Evaluate that polynomial!

Since that $f(n) = np(n)+ f(0)$ and $n|f(n)$ , so $n|f(0)$ for $1\le n \le 6$ . Therefore, $f(0)=k*LCM(1,2,3,4,5,6)=60k$ . Then, the smallest positive value is when $k=1$ and we have: $f(0)=60$

Well known result:

For any polynomial $f(x)$ with integer coefficients, $f(x) - f(y)$ is divisible by $x-y$ for all integers $x,y$ .

Thus, $f(n) - f(0)$ is divisible by $n - 0 = n$ for all $n \in \overline{1,6}$ .

But, since $f(n)$ is divisible by $n$ , we have $f(0)$ is divisible by $n$ for all $n \in \overline{1,6}$ .

Hence, $f(0)$ must be divisible by $\text{lcm}\{1,2,3,4,5,6\} = 2^2 \cdot 3 \cdot 5 = 60$ .

So, if $f(0)$ is positive, then we must have $f(0) \ge 60$ .

The polynomial $f(x) = 60$ has integer coefficients, satisfies the condition that $f(n) = 60$ is divisible by $n$ for all $n \in \overline{1,6}$ , and has $f(0) = 60$ .

Therefore, the smallest possible positive value of $f(0)$ is infact $\boxed{60}$ .

Be $f(x) = a_0x^{n} + a_1x^{n-1} + ... + a_n$ , such that every $a_i$ is an integer. Also, $f(0) = a_n$ , and $1$ divides $f(k)$ for every integer $k$ .

We know that:

• $2 | f(2) = a_02^{n} + a_12^{n-1} + ... + a_n \iff 2 | a_n$
• $3 | f(3) = a_03^{n} + a_13^{n-1} + ... + a_n \iff 3 | a_n$
• $4 | f(4) = a_04^{n} + a_14^{n-1} + ... + a_n \iff 4 | a_n$
• $5 | f(5) = a_05^{n} + a_15^{n-1} + ... + a_n \iff 5 | a_n$
• $6 | f(6) = a_06^{n} + a_16^{n-1} + ... + a_n \iff 6 | a_n$

So, $a_n$ is a positive multiple of $2, 3, 4, 5,$ and $6$ . To minize $a_n$ , we find the least common multiple of these numbers. Therefore, the answer is $60$ .

Note that $f(0)$ is equal to the constant coefficient of the polynomial. And since all the other terms of polynomial will always be multiples of $n$ when substitute $x$ with $n$ , the problem is equivalent to finding the positive least common multiple of $1,2,...,6$ , which is $60$ .

A general expression for $f(x)$ is

$f(x) = a_m x^m + a_{m-1} x^{m-1} + \ldots + a_1 x + a_0$ ,

with $a_0 = f(0)$ . We can rewrite this as

$f(x) = (a_m x^{m-1} + a_{m-1} x^{m-2} + \ldots + a_1)x + a_0$ ,

from which we see that $n$ divides $f(n)$ iff $n$ divides $a_0$ . Therefore, for $f(n)$ to be divisible by $n$ for all $1 \leq n \leq 6$ , $a_0 = f(0)$ must be divisible by all $n$ . The smallest such positive number is $f(0) = \boxed{60}$ .

Let $f(n)=n \times q(n) +a$ .

$n$ must divide $f(n)$ . That can be expressed as $\frac{f(n)}{n}=q(n)+\frac{a}{n}$ . Therefore $n$ must divide $a$ for all values of $n$ . The smallest such $a$ is the lcm of the numbers 1-6. That is 60.

As $f(n)$ is multiple of n then $f(0)$ [constant terms] must be multiple of n. The smallest possible positive value of $f(0)$ is the LCM of the integers $1,2,3,4,5,6$ ,which is $60$ .

For any polynomial $f$ with integer coefficients and any integers $m,n$ : $m-n$ must divide $f(m)-f(n)$ . (This easily follows from the fact that $m-n$ divides $m^k-n^k$ for any $k\geq 0$ .)

