Matt's integer

Let the smallest possible integer required by the question be x

We can construct 3 equations, where a , b , c are three distinct positive integers:

• a + (a + 1) + (a + 2) ... (a + 10) = x

• b + (b + 1) + (b + 2) ... (b + 11) = x

• c + (c + 1) + (c + 2) ... (c + 12) = x

Next we simplify :

• 11a + 55 = x

• 12b + 66 = x

• 13c + 78 = x

Then we factorise :

• 11 (a + 5) = x

• 6 (2b + 11) = x

• 13 (c + 6) = x

Therefore, we can see that x is divisible by 6 , 11 and 13

Finally, the Lowest Common Multiple of these numbers, is found by multiplying all their factors:

2 x 3 x 11 x 13 = 858

For the first sum beginning in x:

N = 11*x + 55

For the second sum beginning in y:

N = 12*y + 66

For the third sum beginning in z:

N = 13*z + 78

then we have the equivalents:

N = 11(x + 5)

N = 6 (2*y + 11)

N = 13(z + 6)

if we make N = 11 * 6 * 13 = 858 (mcm of the three numbers), we have to prove that N between 6 and minus 11 is even, since 132 is even we have that N = 858

Let N be such a number Then N can be expressed as : N = 11a+11*12/2 ( sum of 11 consecutive integers with a as first integer )

N = 12b+12*13/2 ( sum of 12 consecutive integers with b as first integer )

N = 13c+13*14/2 ( sum of 13 consecutive integers with c as the first integer )

Simplifying , we get , 11(a+6)=6(2b+13)=13(c+7) Since 11 , 6 and 13 are mutually prime , we get a+6=13*6=78 thus a = 72 Solving we get N=858

Denote the first of each of the series of consecutive integers as $a$ $b$ $c$ We have these 3 equations

$11a + 55 =n$

$12b + 66 = n$

$13c + 78 = n$

Finding the relationship of $a$ to $b$ and $b$ to $c$ .

$11a = 12b + 11$ which then implies that $b$ is divisible by $11$ $12(b-1)\ = 13c$ also implies that the factor $(b-1)$ is divisible by $13$

We then find the least positive integer for $b$ which is $66$ Substituting $b$ to the 2nd equation we would get $n = 858$

The sum of $N$ consecutive positive integers starting from $n_{0}$ is:

$\sum_{i = n_{0}}^{n_{0} + N - 1} i = \frac{2 n_{0} + N - 1} {2} N$

Let be $x$ the number we are searching for. We have:

$x = ( n_{0} + 5 ) 11$

$x = ( 2 n_{1} + 11 ) 6$

$x = ( n_{2} + 6 ) 13$

with $x, n_{0}, n_{1}, n_{2}$ positive integers.

Hence $x$ must be divisible by $11$ , $6$ and $13$ .

$LCM(6, 11, 13) = 6 \times 11 \times 13 = 858$