Relative of Matt's integer

We let $k$ be such a number.

Since $k$ could be expressed as the sum of six consecutive positive integers, we have $k=b+(b+1)+...+(b+5)=6b+15$ . Thus it is congruent to $3$ modulo $6$ . Hence it is odd.

Since $k$ could be expressed as the sum of eight consecutive positive integers, we have $k=c+(c+1)+...+(c+8)=8c+28$ . Thus it is congruent to $4$ modulo $8$ . Hence it is even.

However, a number cannot be both even and odd, so there are $\boxed{0}$ such numbers.

The sum of six consecutive numbers can be expressed as $6a+15 = 3\left(2a+5\right)$

The sum of eight consecutive numbers can be expressed as $8b +28 = 4\left(2b+7\right)$

Clearly the sum of all the six and the eight consecutive numbers is a multiple of $12$ . This means that $2a+5=4k$ for some integer $k$ . However, $2a$ and $4k$ are even whereas $5$ is odd. We have an odd total on the LHS, and an even term on the RHS. Therefore, both sides must be zero.

If we say that the number $N$ is the sum of six consecutive positive integers, and the sum of eight consecutive positive integers, then:

$N = a + (a + 1) + (a + 2) ... (a + 5) = 6a + 15$

$N = b + (b + 1) + (b + 2) ... (b + 7) = 8b + 28$

where $a$ and $b$ are integers. Clearly, $8b + 28$ will always be even, and $6a + 15$ will always be odd. Therefore, the number $N$ cannot exist, because $N$ cannot be both odd and even.

Supposing the first integer as "a", sum of 6 consecutive positive integers=6×a+15

Supposing the first integer as "b", sum of 8 consecutive positive integers=8×a+23

According to the question, 6×a+15=8×a+23

or, 6×a=8×a+13

or, a=(8×b+13)/6

When "b" is a positive integer, "8×b+13" will always be an odd number, so it will not be divisible by 6.

For integer values of "b" there will be no integer value of "a". So there is no positive integer which can be expressed as the sum of six consecutive positive integers, and as the sum of eight consecutive positive integers.

Therefore, the answer is 0.

The sum of 6 consecutive integers=6/2(2a+5)=6a+13,where a is the starting integer. The sum of 8 consecutive integers=8a'+28,where a' is the starting integer. According to the question 6a+13=8a'+28; i.e,6a-8a'=13; i.e,3a-4a'=6.5. But a and a' are integers.Thus 3a-4a' is also an integer. Thus there is no solutions for this equation.

Suppose $n$ satisfies the problem, then $n = \sum_{i = 0}^{5} (x + i) = 6x + 15$ $n = \sum_{i = 0}^7 (y + i) = 8y + 28$ From first equation $n$ is even and from second equation $n$ is odd, a contradiction! Hence no such $n$ exists, thus the answer is $$\boxed{0}$$

The sum of $6$ consecutive positive integers can be expressed as $6n+15$

and the sum of $8$ consecutive positive integers can be expressed as $8m+28$

Notice that one is always odd and the other always even. Thus there are $\boxed{0}$ integers less than $1000$

If N can be written as a sum of six consecutive positive integers then there exists a positive integer n such that $N=n+(n+1)(n+2)+(n+3)+(n+4)+(n+5)+(n+6)=6n+15$ . As well, if N can be written as a sum of eight consecutive positive integers then there exists a positive integer k such that $N=8k+28$ . Putting the equations together we get:

$6n+15=8k+28 <=> 2(3n-4k)=13$

Now (3n-4k) is an integer so the equation tells us that 2 divides 13 which is a contradiction, so there exists no positive integers that can be expressed as the sum of six consecutive positive integers, and as the sum of eight consecutive positive integers. Hence, the answer 0.

• n = a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5

  - n = b + b + 1 + ... + b + 7

6a + 15 = 8b + 24

But 6a + 15 is odd and 8b + 24 is even so no natural numbers can verify the ecuation

Six Consecutive Positive Integers: E (even) and O (odd)

In the case of 6 consecutive integers there are two combination and they are:

1. E+O+E+O+E+O= 3 \times E+3 \times O= O

2. O+E+O+E+O+E= 3 \times E+3 \times O= O

Therefore, in the case of 6 consecutive integers, the answer is always odd.

