All categories

3 years ago

What is the smallest 6 digit palindrome that is divisible by 99

Mathematics

2 answers:

Scorpion4ik [409]3 years ago

4 0

Answer:

108801

Step-by-step explanation:

6-digit palindrome is the number n the form of:

xyzzyx

This is divisible by 11 by default as the sum of the digits in odd placed is same as sum of the number in even places (remember the divisibility rule by 11):

x + z + y = y + z + x

Now, in order to be divisible by 99, the number must be divisible by 11 and 9.

According to divisibility rule by 9 the sum of all digits must be divisible by 9. You can see In our case we need to have (the minimum):

x + y + z = 9

The smallest number we could get is when x is minimum, y is minimum, so:

x = 1, y = 0, then y = 8

The number we get is:

108801

Proof:

108801/99 = 1099

mezya [45]3 years ago

4 0

Hello,

Let's assume n the palindrome

$n=\overline{abccba}\\\\Since\ 99=9*11:\\\\the \ smallest\ \Longrightarrow \ a=1\\\\a+b+c+c+b+a=9*k\ ,\ k\in \mathbb{N}\\2*(a+b+c)=9*k\ \Longrightarrow \ k\in 2\mathbb{N} , \ : k=2\\\\1+b+c=9 \Longrightarrow \ b=0\ and\ c=8\\\\n=108801\\\\Proof:\\108801=9*12089 =11*9891\\Nota\ bene:(a+c+b)-(b+c+a)=11*p \Longrightarrow \ 0=11*p \Longrightarrow \ p=0$

You might be interested in

A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro

-Dominant- [34]

Answer:

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

The degrees of freedom are given by:

$df =n-1= 32-1=31$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{31}>3.07)=0.0044$

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ( $\alpha=0.01,0.05, 0.1, 0.15$ ).

Step-by-step explanation:

Data given and notation

$\bar X=1.6$ represent the sample mean

$s=0.46$ represent the sample deviation

$n=32$ sample size

$\mu_o =1.35$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:

Null hypothesis: $\mu =1.35$

Alternative hypothesis: $\mu \neq 1.35$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

P-value

The degrees of freedom are given by:

$df =n-1= 32-1=31$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{31}>3.07)=0.0044$

5 0

3 years ago

The matrix method is being used to solve the system of equations x + 2y = 4 and 3x - y = 6.

Ira Lisetskai [31]

Answer: b) multiply row 1 by -3 and add the result to row 2

Step-by-step explanation:

8 0

3 years ago

Which of the following is NOT a linear factor of the polynomial function?

Natalija [7]

Answer:

Among the four choices, $(x + 5)$ is the only one that is not a linear factor of this polynomial function.

Step-by-step explanation:

Let $a$ denote some constant. A linear factor of the form $(x - a)$ is a factor of a polynomial $f(x)$ if and only if $f(a) = 0$ (that is: replacing all $x$ in the polynomial $f(x) \!$ with the constant $a\!$ would give this polynomial a value of $0$ .)

For example, in the second linear factor $(x - 2)$ , the value of the constant is $a = 2$ . Verify that the value of $f(2)$ is indeed $0$ . (In other words, replacing all $x$ in the polynomial $f(x) \!$ with the constant $2$ should give this polynomial a value of $0\!$ .)