I am alone, but still I am. I can't do everything, but I can still do something. And I won't refuse to do the little I can (c)
Number and amount natural divisors natural number
Fundamental theorem of arithmetic. Every natural number n > 1 is either simple or can be represented, and in a unique way - up to the order of the factors, as a product of prime numbers (we can assume that any natural number greater than 1 can be represented as a product of prime numbers , if we assume that this product can contain only one factor).
Among the simple factors present in the expansion `n = p1*p2*...*pk`, there may be identical ones. For example, `24=2*2*2*3`. They can be combined using the exponentiation operation. In addition, prime factors can be ordered by magnitude. The result is a decomposition
`n = p_1^(alpha_1)*p_2^(alpha_2)*.......*p_k^(alpha_k)`, where `alpha_1, alpha_2, ......, alpha_k in NN`
(1)
This representation of a number is called its canonical decomposition into prime factors. For example, canonical representation the number 2 520 has the form 2 520 = 2 3 3 2 5 7.
From canonical expansion numbers, you can easily derive the following lemma: If n has the form (1), then all divisors of this number have the form:
`d = p_1^(beta_1)*p_2^(beta_2)*......*p_k^(beta^k)`, where `0<= beta_m <= alpha_m` (`m = 1,2,..., k`)
(2)
In fact, it is obvious that every d of the form (2) divides a. Conversely, let d divide a, then a=cd, where c is some natural number and, therefore, all prime divisors of the number d are included in the canonical expansion of the number a with exponents not exceeding the corresponding exponents of the number a.
Consider two functions defined on the set of natural numbers:
a) τ(n) - the number of all natural divisors of n;
2) σ(n) the sum of all natural divisors of the number n.
Let n have canonical expansion (1). Let us derive formulas for a number and the sum of its natural divisors.
Theorem 1. Number of natural divisors of n
`tau(n) = (alpha_1 + 1)*(alpha_2 + 1)*.....*(alpha_k + 1);`
(3)
Proof.
Example. The number 2 520 = 2 3 3 2 5 7. has (3+1)(2+1)(1+1)(1+1) = 48 divisors.
Theorem 2. Let n have canonical expansion (1). Then the sum of the natural divisors of n is equal to
`sigma(n) = (1 + p_1 + p_1^2 + ..... + p_1^(alpha_1))*(1 + p_2 + p_2^2 + ..... + p_2^(alpha_2))* ........* (1 + p_k + p_k^2 + .....+ p_k^(alpha_k));`
(4)
Proof.
Example. Find the sum of all divisors of the number 90.
90=2 3 2 5. Then σ(90)=[(2 2 -1)/(2-1)] [3 3 -1)/(3-1)] [(5 2 -1)/(5 -1)]=234
Formula (4) can help find all the divisors of a number. So, for example, to find all the divisors of the number 90, we open the brackets in the following product (without performing the addition operation): (1+2)(1+3+3 2)(1+ 5)=(1+1*3+1*Z 2 +1*2+2*3+2*Z 2)(1+5) = 1+3+Z 2 +2+2*3+2*Z 2 + 5+3*5+Z 2 *5+2*5+2*3*5+2*Z 2 *5 = 1+3+9+2+6+18+5+15+45+10+ 30+90 - the terms are the divisors of the number 90.
Let's solve several problems on the topic "Number and sum of natural divisors of a natural number"
Exercise 1. Find a natural number, knowing that it has only two prime factors, that the number of all factors is 6, and the sum of all factors is 28.
Assignments from the collection TTZ - Unified State Exam 2010. Mathematics. Typical test tasks
Task 2. TTZ.С6.2 Find all natural numbers that are divisible by 42 and have exactly 42 different natural divisors (including one and the number itself).
Task 3. TTZ.С6.9 Find all natural numbers whose last decimal digit is 0 and which have exactly 15 different natural factors (including one and the number itself).
Task 4. SPI.С6.9. The natural number n has exactly 6 divisors. The sum of these divisors is 3500. Find n.
VEk solution:
Tasks for independent work
SR1. Find all numbers that have exactly 2 prime factors, a total of 8 factors whose sum is 60.
SR2. Find natural numbers that are divisible by 3 and 4 and have exactly 21 natural factors.
