Math Problem Statement

Determinați valoarea cifrei x, știind că numărul 50x6 se divide cu 3, dar nu și cu 9. Aflați suma divizorilor naturali ai numărului 24. Câți multipli ai numărului 15 sunt mai mari decât 300, dar mai mici decât 1000?

Solution

Let me help you with some of the problems I can read from the image:

  1. Problem 19: "Determinați valoarea cifrei x, știind că numărul 50×6 se divide cu 3, dar nu și cu 9."

This is a divisibility rule problem. For a number to be divisible by 3, the sum of its digits must be divisible by 3. For it not to be divisible by 9, the sum of its digits should not be divisible by 9.

Given number: 50x650x6

The sum of the digits is 5+0+x+6=11+x5 + 0 + x + 6 = 11 + x.

  • For divisibility by 3: 11+x11 + x must be divisible by 3.
  • For not divisible by 9: 11+x11 + x must not be divisible by 9.

Let’s solve this:

  • 11+x0 (mod 3)11 + x \equiv 0 \ (\text{mod} \ 3) means x1 (mod 3)x \equiv 1 \ (\text{mod} \ 3). So, x=1x = 1 or x=4x = 4 or x=7x = 7.

  • Now check divisibility by 9 for 11+x11 + x. We need to ensure 11+x11 + x is not divisible by 9.

For:

  • x=1x = 1: 11+1=1211 + 1 = 12, which is divisible by 3 but not by 9.
  • x=4x = 4: 11+4=1511 + 4 = 15, which is divisible by 3 but not by 9.
  • x=7x = 7: 11+7=1811 + 7 = 18, which is divisible by 9.

Therefore, the possible values of xx are 1 and 4.


  1. Problem 21: "Aflați suma divizorilor naturali ai numărului 24."

The divisors of 24 are: 1,2,3,4,6,8,12,241, 2, 3, 4, 6, 8, 12, 24.

The sum is: 1+2+3+4+6+8+12+24=60.1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60.

So, the sum of the divisors is 60.


  1. Problem 22: "Câți multipli ai numărului 15 sunt mai mari decât 300, dar mai mici decât 1000?"

We need to find the multiples of 15 in the range (300,1000)(300, 1000).

  • First multiple of 15 greater than 300: 15×21=31515 \times 21 = 315.
  • Largest multiple of 15 less than 1000: 15×66=99015 \times 66 = 990.

Now, the multiples are: 315,330,345,,990315, 330, 345, \dots, 990.

This forms an arithmetic progression with the first term a=315a = 315, common difference d=15d = 15, and last term l=990l = 990.

The number of terms is: n=lad+1=99031515+1=46.n = \frac{l - a}{d} + 1 = \frac{990 - 315}{15} + 1 = 46.

So, there are 46 multiples of 15 between 300 and 1000.


Would you like me to provide details on any of these problems or further explain any concepts?

Here are 5 related questions based on these problems:

  1. How do we determine divisibility by numbers other than 3 and 9?
  2. What is the general method to find the sum of divisors for a number like 36 or 48?
  3. How can you determine the greatest common divisor of two numbers?
  4. How would you solve a similar problem to find multiples of a number like 25 between certain limits?
  5. What is the rule to determine if a number is divisible by both 6 and 9?

Tip: When checking divisibility by 3 or 9, always sum the digits of the number for a quick check!

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Math Problem Analysis

Mathematical Concepts

Divisibility Rules
Divisors
Arithmetic Progressions

Formulas

Divisibility by 3 and 9 rules
Sum of divisors of a number
Formula for arithmetic progression

Theorems

Divisibility Theorems

Suitable Grade Level

Grades 5-8