In our problem it follows that for each $1\leq n\leq 6$ : $f(n)-f(0)$ must be divisible by $n$ . And as each $f(n)$ is divisible by $n$ , $f(0)$ must be divisible by each of the numbers 1 through 6.

The least common multiple of these numbers is $\fbox{60}$ . The constant polynomial $f(x)=60$ shows that this value can indeed be obtained as $f(0)$ .

Let $f(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \ldots + a_1 x + a_0$ , with integer $a_n, a_{n-1}, a_{n-2}, \ldots , a_1, a_0$

Then $f(x) = x ( a_n x^{n-1} + a_{n-1} x^{n-2} + a_{n-2} x^{n-3} + \ldots + a_1 ) + x_0$

Since $2 | f(2)$ , and $f(2)$ is simply a scalar multiple of $2$ plus $a_0$ . Then $2 | a_0$

Similarly, since Since $2 | f(2)$ , and $f(2)$ is simply a scalar multiple of $3$ plus $a_0$ . Then $3 | a_0$

Using the same analogy, we can prove that $4 | a_0, 5 | a_0, 6 | a_0$

Thus $a_0$ is divisible by $1,2,3,4,5,6$ . So the smallest positive value of $f(0)$ is $lcm(1,2,3,4,5,6) = 2^2 \times 3 \times 5 = \boxed{60}$

We have $a-b|f(a)-f(b)$ for all $p(x)$ polynomial with integers coeficients.

So $n|f(n), \forall 1\le n\le 6\Rightarrow n-0|f(n)-f(0)\Rightarrow n|f(0), \forall 1\le n\le 6$

and then $60=lcm(1, 2, ..., 6)|f(0)\Rightarrow f(0)\ge 60$ since that $f(0)>0$ .

Actually, $f(x)=60$ satisfies $n|f(n), \forall 1\le n\le 6$ and $f(0)=60$ .

For a number to be multiple of 1,2,3,4,5,6, it only needs to be a multiple of 3,4 and 5. Why? Why not 1,2 and 6. Because every number is a multiple of 1, and 2 is already contained in a 4 (4=2x2) and 6 is also contained ( and this is a little less obvious) in a 3 and 4. Because 6=2x3 and 4=2x2 and 4x3= 2x(2x3)= 2x6 =12. therefore that 4x3 is a multiple of 6. How do we find the numbers needed to create a multiple of every number. Well all we do is prime factorization of every number. Therefore 1,2,3,4,5,6 is equal to 1,2,3,2x2,5,2x3. Next we choose the number of different numbers we see. We have to choose a 5, a 3 and two 2's because 4 contains two 2's and for a number to be a multiple we must choose two 2's. Therefore the number we choose is 5x3x2x2=60. Now f(n) is a multiple of 60. Therefore all the integer coeffiencents of f(n) are multiples of 60. But we are interested in f(0) namely the constant term. For f(n) to be a multiple of 60 either the constant term is zero (it doesm't exist) or its a multiple of 60. The question says us to find the smallest possible positive value. 0 is not positive, and the smallest value is 60. Therfore 60 is the answer.

constant term has to be divisible by n such that 1<=n<=6. LCM of all possible n's i.e 1,2,3,4,5,6 is 60. f(0) always indicates the constant term. So, the smallest possible value of f(0) is 60

Let $f(x)=a_1x^n+a_2x^{n-1}+a_3x^{n-2}...+a_k$ .

Clearly, $a_1x^n+a_2x^{n-1}+a_3x^{n-2}...$ is divisible by $n$ because $n$ is in it. That means that the last term, $a_k$ must be divisible by the numbers $1,2,3..6$ . This is simply the least common multiple of $1,2,3..6$ , which is $60$ . Plug in $n=0$ to get $\boxed{f(0)=60}$ .

