Tricky summation

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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259 posts

#1 h Apr 2, 2025, 2:15 AM • 3 Y

Y by eg4334, aidan0626, lpieleanu

If $\dots = 7$ , what is the value of $1 + 2 + 3 + \dots + 100$ ?

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#2 h Apr 2, 2025, 2:17 AM

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w problem i cannot lie
113(do i even need to write a solution)

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GallopingUnicorn45

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GallopingUnicorn45

#3 h Apr 2, 2025, 2:19 AM

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how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$

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MathPerson12321

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MathPerson12321

#4 h Apr 2, 2025, 2:32 AM

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GallopingUnicorn45 wrote:

how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$

There's no elementary math school category

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899 posts

#5 h Apr 2, 2025, 2:33 AM

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MathPerson12321 wrote:

GallopingUnicorn45 wrote:

how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$

There's no elementary math school category

I think there's a user created one

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Yiyj1

#6 h Apr 2, 2025, 2:35 AM

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DhruvJha wrote:

MathPerson12321 wrote:

GallopingUnicorn45 wrote:

how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$

There's no elementary math school category

I think there's a user created one

never heard of it, doubt the op would know

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362 posts

#7 h Apr 2, 2025, 2:36 AM • 4 Y

Y by aidan0626, lpieleanu, arfekete, eg4334

GallopingUnicorn45 wrote:

how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$

Clearly sir, you are deeply mistaken.

Solved with resources, greendivisors, eg4334, lpieleanu, SigmaPiE, Arcticturn, and CoolJupiter.

Here, having several continguous characters as a variable name is absurd! A clear counterexample is in programming, a variable name is invalid if it contains spaces. Thus, the only reasonable explanation is a multiplication using the $\cdot$ symbol standard. We want to solve:
$\cdot \cdot \cdot = 7.$ But this is just $\cdot^2 = 7 \implies \cdot = \sqrt{7}$ . Since we are in Middle School Math, we will not consider the case of $\cdot = -\sqrt{7}$ as surely outrage will spark. Now if you are not experienced in the dark arts, a feeble-minded individual would simply plug in $\dots = 7$ and sum it up. How absurd! Instead, we explore the more reasonable path of multiplying the "normal" sum of $5050$ by $\sqrt{7}$ , as every unit in the sum is replaced by the embedded $\cdots$ within the sequence, clearly the intended path of the creator.

Now suppose it is thousands of years ago and we do not have a calculator. We instead use the approximation $\sqrt{7} \approx 2.64575131106$ written by Euclid himself on a humble rock. Multiplying with our fingers, we obtain
$5050 \cdot \sqrt{7} \approx 13361.0441209.$ Since $5050$ has $3$ significant figures, we round our answer accordingly to scientific procedure to obtain $\boxed{1.34 \times 10^4}$ .

This post has been edited 2 times. Last edited by blueprimes, Apr 2, 2025, 2:37 AM

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Yiyj1

#8 h Apr 2, 2025, 2:39 AM

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blueprimes wrote:

GallopingUnicorn45 wrote:

how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$

Clearly sir, you are deeply mistaken.

Solved with resources, greendivisors, eg4334, lpieleanu, SigmaPiE, Arcticturn, and CoolJupiter.

Here, having several continguous characters as a variable name is absurd! A clear counterexample is in programming, a variable name is invalid if it contains spaces. Thus, the only reasonable explanation is a multiplication using the $\cdot$ symbol standard. We want to solve:
$\cdot \cdot \cdot = 7.$ But this is just $\cdot^2 = 7 \implies \cdot = \sqrt{7}$ . Since we are in Middle School Math, we will not consider the case of $\cdot = -\sqrt{7}$ as surely outrage will spark. Now if you are not experienced in the dark arts, a feeble-minded individual would simply plug in $\dots = 7$ and sum it up. How absurd! Instead, we explore the more reasonable path of multiplying the "normal" sum of $5050$ by $\sqrt{7}$ , as every unit in the sum is replaced by the embedded $\cdots$ within the sequence, clearly the intended path of the creator.

Now suppose it is thousands of years ago and we do not have a calculator. We instead use the approximation $\sqrt{7} \approx 2.64575131106$ written by Euclid himself on a humble rock. Multiplying with our fingers, we obtain
$5050 \cdot \sqrt{7} \approx 13361.0441209.$ Since $5050$ has $3$ significant figures, we round our answer accordingly to scientific procedure to obtain $\boxed{1.34 \times 10^4}$ .

wait why am i able to edit ur post

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1311 posts

Yrock

#9 h Apr 2, 2025, 2:40 AM

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#8 nah don't mind it it won't work its just a weird glitch

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Yiyj1

#10 h Apr 2, 2025, 2:42 AM

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Yrock wrote:

#8 nah don't mind it it won't work its just a weird glitch

oh aight chat

edit: one more post away from 1200!

