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I highly doubt this story is true but I heard it from one of the girls I took to the top of the Marriott Marquis, who claimed it happened to one of her friends.
While her friend was also at the top of the Marriott Marquis she felt the need to remove her shoes to let her feet “air out”. Apparently she didn’t realize that the bar revolves and she had to wait 45 minutes or so until she met up with her shoes again.
If it happened to you and you’re reading this please let us know.
Thanks
SJSinNYC
A similar story happened to two friends of mine when we were in yeshiva in Israel.
Both were Dovid and both had dates on the same night through the same shadchan.
I’ll never understand why, but the shadchan set up both dates to meet at the same place in the same hotel. A few minutes in one Dovid realized he was dating the wrong person and went to find the other Dovid (and his date) so they could switch.
Another similar story- one shadchan thought she had a great idea for a guy I knew from yeshiva and wanted to speak it over with another shadchan before she suggested it. The second shadchan thought she was talking about another guy and decided to call him and suggest it before the first one did (since it was such a good idea). The second guy ended up getting engaged to her.
I heard this from a lady that we were introduced to at the wedding of a friend. As far fetched as it sounds I do believe it but you can decide for yourself to believe it or not.
The date took place in downtown Baltimore by the Harbor. The girl realized right away that it wasn’t going to happen but apparently her date did not.
They stopped at a street vendor and he bought her a big yellow helium balloon. Not wanting to offend him and not wanting to have to hold it the whole time, she tied it to her hand and thanked him (while praying that she wouldn’t meet any friends).
While they were walking around and talking she felt him moving uncomfortably close to her and she moved over a few inches. After a couple more times that this happened she lost her balance and fell into the water (she said there were no gates there at the time).
Luckily there was a water taxi right there and she was rescued. For the rest of the date he was complimenting himself on how smart he was for buying her a yellow balloon so that if she fell into the water it would be easy to spot her.
They did not marry each other.
The Big One-
My wife can’t stop laughing at your post. What did I say that made you think I am older than I really am?
This didn’t happen to me but I know the person who it happened to and I assure you it’s not a made up legend.
The guy is driving the girl in a car with a steering wheel on the wrong side (a British car) when the transmission blows and he can’t do more that 25 M.P.H. Many drivers that pass feel the need to honk and make obscene gestures at the person they think is the driver.
The girl is about to lose it but the guy keeps reassuring her that they are honking because they think know her and they are exerting one finger to let her know that she is number 1.
Eventually she lost her cool and started screaming back and letting everyone know that she was not the driver…
For the record they did not get married. (Or go out again.)
I can only try
That’s correct, no matter how many cards are removed (up to 51 and assumning that it is not known which cards are removed) the probability of the picking a heart will be 1/4. Let’s see you prove it for where 32 cards are removed.
(Hint- there are 125,994,627,894,135 different ways that 32 cards can be chosen from a deck of 52 where order doesn’t make a difference. Hope this helps!)
Back in the summer of ’89 the camp I was in took us to a minor league base ball game that was stopped for 45 minutes due to rain. After the game continued the ushers let us all go down to (the now vacant) front row seats.
Later on in the game the catcher struck out and disputed the call with the umpire. The umpire kept quiet while the catcher was using vulgar language and making obscene gestures until he was ejected from the game. Unfortunately, to us young campers, the catcher was our hero.
One of the Rabeim knew he had to make a deep impression on us quick- I’ll never forget his Moshul the next day. “Do you know why the team picked him to be the catcher?, Because the catcher sits behind a cage (his mask) and that player is an animal who needs to be kept behind a cage at all times!”
It’s now almost 20 years later, and when I hear that players name mentioned I don’t think of his major league career or his minor league career, I think of him as animal who needs to be kept behind a cage.
Yes ames you are correct, the answer is 1/4.
Here’s how I solved the problem when I had it on an exam.
There are 1,326 unique ways to choose 2 cards from a standard 52 card deck. (52 x 51 / 2 = 1,326. We divide by two since order doesn’t matter i.e. a seven of spades and a king of clubs is the same group as a king of clubs and seven of spades.)
Of the 1,326 possibilities 78 include 2 hearts (13 * 12 / 2= 78), 507 include one heart and one non-heart ( 2 * (13 * 39 / 2) = 507) and 741 consist of two non-hearts (39 * 38 / 2 = 741). (Please note that 78 + 507 + 741 = 1,326.)