Let $f(x) = a_ix^i + a_{i-1}x^{i-1} + \ldots + a_1x + a_0$ (the general form of a polynomial), so that $f(0) = a_0.$ Then we have that $f(n) = a_in^i + a_{i-1}n^{i-1} + \ldots + a_1n + a_0$ is a multiple of $n,$ for $n = 1, 2, \ldots, 6.$ Because $a_0, a_1, \ldots, a_i$ are integers, we note that all terms of the above sum, except for $a_0,$ are already multiples of $n.$ Thus it must be the case that $f(0) = a_0$ is a multiple of $n$ for $n = 1, 2, \ldots, 6.$

So, we want the smallest multiple of $1, 2, 3, 4, 5, 6,$ which is $\text{lcm}(1, 2, 3, 4, 5, 6) = 60.$ The smallest positive value of $f(0)$ is $\boxed{60}.$

In order for $f(n)$ to be a multiple of $n$ , the constant term $\left(\text{which is }f(0)\right)$ must be a multiple of $n$ , because all other terms are a multiple of $n$ . So, the smallest possible positive value of $f(0)$ is the least common multiple of the integers $1,2,3,4,5,6$ , which is $\boxed{60}$ .

We let $f(x)= a_1 x^n + a_2 x^{n-1} +.....+ a_n$ where $a_1, a_2,...., a_n \neq 0$ are integers.

If we substitute for $n$ it is clear that all the terms except the constant term( $a_n$ ) will be divisible by $n$ . Hence, we must have that $a_n$ is a multiple of $n$ .

Since $a_n$ must be divisible by all $1 \leq n \leq 6$ in the range then to minimize $f(0)$ we minimize $a_n$ .

Thus, the least possible value of $a_n= lcm(1,2,3,4,5,6)=60$

Hence least possible value of $f(0)= \boxed{60}$

Every thing in the polynomial should be divisible by n since we put n in place of x, except the constant in the expression. Hence, to be divisible, the constant must also be divisible by all integers from 1 to 6. Therefore the constant must be the L.C.m of 1,2,3,4,5, and 6 = 60. Hence, the answer is 60.

Write $f(x)=x^n+a_1x^{n-1}+...+a_{n-1}x+a_n$ with $a_i;i=1,2,3,...,n$ is an integer. From here, we can conclude that $x|f(x)-a_n$ . But, we know that $x|f(x)$ for all integers $1\leq x\leq6$ . So, $x|a_n$ for all integers $1\leq x\leq6$ . Since $f(0)=a_n$ , the minimal positive value of $f(0)$ is $LCM(1,2,3,4,5)$ which is equal with $60$ . So, the answer is $60$ .

A polynomial $f(x)$ can be written in the form $a_ix^i+a_{i-1}x^{i-1}+a_{i-2}x^{i-2}+...a_{1}x+a_{0}$ , so $f(n)=a_in^i+a_{i-1}n^{i-1}+...a_{1}n+a_{0}$ , or $n(a_in^{i-1}+a_{i-1}n^{i-2}+...a_1)+a_0$ . Since $n(a_in^{i-1}+a_{i-1}n^{i-2}+...a_1)$ is clearly a multiple of $n$ , for $n$ to divide $f(n)$ , $a_0$ must be a multiple of $n$ . Therefore, for $n$ to divide $f(n)$ for $n$ from $1$ to $6$ , $a_0=lcm(1, 2, 3, 4, 5, 6$ =60). $f(0)$ is clearly just $a_0$ , so our answer is $\fbox{60}$ .

f(n)-f(0) always divides by n, so f(0) should divide by lcm(1, 2, 3, 4, 5, 6)=60
Polinom f(n)=60 works.

