Question

Question 1 ❓

Count the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digits are repeated.

Question 2 ⭐


An integer is chosen at random from the first ten positive integers. Find the probability that it is multiple of three.​

Answer 1

$\Large\textrm{Explanation (Q.1)}$

To determine the number is a multiple of 5, we expect the last digit to be 0 or 5.

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The first digit is 6.

The last digit is 0 or 5.

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The number of the case for the last digit is $\rm2$.

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Case I. Let's consider a case where the last digit is 0.

The number of cases for two digits in the middle can't be 0, 1, or 6.

$\rm1\times8\times7\times1=56$$\;$

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Case II. Lastly, we consider the number of cases where the last digit is 5.

The number of cases for two digits in the middle can't be 1, 5, or 6.

$\rm1\times8\times7\times1=56$

$\;$

Since they are independent cases, taking the sum,

$\rm56+56=112$

The result.

$\cdots\longrightarrow\boxed{\textrm{Such 112 numbers exist.}}$

$\;$

$\Large\textrm{Explanation (Q.2)}$

From 1 to 10, the multiples of 3 are 3, 6, and 9.

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From 1 to 10, there are 10 natural numbers.

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By formula of probability,

$\boxed{\textrm{(Probability)}=\dfrac{\textrm{(Desired outcomes)}}{\textrm{(Total number of outcomes)}}}$

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The result.

$\textrm{ \cdots\longrightarrow\boxed{\textrm{The probability is \rm\dfrac{3}{10} .}} }$

Answer 2

Clearly, thousands digit is 6.

Number of numbers with units digit 0=(1×8×7×1)=56

Number of numbers with units digit 5=(1×8×7×1)=56

Required number of numbers=(56+56)=112.