SR3. Find the smallest natural number that has exactly 18 natural divisors.
SR4. Find the smallest number that is a multiple of 5 and has 18 natural factors.
SR5. Some natural number has two prime factors. Its square has only 15 divisors. How many divisors does the cube of this number have?
SR6. Some natural number has two prime factors. Its square has only 81 divisors. How many divisors does the cube of this number have?
SR7. Find a number of the form m = 2 x 3 y 5 z, knowing that half of it has 30 divisors less, a third has 35 divisors, and a fifth has 42 divisors less than the number itself.
Instructions
Most often, you need to factor a number into prime factors. These are numbers that divide the original number without a remainder, and at the same time can themselves be divided without a remainder only by themselves and one (such numbers as 2, 3, 5, 7, 11, 13, 17, etc.). Moreover, no pattern was found in the series. Take them from a special table or find them using an algorithm called the “sieve of Eratosthenes”.
Numbers that have more than two divisors are called composite numbers. What kind numbers can they be compound?
Because numbers are divisible by 2, then all are even numbers, except numbers 2 will be composite. Indeed, in division 2:2, two is divided by itself, that is, it has only two divisors (1 and 2) and is a prime number.
Let's see if the even one has numbers any other way dividers. Let us first divide it by 2. From the commutative nature of the multiplication operation, it is obvious that the resulting quotient will also be a divisor numbers. Then, if the resulting quotient is an integer, we divide this quotient by 2 again. Then the resulting new quotient y = (x:2):2 = x:4 will also be a divisor of the original numbers. Likewise, 4 will be a divisor of the original numbers.
Continuing this chain, let us generalize the rule: we sequentially divide first and then the resulting quotients by 2 until the quotient becomes equal to an odd number. In this case, all the resulting quotients will be divisors of this numbers. In addition, the divisors of this numbers there will be numbers 2^k where k = 1...n, where n is the number of steps in this chain. Example: 24:2 = 12, 12:2 = 6, 6:2 = 3 is an odd number. Therefore, 12, 6 and 3 are dividers numbers 24. There are 3 steps in this chain, therefore, divisors numbers 24 will also numbers 2^1 = 2 (already known from parity numbers 24), 2^2 = 4 and 2^3 = 8. Thus, numbers 1, 2, 3, 4, 6, 8, 12 and 24 will be divisors numbers 24.
However, not for all even numbers this can give everything dividers numbers. Consider, for example, the number 42. 42:2 = 21. However, as is known, numbers 3, 6 and 7 will also be divisors numbers 42.
There are divisibility into numbers. Let's consider the most important of them:
Test of divisibility by 3: when the sum of the digits numbers divisible by 3 without remainder.
Divisibility test by 5: when the last digit numbers 5 or 0.
Test for divisibility by 7: when the result of subtracting twice the last digit from this numbers Without the last digit it is divisible by 7.
Divisibility test by 9: when the sum of the digits numbers divisible by 9 without a remainder.
Test of divisibility by 11: when the sum of the digits occupying odd places is either equal to the sum of the digits occupying even places, or from it by a number divisible by 11.
There are also signs of divisibility by 13, 17, 19, 23 and others numbers.
For both even and odd numbers, you need to use signs of division by a particular number. By dividing the number, you should determine dividers the resulting quotient, etc. (the chain is similar to the chain of even numbers when dividing them by 2, described above).
Sources:
- Signs of divisibility
Of the four basic mathematical operations, the most resource-intensive operation is division. It can be done manually (in a column), on calculators of various designs, and also using a slide rule.
Instructions
To divide one number by another using a column, write down the dividend first, then the divisor. Place a vertical line between them. Draw a horizontal line under the divider. Consistently, as if removing the low-order digits, you will get a number that is greater than the divisor. Sequentially multiplying the numbers from 0 to 9 by the divisor, find the greatest of numbers, less than that obtained at the previous stage. Write this figure as the first digit of the quotient. Write the result of multiplying this figure by the divisor under the dividend with a shift of one place to the right. Perform the subtraction, and with its result, carry out the same actions until you find all the digits of the quotient. Determine the location of the comma by subtracting the order of the divisor from the order of the dividend.