f(1)= a 1×1^n+a 2×1^(n-1)+..+a n×1 thus a n must be divisible by 1 f(2)= a 1×2^n+a 2×2^(n-1)+..+a n×2^0 thus a n must be divisible by 2 f(3)= a 1×3^n+a 2×3^(n-1)+..+a n×3^0 thus a n must be divisible by 3 f(4)= a 1×4^n+a 2×4^(n-1)+..+a n×4^0 thus a n must be divisible by 4 f(5)= a 1×5^n+a 2×5^(n-1)+..+a n×5^0 thus a n must be divisible by 5 f(6)= a 1×6^n+a 2×6^(n-1)+..+a n×6^0 thus a n must be divisible by 6 Then a n must be divisible by 1,2,3,4,5,6 then the GCD=60,then the smallest a n must be 60 then f(0)= a 1×0^n+a 2×0^(n-1)+..+a n×0^0 thus a 1×0^n+a 2×0^(n-1)+..+a (n-1)×0^1=0 then f(0)=60

For any polynomial with integer coefficients, $f\left(n\right) = a_0n^0 + a_1n^1 + a_2n^2 + ... + a_kn^k$

If $n$ is to be a divisor of $f\left(n\right)$ for $1\leq n\leq6$ then it is clear $a_0$ is divisible by $1,2,3,4,5,6$ . The smallest such positive integer is therefore $1*2*2*3*5 = 60$

Write $f(x) = a_k x^k + a_{k-1} n^{k-1} + \ldots + a_1 n + a_0$ , with $a_i \in \mathbb{Z}$ for all $0 \leq i \leq k$ . If $n$ divides $f(n)$ , then as $n$ divides all the terms on the right-hand side of the previous expression except $a_0$ , $n$ must also divide $a_0$ . Thus, $a_0$ is a common multiple of all the integers $1$ through $6$ , and the least such common multiple is $60$ ; as $f(0) = a_0$ , this solves the problem.

Because any multiple of $n$ added to another multiple of $n$ is also a multiple of $n$ , we can assume that for all integers $n$ , $f(n)$ is a multiple of $n$ as long as there is no constant term in $f(x)$ (i.e., such that the coefficient is multiplied by $n^0$ ).

The problem states that $f(n)$ must be a multiple of $n$ for all integers $1 \leq n \leq 6$ . Zero is not positive, so the solution can't be $f(0) = 0$ . For any polynomial, $f(0) = 0 +$ any constants in the polynomial. Again, because any multiple of n plus another multiple of n is a multiple of n as well, in order for f(n) to be a multiple of $n$ for $n = {1, 2, 3, 4, 5, 6}$ , the constant must be a multiple of all these integers. We are looking for the smallest possible value of n, so the problem is simply to find the lcm(1, 2, 3, 4, 5, 6):

lcm(1, 2) = 2

lcm(2,3) = 6

lcm (6, 4) = 12

lcm (6, 5) = 60

lcm(1, 2, 3, 4, 5, 6) = 60

As given in the question, all coefficients are in integer value, that means it is a multiple of n; to let the funcion be a multiple of n, the constant must also be the multiple of n.

We know that during f(0), all terms with x have the same value 0.

Then, the only positive value is from the constant.

The value of the constant = L.C.M. of (2, 3, 4, 5, 6) = 30

Then f(0) = 30

For every polynomial x, all terms except the last term is divisible by powers of x. Therefore if n divides f(n) for every integer n less greater than or equal to 1 and less than equal to 6, then n must divide the last term of f(x). Getting the LCM of 1, 2, 3, 4, 5, and 6, we will get 60 which is the smallest positive value of f(0) or in other words the last term

We let the polynomial be $a_kx^k+a_{n-1}x^{n-1}+...+a_1$ . It is given that $a_1, a_2, ..., a_k$ are integers. For integer $k$ where $1 \le i \le k$ , all of $a_ix^i$ is divisible by $n$ . Hence, since $f(n)$ is a multiple of $n$ , it is then required that $a_1$ is divisible by all of $1, 2, 3, 4, 5, 6$ . Hence $n$ must be a multiple of the least common multiple of $1, 2, 3, 4, 5, 6$ , which is $60$ . Thus, the smallest possible positive value of $f(0)$ is $\boxed{60}$ .

tle LCM (lowest common multiple) from 1 to 6 is 60

All the non-constant terms of a polynomial $f(x)$ have a variable such that those terms will be a multiple of whatever $n$ is when $n$ is plugged in as the variable. Thus, the only term that determines whether or not $f(n)$ is a multiple of $n$ is the constant term. The LCM of the integers 1-6 is 60, since the next smallest value that works is zero, 60 is the smallest positive value that works.