If two hearts are missing from the deck the probability of picking a heart is 11/50.
If one heart is missing from the deck the probability of picking a heart is 12/50.
If no hearts are missing from the deck the probability of picking a heart is 13/50.
So the probability of two hearts missing (78/1,326) x the probability of picking a heart (11/50) = (78/1,326)*(11/50).
Summing up the probabilities from all three scenarios gives us (78/1,326)*(11/50) + (507/1,326)*(12/50) + (741/1,326)*(13/50) = 1/4.
Special for valentines day;
You have a regular deck of 52 cards with two cards missing (you don’t know which 2).
A card is selected at random, what is the probability that the suit of the card picked is Hearts?
gleeg-
The distance traveled can not be a negative number. If the question was how much higher is the ball when it stops bouncing then -8 would be correct.
This might not count as a Shidduch story since my friend and his wife were engaged at the time but it definitely messed up someone else’s proposal.
This friend goes out to a fancy restaurant to eat with his (at the time) kallah and asks for a table for two. The waiter asks him if his name is Dovid to which he responds, “Yes, Why?”, the waiter just winked at him and said, “Don’t worry everything has been taken care of”.
After they were done eating the waiter asked if they were ready for dessert. A minute later the waiter gave her a cake that said “Will you marry me?”.
They all had a good laugh and he was very thankful it didn’t happen on their first date.
While I was driving back this really nice girl after a date I thought I was going through a yellow light and only realized it must have been red when the flash from the camera lit up the night. I tried to disregard it but she said, “Don’t worry, just send me a copy of the picture and I’ll pay for half of it”. (It never came.)
On a subsequent date we were walking in a park and stopped by an observation point to look at the scenery. Out of the corner of my eye I saw some blond hair walking away and, assuming it was her, started walking next to her. When I started to make conversation about the scenery she said something that the moderators wouldn’t approve of (it had something to do with her lineage). I looked down and it was not her. It was so embarrassing.
I returned to the lookout and there she was standing next to another guy (who was also about 12 inches taller than her and wearing a black suit jacket) telling him about a different time she was in the area. You should have seen the look on her face when she saw me and realized that she was talking to someone else. She apologized and said that, due to the difference in height, when she stood next to me she could only see my arm and shoulder. I assured her not to worry since it could happen to anyone. (I never told her what happened to me just one minute earlier.)
A friend and I planned to switch cars while we were on dates but it didn’t work out.
We both had dates from the same far away city (date # 1 for them, #5 for us), and the girls lived within a block of each other. The plan was that I would rent a standard car and he was going to rent a luxury car. We would park the cars in the same garage, go to the same hotel (but to lounges on different floors) and ask to be “excused” at the same time to take care of business where we would swap receipts. Mt friend backed out at the end since he was reluctant to rent a luxury car for a first date.
That was their first and last date while we ended up getting married.
When I told my wife the plan she said it would have been a waste since neither she or her friend would have noticed. Compared to what her parents drive, any vehicle (even a Dodge Stratus) is considered a luxury vehicle and her friends parents are unable to drive and do not have a car.
I got this idea from a friend of mine (he eventually married the girl he was dating at the time), I started talking in a monotone for 30 minutes about the history of the theorem, the different conjectures used to prove the theorem as well as using my fingers to draw the 3D graphs used to prove the theorem (in the hotel lobby). I still crack up when I think about the scene- some guy in a hat, jacket and tie using both his hands to plot x, y and z points on an imaginary 3D grid, while the girl sitting across from him is nodding reassuringly that she understands everything he is saying.
I couldn’t argue with that.
A few minutes into our first date we sat down on a park bench to drink sodas. I opened mine and said, “look I won a free soda”. “No”, She exclaimed, “we won a free soda”.
moish01- The soda with the same name as me is Dr Pepper (not Dr. since it’s not a real doctor), I hope this removes any confusion.
teen- The problem says “What is the total vertical distance the ball will eventually travel?” not “What is the total vertical distance traveled by the ball when it comes to a complete stop?”. The ball will never come to a complete stop but the vertical distance travelled will eventually reach a limit as the amount of time elapsed approaches infinity, what is this limit?
The answer is 56 feet.
I can only try- 24 feet is the upward distance, multiply that by two for the distance it travelled down each time it bounced up plus the original 8 foot drop gives the correct answer of 56 feet.
