The Present Pyramid

By the sum of squares, we have the formula $\frac{(n)(n+1)(2n+1)}{6}$ for any integer $n$ where $n$ is the square root of the last number in the squaring sequence. Since $\sqrt{10000}$ is equal to $10$ , thus, by plugging $100$ as $n$ , we quickly see that the largest prime factor of $\frac{(n)(n+1)(2n+1)}{6}$ is indeed $n+1$ , which is equal to 101, a prime number. Our answer is $\boxed{101}$ .

We see that $1^2+2^2+...+100^2 = \frac{100*101*201}{6} = 50*101*67$ . Now the greatest prime factor of this number is obviously 101, so our answer is 101

The base of the present pyramid has 10,000 presents.

The second level has 9,801 i.e (10,000-199).

The Third level has 9,604 i.e (10,000-396)

The Third level has 9401 i.e (10,000-591).

199, 396, 591.. etc. can be found to be following the series $200x - x^{2}$ where x is the level.

Thus for a level x, the number of presents is given by $10,000-(200x-x^{2})$

On the top most level, there is one present. thus, $x^{2}-200x+10,000 = 1$

$(x-99)(x-101) = 0$ .

Hence, the number of levels are either 99 or 101 .

Since, we've been asked to find the greatest prime factor of N, the number of levels can be taken as 101 Total No. of Levels $= \sum_{i=1}^{101} (x^{2}-200x+10,000)$

We know that,

$\sum_{i+1}^n x = \frac {n(n+1)}{2}$

And, $\sum_{i+1}^{n^2} x = \frac {n(n+1)(2n+1)}{6}$

$\sum_{i+1}^n 1 = n$

Hence, Total Number of Levels $= \frac{101*102*203}{6} - \frac{200*101*102}{2} + {101*10,000}$

Thus $\boxed{101}$ is the greatest prime factor of N.

We note that the number of presents in each row of the pyramid is a perfect square. The last row consists of $100^2$ presents, the row above that has $99^2$ presents, and all the way up to the first row which has $1^2$ presents. Hence the total number of presents is is the sum all squares from $1$ $\ to$ $\ 99$ which is evaluated to be:

$N=\frac{n(2n+1)(n+1)}{6}=\frac{100 \times 201 \times 101}{6}=50 \times 67 \times 101 = 2 \times 5^2 \times 67 \times101$

From the above prime factorisation of $N$ we see that the largest prime factor is $Y=\boxed{101}$ .

Total number of gifts are sum of squares of first 100 natural numbers. That sum = [N (N+1) (2N+1)]/6 where N = 100 Then sum = 2030100 When we factorize the sum the biggest prime factor we get is 101

• $\displaystyle N=\sum_{n=1}^{100}n^2=\frac16\times100\times101\times201=50\times101\times67$
• $\because$ 101 is a prime.
• $\therefore \boxed{101}$

(Proof of the formula of the summation will be submitted ON REQUEST)

Piles of total gifts.. 100 Square, 99 Square, 98 Square..... 1 Square Total Gifts = 10000 + 9801 + 9604 + ....... + 1 = 338350

Factors of 338350 = 2 X 5 X 5 X 67 X 101

101 is largest prime factor !!

There is a well-known formula for the sum of the first $n$ squares: $1^2+2^2+\ldots+n^2=\frac{n(n+1)(2n+1)}{6}$ . Using this and then using prime factorisations, we're left with $\frac{100\cdot101\cdot201}{6} = \frac{(2^2\cdot5^2)\cdot101\cdot(3\cdot67)}{2\cdot3}=2\cdot5^2\cdot67\cdot101$ . It is then clear that $N=101$ .

We can see the sequence $100^{2} + 99^{2} + 98^{2} + ... + 1^{2}$

We get $\sum_{n=1}^{100} n^{2} = \frac{100(100+1)(2\times 100+1)}{6}$

$= \frac{100\times 101\times 201}{6} = 2\times 5^{2} \times 67\times 101$

Therefore, $Y = \boxed{101}$ .