LCM 1,2,3,4,5,6 = 60

Let $f(x)$ be a polynomial of order $m$ , that is $f(x) = a_0 + a_1x + a_2x^2 + \ldots + a_mx^m$ . Notice that if $n \mid f(n)$ , then $n \mid a_0$ . Because $n \mid f(n)$ for $1 \le n \le 6$ , then we have $n \mid a_0$ for $1 \le n \le 6$ . The smallest $a_0$ therefore is $a_0 = \text{lcm}(1, 2, 3, 4, 5, 6) = 60$ .

Note first that if we write $f(x) = a_k x^k + \ldots + a_1 x + a_0$ , then $f(0) = a_0$ .

We are given that for $1 \leq n \leq 6$ , $n | f(n)$ . Since for each integer $n$ we also have $n | a_k n^k + \ldots + a_1 n$ , we conclude that $n | a_0$ .

It's easy to check that the smallest positive integer that divides $n$ for all $1 \leq n \leq 6$ is $60$ and that any polynomial with integer coefficients where $a_0 = 60$ will satisfy the requirements.

Since $f(x)$ is a polynomial with integer coefficients, it follows that $n \mid (f(0)-f(n))$ for all positive integers $n.$ From the assumption, we have $n \mid f(n)$ for $1 \le n \le 6.$ Therefore $n \mid (f(n)+(f(0)-f(n)) \quad \text{or} \quad n\mid f(0)$ for $1\le n \le 6.$ This implies $f(0)$ is a common multiple of the integers $1,2,\dots,6.$ Hence the smallest possible positive value of $f(0)$ is the least common multiple of the integers $1,2,\dots,6$ , namely $\boxed{60}.$

$f(n)=\sum_{i=0}^{\infty}{a_i n_i}=k n \;\forall n,k,a\in\mathbb{Z},0<n<7$ .

As all the sum terms of order $n$ or higher are divisible by $n$ , and the sum itself is divisible by $n$ , it implies that $a_0$ is also divisible by $n$ . As $f(0)=a_0$ , we seek the minimal positive value of $a_0$ , which is $\mathrm{lcm}(1,2,3,4,5,6)=60$ .

since you have f(x)=a1x^n+a2x^(n-1)....+a (n-2)+a (n-1)x+a n= x[a1x^(n-1+a2x^(n-2)....+a (n-2)x+a (n-1)]+a n, we have f(x) = a_n mod x. So, the constant term has to be divisible by all integer n in [1,6]. The least common multiple of all of these is 60. Since f(0) is the constant term, f(0)=60

its the LCM of all numbers between 1 to 6 = 60...

f(x)>= LCM {1,2,3,4,5,6}=60

the min f(x)=60

The polynomial $f(x)$ , to have the smallest possible value of $f(0)$ , should have a constant K which is the LCM of (1,6).

Since all the terms with the powers of x will cancel in $f(0)$ , the smallest possible value of $f(0)$ is the LCM (1,6) which is 60.

To find $f(0)$ implies finding the value of the constant in the polynomial. Since any term carrying the variable $n$ will essentially be a multiple of $n$ , the smallest possible value of $f(0)$ will be the LCM of $1,2,3,4,5,6$ , namely $60$ .

The LCM of 1 to 6 is 60

It is well known that if $a|b$ and $a|b+c$ , then $a|c$ . We have that for all $1 \le n \le, n|f(n)$ . We express $f(x)$ as a general polynomial: $a_kx^k + a_{k-1}x^{k-1} + ... + a_1x + a_0$ . Consider all the nonconstant terms (ie: all the terms except $a_0$ ). They are always divisible by $x$ because the coefficients are given to be integers. Since $n|f(n)$ and $n$ divides all the nonconstant terms, we must also have $n|a_0$ . Thus, $a_0$ has to be a multiple of all $n$ between $1$ and $6$ . The least possible $a_0$ is $LCM(1, 2, 3, 4, 5, 6) = \boxed{60}$

We have $f(x) = g(x) + c$ , where $g$ is a polynomial divisible by $x$ and $c$ is an integer constant. The problem boils down to finding the smallest possible positive value of $c$ that is divisible by all integers $1 \le n \le 6$ .