I can only try- It’s great to be back here. The family is great thanks for asking. This morning I saw the baby kicking her feet while deep in concentration- it was the cutest thing- I thought she was trying to figure out where her center of gravity was so she could try to roll over. My wife figured out exactly what she was doing and made me change the diaper…
It’s really touching to see that I was missed. I’m afraid to log on at home when I should be helping my wife and at work our department was assigned some challenging projects so I’m a little reluctant to log on over there (I’m still trying to explain to facilities why I had 3,121 coconuts, 4 colleagues and a monkey in my 5′ X 7′ cubicle…).
Hopefully we should be finished the projects by early July and I’ll be able to return to this thread full time.
A boy drops a rubber ball from a height of 8 feet above a concrete floor. Each time the ball hits the ground it bounces up 75% of its previous maximum height (it reaches 6 feet on its first bounce, 4.5 feet on the second bounce…). What is the total vertical distance the ball will eventually travel?
Joseph-
A solar year is approximately 365.2422 days long. The Julian Calendar rounded the value to 365.25 and had a leap year every four years (1/4 = .25). Pope Gregory the 10th, 11th, 12th, 13th or 14th(I forget what number) realized the mistake and tried to fix it.
As of Thursday, October 4th 1582 the Julian Calendar ceased to exist and the next day the Gregorian Calendar took over with the date being Friday, October 15th 1582. (The skipped days compensated for the slight decimal error over close to 16 centuries.)
The new calendar calculates the fractional part of the day as .2425 which is equal to 97/400. Therefore 97 out of 400 years are leap years.
In general a year that is divisible by 4 is a leap year. The exception to that rule is if the year is divisible by 100. The exception to that rule is if the year is divisible by 400.
1892, 1896, 1904 and 2000 were leap years.
1700, 1800 and 1900 were not. 2100 will be the next multiple of 4 that’s not a leap year.
Put them together like a “T”. The middle of a bar magnet will not be magnetized because the magnetic force from the poles are cancelled out in the middle. If they are attracted to each other then the vertical bar is the magnet. If there is no attraction then the horizontal bar is the magnet.
Thanks everyone for the warm Mazel Tov wishes. I wish I could thank each of you personally but as you can imagine I’m very busy now.
May we all continue to see many more simchas in the future.
The correct answer is our baby girl.
Hi everyone,
It’s great to be back, I missed you all!
Here’s a riddle for today:
What’s red and white, 19.5 inches long, weighs 7 lbs. 6 3/4 oz, cries a lot and gives Dr. and Dr. Pepper so much nachas?
Joseph,
“Dr. Pepper: How do you know which post is what post #?”
Easy- find the post, count how many down on the page the post is, multiply the page number by 40 and add the two together.
I can only try-
“Dr. Pepper – don’t be modest now, because I’m really curious – is this level of math expertise common among math majors? Among holders of a math PhD?”
I hate questions like this because I don’t want to appear arrogant. The math used to solve that riddle was basic algebra and arithmetic (the MOD(x,y) function is just a fancy way of getting the remainder of x divided by y).
Let me try to explain the thought process of a mathematician the way I heard it from the son of a math professor (it’s not supposed to be funny so don’t complain that you find it corny):
Q. What do you do if you see a piece of wood on fire and a pail of water next to it?
A. Use the pail of water to put out the fire.
Q. What do you do if you see a piece of wood on fire and there is no pail of water next to it?
A. Go get a pail of water and put out the fire.
Q. What do you do if you have a pail of water?
A. Go find a piece of wood to set on fire so you can pour the bucket of water on it.
The way a mathematician goes about solving a problem is to take it apart and change it around to fit into an equation he knows how to handle.
By the time a math major gets to upper level math courses, even calculus should come naturally. When they are trying to set up a mathematical model to solve some sort of equation on an exam they don’t have time to try to figure out how to differentiate or integrate the equation.
Solution to post # 292 http://www.theyeshivaworld.com/coffeeroom/topic/the-riddle-thread/page/8#post-15915
In the following equation, what Yom Tov does x equal to: 10^x = baomer?