We see the pattern, 10000-9801=199, 9801-9604=197 9604-9409=195 . . So, we can write the answer in the form

$$10000*101-\sum_{i=0}^{100}{(199-2i)(100-i)} $$

Note, 199-200=-1 but 100-100=0 and it wont influence the result.<br/> Simplifying,

$$10000*101-\sum_{i=0}^{100}{19900-399i+2i^2}$$

Then, <br/>

10000x101-[19900x101-(399x100x101)/2+(2x100x101x201)/6]<br/> and

<br/> 101[10000-19900+19950-6700]<br/>

<br> 101x3350<br/>

Since, 101 is already a prime number,<br/> we have to check 3350.<br/> <br/> 3350=3x5x5x67.<br/>

Therefore, 101 is Y which is the answer.<br/> And Mr. Tumnas had 338350 gifts but none reached my home :( :D

This is: 100^2+99^2+98^2+.....+2^2+1^2 Sum of this series is: 100 $\times101$ $\times201$ /6=100 $\times101$ $\times67$ hence,101 is the largest prime factor

Square(1) + Square(2) + ................... Square(100)

= 1/6 x (100) x (100 +1) x (2x100 +1)

= 50x67x101

GPF = 101

Follow me

If we carefully study the question we find that all the numbers of gifts from bottom layer are in pattern of $100^{2}$ , $99^{2}$ , $98^{2}$ , ......, $1^{2}$

If we add all the square numbers from $1$ to $100$ using formula we get the total no. of gifts.

$N=338350$ gifts

Rest, perform the prime factorization and the largest prime factor would be $101$ .

Thus, the answer is $\boxed{101}$

in each layer the no. of present is square of integer so total no of present is \sum_{i=1}^100 {i}^2 = 338350 and it is factorization is 2 * 5 * 5 * 67 * 101 thus, prime factor is '101'

Very simple. The number of presents in each layer is the square of the number of that layer from bottom. The last layer has 10000 presents which is a square. Next layer has presents which is a square of 99 and so on. So the total number of presents would be a multiple of the number of layers. The number of layers are 101(layers FROM 1st layer TO 100th layer). 101 is a prime number, so its greatest prime number is 101.

FIRST OF ALL,WE CAN OBSERVE THE SERIES THAT IS 10,000;9801;9604 WHICH IS THE SERIES OF SQUARES 100,99,97 AND SO ON.NOW WE FIND THE SUM BY THE FORMULA=n(n+1)(2n+1)/6.then we find this exp. equals to 50X101X67.HERE THE LARGEST PRIME FACTOR IS 101.

this pyramid is arranged in the form of first 10000 (100^2) presents then 9801(99^2) presents and so on ... this will give the series 100^2 +99^2+ 98^2+..................+1 the formula for this series is..[ n (n+1) (2n+1)]/6 here puting n=100 we get total presents as (100 101 201)/6 =>50 101 67 clearly the highest prime factor from above is 101 thus Y=101

The boxes descend from 100 to 99 to 98 to.....1. Total boxes are 5050 (I remembered Gauss when I saw this pattern :) ). The largest prime factor is obviously 101.

the series can be written as 100^2 +99^2+98^2.............1^2. this is the sum of all natural numbers upto 100 therefore applying formula sum=100(100+1)(200+1)/6 it comes out to be 338350 finding highest prime factor 338350 = 2 x 5 x 5 x 67 x 101 i.e. 101

1^2+2^2+...........99^2+100^2=N=n(n+1)(2n+1)/6 where n=100 THEREFORE N=50 101 67
by this we can see that greatest prime here is 101

Just look at it. It is a arithmetic series 1^2 + 2^2 + 3^2 + ... + 100^2 = ? I have just computed the sum by using the following equation n (n+1) (2n+1)/6 And finally have computed the greatest prime factor of sum which is 101.

The greatest trick which is kinda hard to notice here is that the sequence in which Mr. Tumnas has arranged the presents is in the sequence of squares of natural numbers i.e $100^{2}$ , $99^{2}$ , $98^{2}$ and so on and so forth.