This is just $\text {lcm}(1,2,3,4,5,6) = \boxed {60}$ .

f(0) is equal to the constant value of the function. If f(n) = n, then the first statement holds, but then f(0) is not positive. So, we need to find gcd(1, 2, 3, 4, 5, 6). This is 60. We know this is the lowest value because if we increase the degree, we will at least be finding the $gcd(1, 2^d, 3^d, 4^d, 5^d, 6^d)$ . So, $f(0) = 60$

We note, f(n) is always divisible by n if there is no constant term. However, this implies f(0) = 0 which is not a positive value. So the constant term must be divisible by all integers from 1 to 6 so that f(n) is divisible by n.

Again, this constant term is to be minimum so we take the LCM of those integers from 1 to 6. We obtain the constant term as 60.

Consider a polynomial $F(n)$ which does not have a constant term. Then, the number $F(n)$ must have n as factor, which implies $F(n)$ is a multiple of n. This further implies that for the given function $f(n)$ to be a multiple of $1-6$ , the constant term must be a multiple of all six numbers. The smallest possible value of this is the absolute value of the $LCM$ , which can be deduced from the prime factorization of the six numbers. Taking the positive number, we get $LCM=60$

Note that except for the constant term, every other term of f(n) is a multiple of (n). Therefore, for f(x) to be a multiple of x where x is 1--6, we need the constant term to be a multiple of all of 1--6. That number is $60$ .

f(0) = LCM(1,2,3,4,5,6) = 60

Imagine the polynomial is a first degree one. Now you are at a huge numbered line where you start on "b" if $a \cdot x + b = k$ . You walk to get to a desired point, and the amout you walk is $a \cdot x$ , the point you will end up is f(x). If x is one, f(x) will always be a multiple of one. If x is two, it will walk $a \cdot 2$ , so you must start on a even number. If x is three, you will walk $a \cdot 3$ , a multiple of three, so, you must start on a multiple of three aswell, to the left or to the right, depending on "a". So "b" must be a multiple of 1, 2, 3, 4, 5 and 6. Since it wants the smallest possible "b", it will be the least common multiple of these numbers.

Consider a general polynomial $a_1 x^n + a_2 x^{n-1} + a_3 x^{n-2} + ...$ .

In the problem statement, it is known that for $1 \le n \le 6$ n divides the polynomial. However notice that all terms except for the constant term of the polynomial includes powers of n and are divisible by n. Thus only the constant term determines whether the polynomial is divisible by the numbers.

Since all numbers between 1 and 6 divide the constant term, it must be multiples of their LCM, or multiples of 60. Note that f(0) is the constant term. Hence the smallest possible value of it is 60.

Pretty simple... Is there any need to explain??? The answer is,

$LCM(1,2,3,4,5,6) = 60$

The End...

The polynomial $f(x)$ has a term independent of $x$ , and this is multiple of $1$ , $2$ , $3$ , ..., $6$ . So its minimum positive value must be $2\cdot 3\cdot 2\cdot 5=60$ , so the minimum positive value of $f(0)$ is $60$

I just assumed f(x) to be constant.60 satisfies the value. But by this method it cant be proved that 60 is smallest value for f (0)

For any polynomial, n|f(n) if n divides each term in the polynomial. It naturally divides each term where the exponent of x is different from 0, therefore we must look for a constant term that is minimal, positive and divisible by all n from 1 to 6. This is the lcm(1,2,3,4,5,6)=60.