If the numbers are not divisible by each other, two situations are possible. In the first of them, one digit or a combination of several digits will be repeated endlessly. Then there is no point in continuing the calculation - it is enough to take this number or a chain of numbers in a period. In the second situation, any regularity in the particular will not be possible. Then stop dividing, having achieved the desired accuracy of the result, and round the last one.
To divide one number by another using an arithmetic calculator (both basic and engineering), press the reset button, enter the dividend, press the division button, enter the divisor, and then press the equal sign button. On a calculator with formula notation, divide in the same way, taking into account that the key with the equal sign can be, for example, Enter or Exe. Modern devices of this type are two-line: typed on the top line, and the result is displayed in larger numbers on the bottom. Using the Ans key, this result can be used in the next calculation. In all cases, the result is automatically rounded within the calculator's digit grid.
On a calculator with reverse Polish notation, first press the reset button, then enter the dividend and press the Enter key (instead of this inscription there may be an upward arrow). The number will end up in the stack cell. Now enter the divisor and press the division key. The number from the stack will be divided by the number that was previously displayed on the indicator.
Use a slide rule in cases where little accuracy is required. Remove from both numbers, and then take the two most significant digits from each of them. On scale A, find the divisor, and then match it with the dividend on scale B. Then find the unit on the latter - right above it on scale A will be located private. Determine the location of the comma in it in the same way as with a column.
Sources:
- Column division order
- private numbers are
Schoolchildren often come across the following formulation among mathematics assignments: “find the least common multiple of numbers.” You definitely need to learn how to do this in order to perform various operations with fractions with unequal denominators.
Finding Least Common Multiple: Basic Concepts
To understand how to calculate the LCM, you must first determine the meaning of the term “multiple”.
A multiple of A is a natural number that is divisible by A without a remainder. Thus, numbers that are multiples of 5 can be considered 15, 20, 25, and so on.
There can be a limited number of divisors of a particular number, but there are an infinite number of multiples.
A common multiple of natural numbers is a number that is divisible by them without leaving a remainder.
The least common multiple (LCM) of numbers (two, three or more) is the smallest natural number that is divisible by all these numbers.
To find the LOC, you can use several methods.
For small numbers, it is convenient to write down all the multiples of these numbers on a line until you find something common among them. Multiples are denoted by the capital letter K.
For example, multiples of 4 can be written like this:
K (4) = (8,12, 16, 20, 24, ...)
K (6) = (12, 18, 24, ...)
Thus, you can see that the least common multiple of the numbers 4 and 6 is the number 24. This notation is done as follows:
LCM(4, 6) = 24
Greatest total divider- this is the maximum number by which each of the proposed numbers can be divided. This term is often used to reduce complex fractions where both the numerator and denominator must be divided by the same number. Sometimes it is possible to determine the greatest common divider by eye, but in most cases, in order to find it you will need to carry out a series of mathematical operations.
You will need
- To do this you will need a piece of paper or a calculator.
Instructions
Break down each complex number into a product of primes or factors. For example, 60 and 80, where 60 is equal to 2*2*3*5, and 80 is 2*2*2*2*5, this can be written more simply using . IN in this case will look like two in the second multiplied by five and three, and the second is the product of two in the fourth and five.
Now write down the common numbers for both. In our version it is two and five. However, in other cases this number can be one, two or three digits or even . Next you need to work. Choose the smallest one for each multiplier. In the example it is two to the second power and five to the first.
Finally, you just need to multiply the resulting numbers. In our case, everything is extremely simple: two in , multiplied by five, is equal to 20. Thus, the number 20 can be called the greatest common divisor for 60 and 80.
Video on the topic
note
Remember that a prime factor is a number that has only 2 divisors: one and the number itself.
Helpful advice
In addition to this method, you can also use the Euclidean algorithm. Its full description, presented in geometric form, can be found in Euclid's book "Elements".
Related article
You can often find equations in which . For example, 350: X = 50, where 350 is the dividend, X is the divisor, and 50 is the quotient. To solve these examples, it is necessary to perform a certain set of actions with the numbers that are known.
You will need
- - pencil or pen;
- - a sheet of paper or notebook.