Take logs of both sides Log(10^x) = x = Log(baomer)
squeak
brute force method using Excel: (only copy and paste what is between the quotation marks)
Cell A1 “1”, Cell B1 “=((((((A1)*5/4+1)*5/4+1)*5/4+1)*5/4+1)*5/4+1)”, Cell C1 “=B1-INT(B1)”
Select Cells A1 – A3 and move the cursor to the bottom right hand corner of Cell C1 until it turns into a + sign. Now click on the + sign and drag the formulas down a few thousand rows.
Select column “C” and on the “Standard” toolbar click “Sort Ascending” (The icon has an A on top of a Z with a down arrow on the right side). Select to “Expand the Selection” and click “Sort”. The rows where the entry in column C is 0 are possible solutions.
Here’s somewhat of a riddle for the engineers following the thread.
Why are the towers of the Verrazano-Narrows Bridge 1.625 inches further apart from each other at the tops than at the bases? (Both towers stand perfectly plumb.)
squeak
This brings back memories from when I first solved this problem on my abacus… If you read post # 203 http://www.theyeshivaworld.com/coffeeroom/topic/the-riddle-thread/page/6#post-12885 in this thread you’ll understand how I remember the question.
Anyway- what’s “clam stuff”?
And by the way- I am not that old, I was born after the 50s.
squeak
This is not as innocent as it looks.
Let the original pile = y and the final pile (before it’s divided into 5) = x.
We know that x must be divisible by 4 and 5 (the last person split it into 4 even groups and the 5 people split it into 5 even groups).
y = ((((((x)*5/4+1)*5/4+1)*5/4+1)*5/4+1)*5/4+1) which simplifies to (3125*x)/1024 + (8404/1024).
(8404/1024) = 8 + 53/256 therefore the fractional part of (3125*x)/1024 must equal 203/256 (so we are not left with a fraction at the end).
=> (3125*x)/1024 = an integer (henceforth denoted as “I”) + 203/256
=> (3125*x)/1024 = I + 203/256
=> 3125*x = 1024*I + 812
So we need MOD(3125*x,1024) = 812
MOD(3125,1024) = 53
MOD(812,53) = 17
MOD(53,17) = 2
=> x = 4 * 5 * MOD(812,53) * (MOD(53,17) + 1) = 4 * 5 * 17 * 3 = 1020
=> x = 1020 and y = 3121
If we weren’t looking for the minimum value of x the correct answer would be 1020 + 1024c where c is a non-negative constant.
Joseph,
I read the question too fast too late at night. I thought it said that half was remaining, then one third was remaining, then one quarter was remaining…
I can only try
Andrew Wiles announced a proof in 1993 which was almost complete but a colleague of his, I think his last name is Katz, found a “hole” in the proof. Andrew Wiles spent about another year with one of his students “plugging the hole” and I far as I know the proof has been accepted.
This might be a legend but I heard that in the 1800s and early 1900s there were large rewards set aside for the first one to prove (or disprove) the theorem. Unfortunately for Wiles the rewards were in German Marks held in German banks and after the record inflation Germany suffered after World War I, the rewards were practically worthless.
Joseph
15,554,
7777 * 4 * 3 * 2 =186,648
186,648 / 12 = 15,554.
I can only try
There are actually an infinite amount of integers that will satisfy the equation a^2 + b^2 = c^2, another common example is 5, 12 & 13. There are no integers for a, b, c and n such that a^n + b^n = c^n for n > 2. (Please don’t ask me to prove Fermat’s Last Theorem here.)
Of the many uses of Pythagorean Triples, as they are called, the most common one is finding the length of the hypotenuse of a right triangle given the length of the legs. If the legs measure 3 and 4 respectively then the hypotenuse will measure 5.
I can only try
(X + 2 * Y = 98) minus (X + Y = 74) => Y = 24.
We are subtracting one equation from another. (Cramer’s rule would also work.)
3^2 + 4^2 = 5^2 (is that what you were thinking of?) there are an infinite more sets of three integers that will fit that relationship.
The dimples in Golf balls allow the balls to travel further by trapping air inside of them and allowing the ball to ride or “float” on the air. A more detailed explanation is beyond the scope of this thread.
Here’s another one from an upper level math course (that can be answered in English with basic mathematics skills):
Does there exist a prime number “P” such that there is no prime number greater than “P”? (Is there a highest prime number?)
If “P” exists what is it, if not why does “P” not exist?
SJSinNYC
My Ph.D. is not in engineering but I did take some engineering courses. If you have some questions I’ll be happy to try to help you.
anon for this
(x+0.5)*(x+0.5) = x^2 +.5x + .5x +.25 = x^2 + x +.25.