Let the polynomial be $f(x)=a_0 + a_1 x + \cdots + a_m x^m$ , if $f(n)$ is multiple of $n$ then it must be divisible by $n$ . So, $f(n)=a_0 + a_1 n + \cdots + a_m n^m$ . From, $f(n)$ we can see that $a_1 n+a_2 n^2 +\cdots + a_m n^m$ is divisible by $n$ so we need to see the divisibility of $a_0$ . As, question seeks for smallest positive values of $f(0)$ so it will obviously be equal to $\text{L.C.M.} of 1, 2, 3, 4, 5, 6$ which is equal to $\boxed{60}$

$LCM of (1,2,3,4,5,6) = 60$

To make the value of f(n) as small as possible, we let f(n)=an+a, where a is a positive integer. In that case, f(1) divides 2a, f(2) divides 3a, f(3) divides 4a, f(4) divides 5a, f(5) divides 6a and f(6) divides 7a. By observation, the value of a is minimum when it is lcm(1,2,3,4,5,6), thus a is 60. Then, it follows that f(n)=60n+60. Thus, the smallest possible positive value of f(0) is 60.

Answer = LCM(1 2 3 4 5 6)

The question asks for smallest possible value of $f(0)$ i.e. constant term of the polynomial. All non-constant terms of $f(x)$ are divisible by $x$ , $n | f(n)$ iff $n | a_0$ , the constant term. So required smallest positive term is the LCM of 1 to 6 i.e., $60$ .

This problem must have fallen through from level 2...:)

All terms in a polynomial are powers of $n$ , except the constant term. Since the coefficients are integers, $n$ must divide all of the non-constant terms. Therefore, if $n$ divides $f(n)$ , then $n$ must divide the constant term.

So the smallest possible positive constant term is the LCM of $(1, 2, 3, 4, 5, 6)$ , i.e. $60$ .

Let $f(x) = a_{n}x^{n} + a_{n - 1}x^{n - 1} + ... + a_{0}$ .  Note that $f(0) = a_{0}$ . Note that for $f(x)$ to satisfy the problem condition, $a_{0}$ must be a multiple of $1, 2, 3, 4, 5$ and $6$ . So, the least positive possible value for $f(0) = a_{0}$ is $LCM(1, 2, 3, 4, 5, 6) = 60$ .

Let, the polynomial is
$f$ ( $x$ ) = $a_{n}$ $x^{n}$ + $a_{n-1}$ $x^{n-1}$ +........+ $a_{i}$ $x^{i}$ + ........+ $a_{1}$ $x^{1}$ + $a_{0}$ $x^{0}$ ...............(1)
So,

$f$ ( $1$ ) = $a_{n}$ + $a_{n-1}$ +..........+ $a_{i}$ +..........+ $a_{0}$ .............(2)
$f$ ( $2$ ) = $a_{n}$ $2^{n}$ + $a_{n-1}$ $2^{n-1}$ +........+ $a_{i}$ $2^{i}$ + ........+ $a_{1}$ $2^{1}$ + $a_{0}$ ..........(3)
$f$ ( $3$ ) = $a_{n}$ $3^{n}$ + $a_{n-1}$ $3^{n-1}$ +........+ $a_{i}$ $3^{i}$ + ........+ $a_{1}$ $3^{1}$ + $a_{0}$ ..........(4)
$f$ ( $4$ ) = $a_{n}$ $4^{n}$ + $a_{n-1}$ $4^{n-1}$ +........+ $a_{i}$ $4^{i}$ + ........+ $a_{1}$ $4^{1}$ + $a_{0}$ ..........(5)
$f$ ( $5$ ) = $a_{n}$ $5^{n}$ + $a_{n-1}$ $5^{n-1}$ +........+ $a_{i}$ $5^{i}$ + ........+ $a_{1}$ $5^{1}$ + $a_{0}$ ..........(6)
$f$ ( $6$ ) = $a_{n}$ $6^{n}$ + $a_{n-1}$ $6^{n-1}$ +........+ $a_{i}$ $6^{i}$ + ........+ $a_{1}$ $6^{1}$ + $a_{0}$ ..........(7)