Instructions
Make up a simple equation where the unknown, i.e. X is the number of children, 5 is the number of sweets each child received, and 30 is the number of sweets that were purchased. Thus you should get: 30: X = 5. In this mathematical expression, 30 is called the dividend, X is the divisor, and the resulting quotient is 5.
Now start solving. It is known: to find a divisor, you need to divide the dividend by the quotient. It turns out: X = 30: 5; 30: 5 = 6; X = 6.
Check by substituting the resulting number into the equation. So, 30: X = 5, you have found the unknown divisor, i.e. X = 6, thus: 30: 6 = 5. The expression is correct, and from this it follows that the equation is solved. Of course, when solving examples that involve prime numbers, the check is not necessary. But when equations from , three-digit, four-digit, etc. numbers, be sure to check yourself. After all, it does not take much time, but gives absolute confidence in the result obtained.
note
ARITHMETIC
Arithmetic is the queen of mathematics, and everyone will find suitable problems here - from a first-grader to an academician.
Wonderful numbers
Let us call a natural number “remarkable” if it is the smallest among all natural numbers with the same sum of digits. For example, the number 1 is remarkable because it is the smallest of the numbers 1, 10, 100, 1000, and so on. 1 is the first remarkable number. Find the second remarkable number. Describe all numbers whose sum of digits is the same. The same for the third, tenth, wonderful number of 2010.
Find the largest two-digit remarkable number. What's his number?
Rectangles with a given area
On checkered paper, draw all the rectangles whose area is 24 squares. (The sides should follow the boundaries of the cells.) How many such rectangles will there be?
For which areas is there only one rectangle? Which ones require two different rectangles? Three different rectangles? How does the number of options depend on the area?
Of all the rectangles with the same area, find the one with the smallest perimeter.
Number expansion
The number 15 can be represented in three ways as the sum of consecutive natural numbers: 15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5. How many such ways are there for the number 115? How to find the number of ways for an arbitrary number?
Supercomputer
A supercomputer can perform only one operation - the operation of mixing two numbers: from the numbers m, n, the computer obtains the number (m+n) /2. If m+n is odd, then the computer freezes. All received numbers are stored in memory. Let us be given three numbers, one of which is zero, and the other two are natural and not equal to each other. For what numbers m and n can one be obtained on a supercomputer?
Diagonals of rectangles
A rectangle measuring 199 x 991 cells was outlined on a sheet of paper. How many nodes (i.e., vertices of cells) does the diagonal pass through? How many cells does the diagonal of this rectangle intersect? Try to give an answer for an arbitrary size rectangle - M x N cells in size.
Note. A diagonal intersects a cell if it goes “inside” that cell, rather than just passing through the top.
Exchange problem
What amounts can be paid in coins of 3 and 5 rubles? Generalization: what numbers are expressed by the combination ax+by, where a and b are given natural numbers, x and y are arbitrary non-negative integers.
7. Skl A bottom squares
Skl A given numbers are numbers whose square ends in the same number. For example:
5 2 =25 ; 6 2 =36 ; 25 2 = 625 .
"Five five- twenty five", "six six- thirty six».
Find as many folding numbers as possible; find a way to find all such numbers.
Finding numbers with a given number of divisors
There is only one number that has exactly one divisor - one. All prime numbers have exactly two divisors. For example, the numbers 4 and 9, which are squares of prime numbers, have exactly three divisors. Do all numbers that have exactly three divisors have this property? What can be the form of a number that has exactly 4 divisors? 5 divisors? For a given natural number N describe all natural numbers having exactly N dividers.
Fraction expansions
, 
, 
, …
For the number 1/7, the expansion into a decimal fraction is periodic and consists of six digits, and for 2/7, 3/7, ..., 6/7 - from the same six digits in a different order (check!). But for the number 1/13 and 2/13 the sets of numbers are different. Explore the expansions of these numbers and numbers of the form 1/p, 2/p, ..., (p-1)/p, for p = 17, 19, 41, 47 and other prime numbers, and figure out what cycles there are.
A variety of integer problems use basic concepts and theorems related to divisibility. Let's list some of them.
