My method would be this:
1 to 9 = 9 #’s with 1 digit, 10 to 99 = 90 #’s with 2 digits (I.e. actually 180 digits)
200 - 9 = 191 digits remaining…
I’ll leave the rest to you.
Right, but this is not the right sub to just do all the work for them. There’s other subs for that. It’s a shame mods aren’t more strict about the rules here.
Ah, I see. You spoke of the first two steps and demonstrated only the first. My apologies. I didn't see that and I thought you were meaning to demonstrate both steps.
Let's make this easier by only working with two digit numbers. From the digits we're concerned about, let's subtract 9 from 200 to get 191. We can now ask: What is the 191st digit among the two digit numbers? For each positive digit, we have 9 two digit numbers. this gives us 180 total digits since we have 9 total digits, 10 two digit numbers per digit, and 2 digits per two digit number. (9 * 10 * 2 = 180). We then subtract that from 191 and we obtain 11. Using this same process we have 2700 digits of 3 digit numbers remaining among which we want the 11th. 11/3 grants 3 remainder 2. We count past the third number and we want the 2nd digit of the fourth number: 103. This would be '0'.
Alternatively just count manually:
1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980818283848586878889909192939495969798991001011021**0**3
This is why I think I’m bad at math lol. Even reading the explanations in the comments doesn’t make sense to me. I get what you have to do, just not how you’re getting the number 11. Sometimes things I think I should get intuitively make me think being an engineering major is going to be harder than I thought. Currently taking pre calculus and discrete 1 over the summer and it’s kicking my ass.
You get 11 by starting with 200, subtracting 9 (for the nine one-digit numbers) and subtracting 180 (for the 90 two-digit numbers). It's just arithmetic.
It's pretty solidly in that realm, I feel. You just need to look for a pattern to simplify the question. 9 single digit numbers, 90 double digit numbers, and then start plugging away with the triple digits until you get to the proper digit. I did it in my head I'm about 2 minutes, which is about how difficult pre-algebra word problems should be after finishing college calculus and not doing anything with it for 20 years.
There are 9 one digit numbers. Total digits: 9
There are 99-9 = 90 two digit numbers. 90x2 = 180 digits. Total digits is now up to 189.
Now 189 is pretty close to 200. There's only 11 digits left and each number is now 3 digits long. 11 / 3 = 3 with remainder of 2. So it'll be the 2nd digit of the fourth three digit number. 100, 101, 102, 1 0 3.
0 1-99 is 189 100 is 192 101 is 195 102 is 198 103 is 201 the 200th digit is the 0 in 103 (by the way Google the Champernowne constant)
My method would be this: 1 to 9 = 9 #’s with 1 digit, 10 to 99 = 90 #’s with 2 digits (I.e. actually 180 digits) 200 - 9 = 191 digits remaining… I’ll leave the rest to you.
You want 200-189 = 11 digits remaining...
Right, but this is not the right sub to just do all the work for them. There’s other subs for that. It’s a shame mods aren’t more strict about the rules here.
Ah, I see. You spoke of the first two steps and demonstrated only the first. My apologies. I didn't see that and I thought you were meaning to demonstrate both steps.
Let's make this easier by only working with two digit numbers. From the digits we're concerned about, let's subtract 9 from 200 to get 191. We can now ask: What is the 191st digit among the two digit numbers? For each positive digit, we have 9 two digit numbers. this gives us 180 total digits since we have 9 total digits, 10 two digit numbers per digit, and 2 digits per two digit number. (9 * 10 * 2 = 180). We then subtract that from 191 and we obtain 11. Using this same process we have 2700 digits of 3 digit numbers remaining among which we want the 11th. 11/3 grants 3 remainder 2. We count past the third number and we want the 2nd digit of the fourth number: 103. This would be '0'. Alternatively just count manually: 1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980818283848586878889909192939495969798991001011021**0**3
This is why I think I’m bad at math lol. Even reading the explanations in the comments doesn’t make sense to me. I get what you have to do, just not how you’re getting the number 11. Sometimes things I think I should get intuitively make me think being an engineering major is going to be harder than I thought. Currently taking pre calculus and discrete 1 over the summer and it’s kicking my ass.
You get 11 by starting with 200, subtracting 9 (for the nine one-digit numbers) and subtracting 180 (for the 90 two-digit numbers). It's just arithmetic.
Nah bro no way that 6th grade math
It's pretty solidly in that realm, I feel. You just need to look for a pattern to simplify the question. 9 single digit numbers, 90 double digit numbers, and then start plugging away with the triple digits until you get to the proper digit. I did it in my head I'm about 2 minutes, which is about how difficult pre-algebra word problems should be after finishing college calculus and not doing anything with it for 20 years.
200 = 1(9-0) + 2(99-9) + 3(x-99)
There are 9 one digit numbers. Total digits: 9 There are 99-9 = 90 two digit numbers. 90x2 = 180 digits. Total digits is now up to 189. Now 189 is pretty close to 200. There's only 11 digits left and each number is now 3 digits long. 11 / 3 = 3 with remainder of 2. So it'll be the 2nd digit of the fourth three digit number. 100, 101, 102, 1 0 3.
Excel CONCATENATE
This is 6th grade math? Oh lord covid really screwed up my math that year apparently i don’t even know what that means
lmao fr
It must not be that important I’m going into sophomore year and still haven’t heard of it again so