=> 4.5 * 4.5 = 4^2 + 4 + .25 = 20.25
I agree that this will work but I found that my students had a hard time performing more than two mathematical operations mentally. (x+0.5)*(x+0.5) = x^2 +.5x + .5x +.25 reduces to three terms x^2 + x +.25 , by (x – 2)*(x + 2), the two middle terms cancel out so there are only two terms left, x^2 – 4.
If you (or your daughter) have no issue with this consider yourself gifted.
Mrs. “anon for this”,
Here’s how it works,
Instead of x and x + 2, let’s use (x – 1) and (x + 1). Now FOIL (Firsts, Outers, Inners and Lasts) gives us (x – 1)*(x + 1) = x^2 + x – x – 1^2 = x^2 -1.
So 6 x 8 = (7 – 1)*(7 + 1) = 7^2 – 1 = 48.
This will also work for (x – 2)*(x + 2) = x^2 – 4 and so on.
Here’s a trick I figured out for squaring an integer where the units digit is 5. Let’s square 75. Take all the digits besides for the last one (in this case it’s just the number 7) multiply it by by that number + 1 (7 * 8 = 56) and attach the numbers 25 at the end (5625). Please check that 75 * 75 = 5625.
Proof (10x + 5)^2 = 100x^2 + 50x + 50x + 25 = 100x^2 + 100x + 25 = 100(x^2 + x) + 25 = 100 * x * (x + 1) + 25.
Now try it for 45 and move the decimal two spots to the left for 4.5 ^ 2. You should get 20.25.
Joseph,
-40 degrees Celsius = -40 degrees Fahrenheit
I think the North Pole would be one place.
Joseph,
Let X = humans and Y = Horses
X + Y = 74
2 * X + 4 * Y = 196 => X + 2 * Y = 98 (divide both sides by 2)
(X + 2 * Y = 98)
-(X + Y = 74)
=> Y = 24 => X = 50
50 Humans and 24 Horses.
I can only try
I actually thought I got it after the first line but I was wrong.
I did 1 x 5 x 8 x 1 x 2 = 80.
But no!
Next I found the prime factorizations and got almost nowhere on the first line (except that I found many instances of the number 2) but on the second line there was one 7 on the left side and two on the right so I knew that one of the terms was squared.
The rest was smooth sailing.
Here’s an old one:
In the following equation, what Yom Tov does x equal to: 10^x = baomer?
(It was taken from the secular riddle 10^x = cabin.)
I can only try
Why the extra steps?
This is how I would do it:
15, 8, 12, 80 ==> 15 x 8^2 / 12 = 80
18, 7, 6, 147 ==> 18 x 7^2 / 6 = 147
19, 6, 12, ? ==> 19 x 6^2 / 12 = 57
Nobody
If there’s something you don’t understand please ask. I’m here to help. I understand that it may be hard to follow because of the formatting but I wouldn’t consider this example challenging.
I can only try
0 = 2*0*0*9
1 = 2^0+0*9
2 = 2+0*0*9
3 = 2*0*0+?9
4 = 2^0+0+?9
5 = 2+0+0+?9
6 = (2+0+0)*?9
7 = -2+0+0+9
8 = -(2^0)+0+9
9 = 2*0*0+9
10 = 2^0+0+9
I probably should have defined some terms.
A matrix is a rectangular array (m by n) of numbers. It can be a single number (a 1 by 1 matrix) or any other dimension (m and n must be positive integers). A matrix is denoted by brackets on both sides.
A determinant is a value only given to a square matrix (where m = n). For a 1 by 1 matrix the determinant is the lone entry. For a 2 by 2 matrix
a b
c d
the determinant is (a)*(d)-(c)*(b).
The formula for a 3 by 3 matrix was mentioned above in my previous post.
Please don’t ask me to post the formulas for anything greater than m = n = 3, they are very long.
The determinant of a matrix is denoted by vertical bars on both sides. For example if we have a matrix named “A” then the determinant of “A” is denoted by |A|. There are also text books that put the vertical bars on both sides of the whole array to denote the determinant.
I apologize if I offended anyone, with out even thinking I put a title of Reb out of respect for my fellow Yidden, but I’ll stop.
“I can only try”
In the future feel free to ask any math questions, and please don’t think you are taking advantage- I’m here to help.
