 digit multiples can not be divided by 5

Mathematics
2 answers:
Advocard [28]3 years ago
8 0

Answer:

1800

Step-by-step explanation:

Let us solve this question by Modular Airthmatic.

Let the lowest 4 digit number be X.

X= 1 (mod 5)

X =2 (mod 5)

X=3 (mod 5)

X =4 (mod 5)

X=1+5 t where t is any integer.

1+5 t=2 (mod 5)

5 t=1 (mod 5 )

5 is the Modular Inverse of modulo 5

t=1 (mod 5)

t =1+5 u : u= any integer.

X becomes 1+ 5 (1+5u )=6+25 u

6+25u =3 (mod 5)

25 u= -3 (mod 5)=2 (mod 5)

25 is Modular Inverse of Modulo 5

u=2 (mod 5)

u =2+5 v : v= any integer.

X becomes 6 +25 (2+5 v)=56+125 v .

56 +125 v=4 (mod 5)

125 v=-52 (mod 5)=3 (mod 5)

125 is Modular Inverse of modulo 5

v = 3 (mod 5)

v =3 +5 w where w is any integer.

For lowest 4 digit number, take

w=200

X=1003 : Lowest 4 digit number.

Take w=201

X=208 : Next number.

For 4 digit largest number, take

w=1995

X= 9978 : Largest number.

No. of Numbers =(1995 - 200 )+1=1800 □Ans.

Anna [14]3 years ago
5 0

Answer:

1800 ways.

Step-by-step explanation:

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a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

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