Problems with solutions
1. How many natural numbers are there less than 1000 that are not divisible by either 5 or 7?
From 999 numbers less than 1000, we cross out numbers that are multiples of 5: there are 199 of them. Next, we cross out numbers that are multiples of 7: there are 142 of them. But among the numbers that are multiples of 7, there are = 28 numbers that are also multiples of 5; they will be crossed out twice. In total, we must cross out 199+142–28=313 numbers. That leaves 999–313=686.
Answer: 686 numbers.
2. Bus ticket number – a six-digit number. A ticket is called lucky if the sum of the first three digits of the number is equal to the sum of the last three digits. Prove that the sum of all lucky ticket numbers is divisible by 13.
If a lucky ticket has number A, then a ticket with number B = 999999–A is also lucky, while A and B are different. Since A+B=999999=1001·999=13·77·99 is divisible by 13, then the sum of the numbers of all lucky tickets is divisible by 13.
3. Prove that the sum of the squares of three integers cannot leave a remainder of 7 when divided by 8.
Any integer when divided by 8 has a remainder of one of the following eight numbers 0, 1, 2, 3, 4, 5, 6, 7, so the square of an integer has a remainder of one of the three numbers 0, 1, 4 when divided by 8. In order for the sum of the squares of three numbers to have a remainder of 7 when divided by 8, it is necessary that one of two cases be true: either one of the squares or all three have odd remainders when divided by 8.
In the first case, the odd remainder is 1, and the sum of two even remainders is 0, 2, 4, that is, the sum of all remainders is 1, 3, 5. The remainder 7 in this case cannot be obtained. In the second case, three odd remainders are three 1s, and the remainder of the entire sum is 3. So, 7 cannot be a remainder when dividing by 8 the sum of the squares of three integers.
4. Prove that for any natural number n:
a) the number 5 5n+1 + 4 5n+2 + 3 5n is divisible by 11.
b) the number 2 5n+3 + 5 n ·3 n+2 is divisible by 17.
a) Initially, we perform the following transformation of the given expression:
5 5n+1 +4 5n+2 +3 5n = 5(3125) n + 16(1024) n + (243) n = 5(11 284+1) n + 16(11 93+1) n + (11·22+1) n .
Taking into account the Newton binomial of the nth degree, we can write: (x+1) n = Ax+1, where A is some integer for integer x. Then the above expression becomes 11B+5+16+1 = 11C, obviously divisible by 11, where B and C are some integers.
b) Let us carry out the following transformations, from which the statement being proved follows:
2 5n+3 + 5 n 3 n+2 = 8 32 n + 9 15 n = 8(17+15) n + 9 15 n = 17A + 8 15 n + 9 15 n = 17A + 17·15 n = 17V,
where A, B are positive integers.
5. Prove that:
a) if x 2 + y 2 is divisible by 3 and the numbers x and y are integers, then x and y are divisible by 3;
b) if the sum of three integers is divisible by 6, then the sum of the cubes of these numbers is divisible by 6;
c) if p and q are prime numbers and p>3, q>3, then p 2 –q 2 is divisible by 24;
d) if a, b, c are any integers, then there are coprime k and t such that ak+bt is divisible by c.
a) Let x = 3a + r 1, y = 3b + r 2, where r 1 and r 2 are the remainders from division by 3, that is, some of the numbers 0, 1, 2. Then x 2 + y 2 = 3(3a 2 +3b 2 +2аr 1 +2br 2)+(r 1) 2 +(r 2) 2. Since x 2 +y 2 is divided by 3, the first term of the last sum is divided by 3, then (r 1) 2 + (r 2) 2 is divided by 3, which is possible, taking into account the above, only if r 1 = r 2 = 0.
Thus, x = 3a and y = 3b, that is, x and y are divisible by 3, which is what needed to be proven.
b) It is enough to show that x 3 +y 3 +z 3 –(x+y+z) is divisible by 6. This is true, because each of the terms x 3 –x, y 3 –y and z 3 –z is divisible by 6, since a 3 –a=a(a–1)(a+1) – product of three consecutive integers, which is necessarily divisible by 2, 3, and, therefore, 6.
c) The multiplicity of p 2 –q 2 to the number 3 can be proven as follows. When divided by 3, the squares of integers give remainders 0 or 1. Since p and q are prime numbers greater than 3, then p 2 and q 2 when divided by 3 have the same remainders - one. Then p 2 –q 2 is divisible by 3.