Hence the total number of presents *N = $100^{2} + 99^{2}+98^{2}.....+2^{2}+1^{2}$ *

This sum can also be expressed in the form *N = $\frac{n(n+1)(2n+1)}{6}$ *

[Note : The above sum can be proved by Induction]

Since we have n = 100 , plugging in the values we get :

*N = $\frac{100(101)(201)}{6} = 338350$ *

Factorizing this, we get :

$338350 = 5*67670 = 5*67*1010 = 5*67*101*10 = 5*67*101*5*2$

And finally we get to the value of Y = 101

N=(1)^2 + (2)^2 +...+(100)^2

N=(100)(101)(201)/6=(50)(101)(67)

So, Y=101

Observe that 10,000=100^2, 9801=99^2, 9604=98^2 ... 1=1^2. So, an 'x'th level (from top) will have 'x^2' number of boxes.

Hence N=sum(x^2) for x=[1,100].

Wkt, sum of squares of first n numbers is n(n+1)(2n+1)/6 Hence N = 100 * 101 * 201 / 6 = 50 * 101 * 67 = 2 * 5 * 5 * 67 * 101

Clearly, the greatest prime factor is '101', which is the answer of this question.

nth layer has (n^2 -202n + 10201) presents. Thus, on equating it to 1 to find the last layer, we get n=100 or 102. But for n=101, presents are zero. Thus, n can not be greater or equal to 101. Thus, n=100. Now, on simple adding, we get total as (50 * 101 * 67) presents and we can say, 101 is the greatest prime factor.

Notice that 10000= $100^{2}$ , 9801= $99^{2}$ , 9604= $98^{2}$ , ...... , 1= $1^{2}$ , so we knew there are 100 terms in total.

Those terms sum up to be
$\sum$ $n^{2}$ = $\frac{n(n-1)(2n-1)}{6}$ = $\frac{100(101)(201)}{6}$ =2. $5^{2}$ .101.67

Obviously the greatest prime factor is $\boxed{101}$

That is equal to $1^{2} + 2^{2} + \ldots + 100^{2}$ .

Applying that

$\displaystyle \sum_{i=1}^{n} i^{2} = \frac {(n)(n+1)(2n+1)}{6}$

Then the answer is

$\displaystyle \sum_{i=1}^{100} i^{2} = \frac {(100)(101)(201)}{6} = 2\cdot5\cdot5\cdot67\cdot101$

And the greatest prime here is $\boxed {101}$ .

We know that $10000$ is $100^2$ and $9801$ is $99^2$ , this pattern continue until it reach $1$ which is $1^2$ , so the problem wants us to count the sum of all quadratic number from $1$ until $100$ . The easy way to sum it is using the well-known formula $1^2 + 2^2 + 3^2 + ... + n^2 = \frac{1}{6} \times n (n+1) (2n+1)$ hence the answer is $\frac{1}{6} \times 100 \times 101 \times 201 = 338350$ . The prime factorization of $338350$ is $101 \times 67 \times 2 \times 5^2$ so the greatest prime factor Y is $\boxed{101}$

First of all, let us notice that the number of presents on each layer represents the number of the given layer squared (if $n$ is the corresponding layer number then the number of presents on that layer is $n^{2}$ ). This is of course done counting from the top of the pyramid, seeing that the first layer has the most presents.

Therefore, the total number of presents is the sum of the first $n$ squares, which can be written as $\frac{n(n+1)(2n+1)}{6}$ , where $n$ is the total number of layers. Considering that the first layer (that we will consider to be the "last" for easier manipulation) has $10000$ presents, which is actually $100^{2}$ , we may conclude that the pyramid has $100$ layers.

Now, all we need is to calculate the sum $\sum_{i=1}^{100} i^{2}$ , which is actually

$\frac{100\cdot101\cdot201}{6}=50\cdot101\cdot67=2\cdot5^{2}\cdot67\cdot101$

Therefore, we find that $N=2\cdot5^{2}\cdot67\cdot101$ , meaning that $\boxed{Y=101}$

Note: The above given formula can be easily proved with induction if necessary. The general form of it would be

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

I did not feel the need to do that since I consider it to be quite well known.

There are $100^2$ presents on the bottom row, $99^2$ presents on the second bottom row, all the way up to $1^2$ presents on the top row. The total number of presents therefore is: $\sum_{n=1}^{100}{n^2}=\frac{1}{6}n(n+1)(2n+1)=\frac{1}{6}\times 100\times (100 + 1)(2\times 100+1)=338350$

Prime factorization: $338350 = 2\times 5^2\times 67\times 101$

Greatest prime factor, $Y$ , therefore is $\boxed{101}$