Often, when solving problems, you need to find out if a given number is divisible by a given digit without a remainder. But each time it takes a very long time to share it. In addition, there is a high probability of making a mistake in the calculations and getting away from the correct answer. In order to avoid this problem, signs of divisibility into basic prime or single-digit numbers were found: 2, 3, 9, 11. But what if you need to divide by another, larger number? For example, how to calculate the sign of divisibility by 15? We will try to find the answer to this question in this article.
How to formulate the test for divisibility by 15?
If the signs of divisibility are well known for prime numbers, then what to do with the rest?
If the number is not prime, then it can be factored. For example, 33 is the product of 3 and 11, and 45 is 9 and 5. There is a property according to which a number is divisible by a given number withoutremainder if it can be divided by both factors. This means that any large number can be represented in the form of primes, and based on them, we can formulate the sign of divisibility.
So, we need to find out if this number can be divided by 15. To do this, let's look at it in more detail. The number 15 can be represented as a product of 3 and 5. This means that in order for a number to be divisible by 15, it must be a multiple of both 3 and 5. This is the sign of divisibility by 15. In the future, we will consider it in more detail and formulate it more precisely.
How do you know if a number is divisible by 3?
Recall the test for divisibility by 3.
A number is divisible by 3 if the sum of its digits (the number of ones, tens, hundreds, and so on) is divisible by 3.
So, for example, you need to find out which of these numbers can be divided by 3 without a remainder: 76348, 24606, 1128904, 540813.
Of course, you can just split these numbers into a column, but that will take a lot of time. Therefore, we will use the criterion of divisibility by 3.
- 7 + 6 + 3 + 4 + 8=28. The number 28 is not divisible by 3, so 76348 is not divisible by 3.
- 2 + 4 + 6 + 0 + 6=18. The number 18 can be divided by 3, which means that this number is also divisible by 3 without a remainder. Indeed, 24 606: 3=8 202.
Analyze the rest of the numbers in the same way:
- 1 + 1 + 2 + 8 + 9 + 4=25. The number 25 is not divisible by 3. So 1,128,904 is not divisible by 3.
- 5 + 4 + 0 + 8 + 1 + 3=21. The number 21 is divisible by 3, which means that 540,813 is divisible by 3. (540,813: 3=180271)
Answer: 24 606 and 540 813.
When is a number divisible by 5?
However, the sign that a number is divisible by 15 also includes not only divisibility by 3, but also a multiplicity of five.
The sign of divisibility by 5 is as follows: a number is divisible by 5 if it ends in 5 or 0.
For example, you need to find multiples of 5: 11 467, 909, 670, 840 435, 67 900
The numbers 11467 and 909 are not divisible by 5.
The numbers 670, 840 435 and 67 900 end in 0 or 5, which means they are multiples of 5.
Examples with solution
So, now we can fully formulate the sign of divisibility by 15: a number is divisible by 15 when the sum of its digits is a multiple of 3, and the last digit is either 5 or 0. It is important to note that both of these conditions must be met simultaneously. Otherwise, we will get a number that is not a multiple of 15, but only 3 or 5.
The sign of divisibility of numbers by 15 is very often needed to solve control and examination tasks. For example, often in the basic level of the exam in mathematics there are tasks based on an understanding of this particular topic. Consider some of their solutions in practice.
Task 1.
Among the numbers, find those that are divisible by 15.
9 085 475; 78 545; 531; 12,000; 90 952
So, to begin with, we will discard those numbers that obviously do not meet our criteria. These are 531 and 90,952. Despite the fact that the sum 5+3+1=9 is divisible by 3, the number ends in one, which means it doesn't fit. The same goes for 90952, whichends in 2.
9 085 475, 78 545 and 12 000 satisfy the first criterion, now let's check them against the second one.
9+0+8+5+4+7+5=38, 38 is not divisible by 3. So this number is extra in our series.
7+8+5+4+5=29. 29 is not a multiple of 3, does not meet the conditions.
But 1+2=3, 3 is evenly divisible by 3, which means that this number is the answer.
Answer: 12,000
Task 2.
Three-digit number C is greater than 700 and divisible by 15. Write down the smallest such number.
So, according to the criterion of divisibility by 15, this number should end in 5 or 0. Since we need the smallest possible, take 0 - this will be the last digit.
Since the number is greater than 700, the first number can be 7 or greater. Keeping in mind that we should find the smallest value, we choose 7.
For a number to be divisible by 15, the condition 7+x+0=a multiple of 3, where x is the number of tens.
So, 7+x+0=9
X=9 -7
X=2
The number 720 is what you are looking for.
Answer: 720
Problem 3.
Delete any three digits from 3426578 so that the resulting number is a multiple of 15.
Firstly, the desired number must end with the number 5 or 0. So, the last two digits - 7 and 8 must be crossed out immediately.
34265 left.
3+4+2+6+5=20, 20 is not divisible by 3. The nearest multiple of 3 is 18. To get it, you need to subtract 2. Cross out the number 2.
It turns out 3465. Check your answer, 3465: 15=231.
Answer:3465
In this article, the main signs of divisibility by 15 were considered with examples. This material should help students with solving tasks of this type and similar ones, as well as understand the algorithm for working with them.
