Too Many Oranges

Relevant wiki: Hockey Stick Identity

According to Hockey Stick Identity , we can rewrite the number of oranges as:

$\binom{2}{2} + \binom{3}{2} + \binom{4}{2} + \cdots + \binom{200}{2} + \binom{201}{2} = \binom{201+1}{2+1} = \binom{202}{3}$

By using Lucas's Theorem , we can find the remainder of modulus $7$ by rewriting the binomial coefficients as number base $7$

$202 = {406_{7}}$

$3 = {3_{7}}$

$\binom{202}{3}= \binom{406_{7}}{3_{7}} \equiv \binom{4}{0} \binom{0}{0} \binom{6}{3} \equiv \binom{6}{3} \equiv 20 \equiv 6 \pmod 7$

Therefore, there will be $\boxed{6}$ oranges left as a remainder.

The number of oranges in $k$ th layer is the triangular number $T_k$ . Therefore, the sum of all oranges in $n$ layers is given by:

$\begin{aligned} S_n & = \sum_{k=1}^n T_k \\ & = \sum_{k=1}^n \frac {k(k+1)}2 \\ & = \frac 12 \sum_{k=1}^n (k^2 +k) \\ & = \frac 12 \left(\frac {n(n+1)(2n+1)}6+\frac {n(n+1)}2 \right) \\ & = \frac 12 \left(\frac {n(n+1)(2n+4)}6\right) \\ & = \frac {n(n+1)(n+2)}6 \end{aligned}$

$\begin{aligned} \implies S_{2 00} & \equiv \frac {2 00(201)(202)}6 \text{ (mod 7)} \\ & \equiv 100(67)(202) \text{ (mod 7)} \\ & \equiv 2(4)(6) \text{ (mod 7)} \\ & \equiv 48 \text{ (mod 7)} \\ & \equiv \boxed {6} \text{ (mod 7)} \end{aligned}$

The number of oranges in each layer are triangular numbers:

$T_n = 1+2+...+n = \frac {n(n+1)}{2} = \frac {1}{2} (n^2+n)$

The sum of the oranges in all layers is a tetrahedral number:

$P_n = T_1 + T_2 + ... + T_n = \frac {1}{2} ( (1^2 + 2^2 + ... + n^2) + (1+2+...+n) ) =$

$= \frac {1}{2} ( \frac {n(n+1)(2n+1)}{6} + \frac {n(n+1)}{2} ) = \frac {n(n+1)}{4} ( \frac {(2n+1)}{3} + 1) =$

$= \frac {n(n+1)}{4} ( \frac {(2n+4)}{3} ) = \frac {n(n+1)(n+2)}{6}$

In our case n = 200 :

$P_{200} = \frac { 200 × 201 × 202}{6} = 1353400$

$1353400 \equiv \boxed {6} \text { (mod 7) }$

Let us consider this as summing the series f(n+1)=f(f(n)+n+1) in order up to 200 iterations. We start with 0 before any iteration. Counting this series, we may soon realize discover that the sum is 0 (mod 7) at the 7th iteration. Since f(x)=0, and because of the recursive nature of this function which starts at 0, we can start a new. floor(200/7)=28 so we can fit 28 of the length 7 iterations cycles in the 200 iterations, and since 200-7×28=4, we simply need to calculate the sum for 4 iterations to get the answer, this is 6.

• link text One can easily observe that

1st row has 1*2/2 oranges

2nd row has 2*3/2 oranges

. .. ........ nth row has n(n+1)/2 oranges

So, Total no.of oranges available = 1/2(1 2 +2 3 +3 4 +.........................+ 200 201)

On adding...using general term r(r+1)/2

it is found out to be.. 50 * 67 * 404

to find the remainder on dividing by 7

(50 67 404)mod7=(50 mod7 *67 mod 7 *404 mod7 )mod7=4