Eight Consecutive Positive Integers:

In the case of eight consecutive integers there are two combination and they are:

1. E+O+E+O+E+O+E+O= 4 \times E+4 \times O= E

2. O+E+O+E+O+E+O+E= 4 \times E+4 \times O= E

Therefore, in the case of 8 consecutive integers, the answer is always even

For the reasons above, there exists no integers that can be expressed as both the sum of six consecutive positive integers and the sum of eight consecutive positive integers.

Assume that the number can be written in the form a + a+1 ...+ a+5 = 6a+15 . It can also be written as b + b+1 ... b+7 = 8b+28 . Notice that in terms of a , the number will always be odd but in terms of b , the number will always be even. Reaching this contradiction shows there is no value.

let x and y be the two positive integer numbers. So the sum of six consecutive numbers from x would be=6x+15 and similarly for eight starting from y would be=8y+28. so we have equation 6x+15=8y+28 which is 6x=8y+13. Now the R.H.S is even and L.H.S is odd and there is no solution for integer x and y. Hence the ans is 0

A soma dos 6 primeiros números é igual a 21 (somando a partir de 1), (somando a partir de 2 ) é igual 27 , (somando a partir de 3) é igual 33 . Nessa soma encontra-se um padrão onde sempre o próximo número da soma é igual ao valor atual somado mais 6. Vale ressaltar que a (soma de um número impar mais um número par terá sempre com resultado uma impar) , por lógica então todos os valores da soma de 6 serão impares.

A soma dos 8 primeiros números é igual a 36 (somando a partir de 1), (somando a partir de 2 ) é igual 44 , (somando a partir de 3) é igual 52 . Nessa soma encontra-se um padrão onde sempre o próximo número da soma é igual o valor atual somado mais 8. Vale ressaltar que a (soma de um número par mais um número par terá sempre com resultado uma par) , por lógica então todos os valores da soma de 8 serão pares.

Como todos os valores somados de 6 serão número impares e todos os valores somados de 8 serão pares , a interseção da soma de 6 com a soma de 8 será de 0 valores.

Let $x$ be the number which can be expressed as sum of the six consecutive number as well as sum of eight consecutive numbers

hence, $a + (a + 1) + (a + 2) + (a + 3) + (a + 4) + (a + 5) = x$ where $a$ is a positive integer

$6a + 15 = x$ ................(1)

$3(2a + 5) = x$

similarly, $b + (b + 1) + (b + 2) + (b + 3) + (b + 4) + (b + 5) + (b + 6) + (b + 7) = x$ where $b$ is a positive integer

$8b + 28 = x$

$4(2b + 7) = x$

hence, $x$ is a multiple of 4,3 and two other odd munbers.

Hence, x is an even no

From (1),

$6x + 15 = even No.$

$6x = even No. - 15$

$6x= odd No.$

$6x$ is an even number

Hence there is no integer which satisfies the properties of $x$

hence, number of values of $x$ = 0

x = a + (a + 1) + (a + 2) + (a + 3) + (a + 4) + (a + 5) = 6a + 15 x = b + (b + 1) + (b + 2) + (b + 3) + (b + 4) + (b + 5) + (b + 6) + (b + 7) = 8b + 28

6a + 15 = 8b + 28 <=> 6a = 8d + 13, being "a" and "d" natural numbers. For any chosen "d" there is no integer "a" because 8d + 13 will end with the algarisms 3, 1, 9, 7 and 5 for any "d"; and 6a will end with the algarisms 0, 2, 8 and 4 for any "a". As there are no numbers which satisfy both expressions, 8d + 13 is always different from 6a, so there are no numbers that can be written in the form presented in the question.

Resp.: 0

X = 3 * (2 * a + 5) (i)

X = 4 * (2 * b + 7) (ii)

Of (ii) it can be seen that must be even and divisible by 4, also it must be divisible by 3.

But any even number divided by 3 gives another even number,

Suppose X = 2 * k, where k must be divisible by 3 to X be too, therefore:

(2 * k) / 3 is too and even number, but of (i) if we rest 5 to an even number we get an odd number

2 * y - 5 = 2(y - 2) - 1 wich clearly is and odd number.

But to get the value of 'a' in (i) we need and even number since we have to divide it by 2.

Therefore there is no number that have this caracteristics.