This post has been edited 1 time. Last edited by Yiyj1, Apr 2, 2025, 2:43 AM

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#11 h Apr 2, 2025, 2:46 AM • 1 Y

Y by Amkan2022

blueprimes wrote:

GallopingUnicorn45 wrote:

how is this even in middle school math category

$1 + 2 + 3 + ... + 100 = 1 + 2 + 3 + 7 + 100 = \boxed{113}$

Clearly sir, you are deeply mistaken.

Solved with resources, greendivisors, eg4334, lpieleanu, SigmaPiE, Arcticturn, and CoolJupiter.

Here, having several continguous characters as a variable name is absurd! A clear counterexample is in programming, a variable name is invalid if it contains spaces. Thus, the only reasonable explanation is a multiplication using the $\cdot$ symbol standard. We want to solve:
$\cdot \cdot \cdot = 7.$ But this is just $\cdot^2 = 7 \implies \cdot = \sqrt{7}$ . Since we are in Middle School Math, we will not consider the case of $\cdot = -\sqrt{7}$ as surely outrage will spark. Now if you are not experienced in the dark arts, a feeble-minded individual would simply plug in $\dots = 7$ and sum it up. How absurd! Instead, we explore the more reasonable path of multiplying the "normal" sum of $5050$ by $\sqrt{7}$ , as every unit in the sum is replaced by the embedded $\cdots$ within the sequence, clearly the intended path of the creator.

Now suppose it is thousands of years ago and we do not have a calculator. We instead use the approximation $\sqrt{7} \approx 2.64575131106$ written by Euclid himself on a humble rock. Multiplying with our fingers, we obtain
$5050 \cdot \sqrt{7} \approx 13361.0441209.$ Since $5050$ has $3$ significant figures, we round our answer accordingly to scientific procedure to obtain $\boxed{1.34 \times 10^4}$ .

Best solution so far but this makes a slight assumption which seems trivial but is actually incorrect. However, this would probably still get partials.

Intended sol (according to some moppers): Click to reveal hidden text

Remark: I don't know how it would be expected in contest for anyone to actually be able to evaluate $1 + 2 + 3 + 100$ within a reasonable timing even after finding the (already hard) cruxes of considering $G$ and finding $\cdot = \sqrt{7}$ , so this problem is probably best just to be posted here for us to speculate and not used within a timed contest.

This post has been edited 6 times. Last edited by arfekete, Apr 2, 2025, 2:53 AM

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#12 h Apr 2, 2025, 2:37 PM

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ithinkaboveiswrong?

For clarity, we will write any " " in our math as "space". Then space $\cdot$ space $\cdot$ space $\cdot$ space $=7,$ so space= $\sqrt[4]{7}.$

We aim to compute space $1$ space $+$ space $2$ space $+$ space $3$ space $+$ space $\cdot$ space $\cdot$ space $\cdot$ space $+$ space $100.$ This is simply:
$\sqrt[4]{7}1\sqrt[4]{7}+\sqrt[4]{7}2\sqrt[4]{7}+\sqrt[4]{7}3\sqrt[4]{7}+\sqrt[4]{7}\cdot\sqrt[4]{7}\cdot\sqrt[4]{7}\cdot\sqrt[4]{7}+\sqrt[4]{7}100=\sqrt[4]{71}\sqrt[4]{7}+\sqrt[4]{72}\sqrt[4]{7}+\sqrt[4]{73}\sqrt[4]{7}+\sqrt[4]{7+\sqrt[4]{7+\sqrt[4]{7+\sqrt[4]{7}}}}+\sqrt[4]{7100}.$ We will now estimate to the nearest integer, because every number in the problem is an integer. we have 1.6^4=6.5536<7 but 1.7^4=8.3521 so $\sqrt[4]{7}\approx1.61.$ Similarly, $\sqrt[4]{71}\sqrt[4]{72}\sqrt[4]{73}\approx3\sqrt[4]{72}\approx8.7.$ Thus, the first part is $8.7\cdot1.61\approx14.$

for the second part, finitely many nested roots bad. infinitely many better. assume infinitely many. let it be $x.$ then $x=\sqrt[4]{7+x},$ so $x^4=x+7.$ Now, use newton's method on $f(x)=x^4-x-7.$ Guess $x_0=2.$ Then $x_1=x_0-\tfrac{f(x_0)}{f'(x_0)}=2-\tfrac{7}{31}=\tfrac{55}{31}.$ Close enough.

Finally, $\sqrt[4]{7100}\approx\sqrt[4]{6561}=9.$ Our sum is $14+9+\tfrac{55}{31}\approx\boxed{25},$ which fittingly enough is the last two digits of the year. Also, the sum of the first two parts and the last part are, when rounded, are the two squares that when combined with the three in the date, make the first five squares, which is a beautiful easter egg in memorium for easter being in (last two digits of year)-(month number) days. $\blacksquare$

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#17 h Apr 3, 2025, 6:17 AM

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In the sum at the ... there are not present all the missing numbers?

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