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# Write the following numbers in generalised form:$39,52,106,359,628,3458,9502,7000$  Verified
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Hint: We have given two-digit, three-digit and four-digit numbers. It can be represented in a generalised form; if it is expressed as the sum of the product of its digits with their respective place values. So, a two digit number having $a$ and $b$ as its digits at tens and the ones places respectively is written in the form as $10a + b$, where $a$ can be any of the digits from $1$ to $9$ and $b$ can be any of the digits from $0$ to $9$.

We can solve this question with three simple steps, which are given below:
Step 1: $39$ is a two-digit number where digits $9$ on unit place and $3$ on tens place. So, it can be written in generalised form i.e., $39 = 10 \times 3 + 9$.
Similarly, $52$ is two-digit number, so it can be written in generalised form i.e. $52 = 10 \times 5 + 2$

Step 2: $106$ is a three-digit number. So, it can be written in generalised form i.e.$106 = 100 \times 1 + 10 \times 0 + 6$.
$359$ is a three-digit number. So, it can be written in generalised form i.e., $359 = 100 \times 3 + 10 \times 5 + 9$
$628$ is a three-digit number. So, it can be written in generalised form i.e., $628 = 100 \times 6 + 10 \times 2 + 8$

Step 3: $3458$ is a four-digit number. So, $3458 = 1000 \times 3 + 100 \times 4 + 10 \times 5 + 8$ is a generalised form.
Similarly, $9502 = 1000 \times 9 + 100 \times 5 + 10 \times 0 + 2$
and $7000 = 1000 \times 7 + 100 \times 0 + 10 \times 0 + 0$ is a generalised form.

Note: Playing with numbers using any of the digits from $0$ to $9$ in generalised form of number. In four-digit number, we can write as $1000a + 100b + 10c + d$, where $a$ be any number from $1$ to $9$ while $b,c$ and $d$ be any number from $0$ to $9$.
Similarly, a three-digit number can be written as $100a + 10b + c$ in generalised form where $a$ be any number from $1$ to $9$ while $b$ and $c$ be any number from $0$ to $9$.