On the other hand, p 2 –q 2 =(p+q)(p–q). Since p and q are odd and when divided by 4 they have remainders of 1 or 3, the expression in some brackets is divisible by 4, and in others by 2, and the difference of the squares of p and q is divided by 8.
Since p 2 –q 2 is divisible by relatively prime numbers 3 and 8, then p 2 –q 2 is divisible by 3·8=24, which is what needed to be proven.
d) Let the largest common divisor numbers b and c–a is equal to d, b=k·d and c–a=t·d. Then the numbers k and t are coprime.
So a·k+b·t is divided by c.
6. Find:
a) the greatest common divisor of the numbers 2n+3 and n+7;
b) all pairs of natural numbers x, y such that 2x+1 is divisible by y and 2y+1 is divisible by x;
c) all integers k for which k 5 +3 is divisible by k 2 +1;
d) at least one natural number n such that each of the numbers n, n+1, n+2, ..., n+20 has a common divisor with the number 30030=2·3·5·7·11·13, greater than one.
a) Note that if m > n, then GCD (m; n) = GCD (m – n; n).
In other words, the greatest common divisor of two natural numbers is equal to the greatest common divisor of the modulus of their difference and smaller number. It's easy to prove this property.
Let k be the common divisor of m u n (m > n). This means that m = ak, n = bk, where a, b are natural numbers, and a > b. Then m – n = k(a – b), which means that k is a divisor of the number m – n. This means that all common divisors of the numbers m and n are divisors of their difference m – n, including the greatest common divisor.
Let's use the above:
GCD (2n+3; n+7) = GCD (n+7; 2n+3 – (n+7)) = GCD (n+7; n–4) = GCD (n–4; 11).
Since 11 is a prime number, the required greatest common divisor is 1 or 11. If n–4 = 11d, that is, n = 4+11d, then the greatest common divisor is 11, otherwise it is 1.
Answer: GCD (2n+3; n+7) = 11, with n equal to 4+11d; GCD (2n+3; n+7) = 1, when n is not equal to 4+11d.
b) The number 2x+1 is odd and is divisible by y, so y is also odd. Similarly, x is odd.
The numbers x and y are relatively prime. Indeed, let k be a common divisor of x and y, then 2x is divisible by k, and (2x+1) is also divisible by k (k is a divisor of y, and y is a divisor of 2x+1). This means that 1 is divisible by k, that is, k=1.
The number 2x+2y+1 is divisible by both x and y, which means it is divisible by xy. Then 2x+2y+1 is not less than xy.
Answer: (1; 1), (1; 3), (3; 1), (3; 7), (7; 3).
c) Since k 5 +3 = (k 3 –k)(k 2 +1) + (k+3), then k 5 +3 is divisible by k 2 +1 if k+3 is divisible by k 2 +1 . When it's possible? Let's consider the options:
1) k+3 = 0, which means k = –3;
2) k+3 = k 2 +1; solving, we find k = –1, k = 2;
3) check the integers k for which k+3 > k 2 +1; after checking: k = 0, k = 1.
Answer: –3, –1, 0, 1, 2.
d) let m = 2·3·5·7·k. Selecting k so that m–1 is divisible by 11, and m+1 is divisible by 13, we obtain that the number n = m–10 satisfies the conditions of the problem.
Answer: for example, 9440.
7. Is there a ten-digit number divisible by 11 in which each digit appears once?
Method I When writing out three-digit numbers divisible by 11, you can find three numbers among them, the recording of which involves all the numbers from 0 to 9. For example, 275, 396,418. Using them, you can create a ten-digit number divisible by 11. For example:
2753964180 = 275 10 7 + 396 10 7 + 418 10 = 11 (25 10 7 + 36 10 4 + 38 10).
II method. To find the required number, we will use the criterion of divisibility by 11, according to which the numbers n=a 1 a 2 a 3 ...a 10 (in this case, a i are not factors, but digits in the notation of number n) and S(n)=a 1 –a 2 +a 3 –…–a 10 are simultaneously divisible by 11.