Here’s how the rule works, (if you know of anyway that I can attach a PDF file to a post or somehow upload a file somewhere that I can link here please let me know. This is hard to follow without mathematical formatting).
2x + y + z = 3
x + 2y + z = 0
Given the general form of a system of three linear equations with three unknown variables:
ax + by +cz = j,
da + ey + fz = k,
ga + hy + iz = l
put this in matrix form:
a b c
d e f
g h i
(3 x 3 matrix)
multiplied by
x
y
z
(3 x 1 matrix)
equals
j
k
l
(3 x 1 matrix)
x =
the determinant of:
j b c
k e f
l h i
divided by
the determinant of:
a b c
d e f
g h i
(The determinant of a 3 x 3 matrix
a b c
d e f
g h i
For our example:
a = 2, b = 1, c = 1, j = 3
d = 1, e = -1, f = -1, k = 0
g = 1, h = 2, i = 1, l = 0
therefore the determinant of
j b c
k e f
l h i
is the determinant of
3 1 1
0 -1 -1
0 2 1
= 3
the determinant of
a b c
d e f
g h i
is the determinant of
2 1 1
1 -1 -1
1 2 1
= -2 -1 + 2 +1 +4 -1
= 3
therefore x = 3/3 = 1.
Similarly
y =
the determinant of:
a j c
d k f
g l i
divided by
the determinant of:
a b c
d e f
g h i
and
z =
the determinant of:
a b j
d e k
g h l
divided by
the determinant of:
a b c
d e f
g h i
The final results are x = 1, y = -2 and z = 3.
I hope this helps.
Reb “anon for this”,
I never looked at it that way, but I guess it would be easier to program a computer to calculate the unknown variables as the ratio of two determinants than to pivot the equations using the Gaussian elimination method.
At first I also assumed that Cramer was the Vilna Goan. When I asked someone in yeshiva if “Cramer was the Vilna Goan”? He looked at me like I was on Coke and said “From Seinfeld”? (From then on I remembered that Gabriel Cramer spells his name with a C.)
This sounds like Cramer’s rule. It is very efficient for solving systems of differential equations, but I guess it could be used for linear equations (although there are easier methods that do not require computing the determinant).
In any case the rule is named after Gabriel Cramer who lived at the same time as the Vilna Goan, but was a different person.
I can only try
Dr. Pepper-
Pick the one that you like best (if any):
1) Sorry Dr., but that answer grades “incomplete”
2) Now say that ten times fast
3) I’m NOT chipping in to replace your worn-out keyboard
4) I think the eleventh zero on line 5071 should be a nine. Please double check.
5) Anybody could’ve done that. Let’s see you say it backwards.
6) This actually makes more sense than most extra-long posts.
7) YWN’s server now needs a new batch of 0’s and 1’s delivered.
(Did you ever read “The Code Book”? I think you’d enjoy it.)
I’ll pick number six.
What I like about this thread is the amount of respect everyone has for each other and there is no fighting. Can anyone else think of another topic where we can schmooze like human beings? The moderators would like it.
noitallmr
Dr. Pepper- Your a genius!
Have you ever taken an I.Q test?????
Thanks for the compliment! But most of the stuff I posted is high school level.
Hi Everyone,
The swelling in my fingers from my post on Sunday finally went away and I can type again.
Getting back to that riddle with the teacher- when I was in 10th grade we were learning ???? ????? and in ??? ?? the end of second Mishna says ?????? ???? ?? ???????, ????? ???? ?? ??????–???? ???? ?? ????? ??????; ??? ????, ???? ????? ?? ??????? ???? ??? ?????. So our Rebbe asked us to add up the total amount of 1 + 2 + 3 + … + 60.
The plan I came up with was to add the numbers 1 through 10 (I used a shortcut for this, I already knew from the boxes of Chanukah candles that Telze sends out that the sum of the integers of 2 through 9 is 44 and 44 + 1 + 10 = 55), the sum from 11 to 20 is just 100 more and from 21 to 30 is just 100 more than that. Continuing we have 55 + 155 + 255 + 355 + 455 + 555 = 50 + 150 + 250 + 350 + 450 + 550 + 6×5 = 1800 + 30 = 1830.
It’s not as fast as the other one I posted and is impractical for large numbers and numbers not ending in 0, but it does work.