1. Find all multiples of $4$ that lie between ... 2. I am a number less than $10$ . Two of my factors is $1$ and the sum of my digits is $8$ . Who am I? 3. I am a two-digit number less than $100$ . Two of factors are $2$ and $5$ . One of my digits is $7$ more than the other. 4. A number for which the sum of its factors is equal to twice the number is called a perfect number. The number $28$ is a perfect number. Its factors are $1, 2, 4, 7, 14$ and $28$ . Their sum is $56$ , which is twice $28$ . Find a perfect number $\le 10$ and $2$ . Ans: Factor of $6 = 1, 2, 3, 6$ Sum of factors $= 1 + 2 + 3 + 6 \rightarrow 12$ So, number is $\rightarrow 6$ . 5. ... $35$ and $50$ ...

Answer: 70

Step-by-step solution

Step 1: Determine the units digit using divisibility

Since the number has both $2$ and $5$ as factors, it must be a multiple of their least common multiple, which is $10$ . Any positive integer divisible by $10$ must end in $0$ , so the units digit is $0$ .

Step 2: Find the tens digit using the difference condition

The problem states that one of the digits is $7$ more than the other. Since the units digit is $0$ , the tens digit cannot be $0 - 7$ because digits must be non-negative. Therefore, the tens digit must be $0 + 7 = 7$ .

Step 3: Verify the two-digit number

With a tens digit of $7$ and a units digit of $0$ , the number is $70$ . We verify that $70$ is a two-digit number less than $100$ , and has both $2$ and $5$ as factors ( $70 = 2 \times 5 \times 7$ ).

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