Let A be the sum of digits included in S(n) with a “+” sign, B – the sum of digits included in S(n) with a “–” sign. The number A–B, according to the conditions of the problem, must be divisible by 11. Let’s put B–A=11, in addition, obviously, A+B=1+2+3+…+9=45. Solving the resulting system B–A=11, A+B=45, we find A=17, B=28. Let's select a group of five different numbers with a sum of 17. For example, 1+2+3+5+6=17. Let's take these numbers as odd-numbered numbers. As even-numbered digits, we’ll take the remaining ones – 4, 7, 8, 9, 0.
We see that, for example, the number 1427385960 satisfies the conditions of the problem.
8. Two two-digit numbers written one after the other form a four-digit number, which is divided by their product. Find these numbers.
Let a and b be two two-digit numbers, then 100a+b is a four-digit number. By condition, 100a+b = k·ab, hence b = a(kb–100), that is, b is divided by a.
So b = ma, but a and b are two-digit numbers, so m is single-digit.
Since 100a+b = 100a+ ma = a(100+m) and 100a+b = kab, then a(100+m) = kab,
that is, 100+m = kb or 100+m = kma, whence 100 = m(ka–1).
Thus, m is a divisor of the number 100, in addition, m is single digit number, which means m = 1, 2, 4, 5.
Since ka = 1+100/m, and a is two-digit, then for m the values 1 and 5 disappear, because
when m = 1 the number 100/1+1 = 101 is not divisible by any two-digit number A;
with m = 5 the number is 100/5+1 = 21 and we have a = 21, for which b = ma = 5·21 is a three-digit number.
For m = 2 we have, ka = 51, a = 17, b = 17 2 = 34;
with m = 4 we have, ka = 26, a = 13, b = 13 4 = 52.
Answer: 17 and 34, 13 and 52.
9. Prove that for any natural k and n the number 1 2k+1 + 2 2k+1 + . . . + n 2k+1 is not divisible by n + 2.
Let us take advantage of the fact that the sum of equal odd degrees two numbers is divided by the sum of these numbers, which follows from . You can write:
2 2k+1 + n 2k+1 = (2 + n) A 1,
3 2k+1 + (n – 1) 2k+1 = (3 + (n – 1)) A 2 = (2 + n) A 2,
4 2k+1 + (n – 2) 2k+1 = (4 + (n – 2)) A 3 = (2 + n) A 3 and so on, where A i are some integers.
Depending on the parity of n, there may be a shortage of numbers to form the last pair; this can be avoided by multiplying by 2, considered in the sum condition. So,
2(1 2k+1 + 2 2k+1 +...+n 2k+1) = 2 1 2k+1 + (2 2k+1 + n 2k+1) + (3 2k+1 + (n – 1) 2k+1) +...+ (n 2k+1 + 2 2k+1) =
2 + (n + 2) A, where A is some integer.
One of the terms of the last sum is divisible by n + 2, the other is not divisible for any natural n. So, the sum considered in the condition is not divisible by n for any natural n and k.
10. Prove that for any prime number p > 2 numerator m fraction
is divisible by p.
Note that the number p–1 is even, and we transform the fraction m/n to the form
Bringing the resulting expression to a common denominator
we get the relation
from which the equality m(p–1)!=pqn follows. Since none of the numbers 1, 2, 3, ..., p–1 is divisible by a prime number p, the last equality is possible only if m is divisible by p, which is what needed to be proven.
Problems without solutions
1. Prove that for any natural number n:
a) the number 4n + 15n – 1 is divisible by 9;
b) the number 3 2n+3 + 40n – 27 is divisible by 64;
c) the number 5 n (5 n + 1) – 6 n (3 n + 2 n) is divisible by 91.
2. Find:
A) natural values n such that n 5 – n is divisible by 120;
b) the smallest natural number n such that n is divisible by 19 and n + 2 is divisible by 82.
3. Let m, n be different natural numbers, and m is odd. Prove that 2 m –1 and 2 n +1 are relatively prime.
4. Four different integers three-digit numbers, starting with the same digit, have the property that their sum is divisible by three of them without a remainder. Find these numbers.
5. Prove that for every natural number n > 1, the number n n – n 2 + n – 1 is divisible by (n – 1) 2 .
