Current Events > 12! + 10! = 4a26b04c0

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FLOUR 08/25/23 5:40:02 PM #1:	Supposed you need to find the unknown digits but didn't want to compute this disgusting number by hand. How would you go about finding the digits? What are they? This would be an ideal number theory test question. --- You wish to fill Lionel's coffers? Yes, ha, ha, ha, yes!!! <cite>FLOUR posted [p:975662350] in 12! + 10! ... [t:80546597]...</cite> <quote>Supposed you need to find the unknown digits but didn't want to compute this disgusting number by hand. How would you go about finding the digits? What are they? This would be an ideal number theory test question. </quote> ... Copied to Clipboard!
Serious_Cat 08/25/23 5:59:54 PM #2:	I feel like I'm missing something here. 12! would include another 10 and both have a 5 and 2 to multiply so the third digit from the left would be another 0 rather than 4. --- I are Serious Cat This is serious thread <cite>Serious_Cat posted [p:975662708] in 12! + 10! ... [t:80546597]...</cite> <quote>I feel like I'm missing something here. 12! would include another 10 and both have a 5 and 2 to multiply so the third digit from the left would be another 0 rather than 4. </quote> ... Copied to Clipboard!
R1masher 08/25/23 6:01:08 PM #3:	Same way you did, ask somebody --- R1R1R1R1R1R1 <cite>R1masher posted [p:975662733] in 12! + 10! ... [t:80546597]...</cite> <quote>Same way you did, ask somebody </quote> ... Copied to Clipboard!
Trumble 08/25/23 6:01:39 PM #4:	Serious_Cat posted... I feel like I'm missing something here. 12! would include another 10 and both have a 5 and 2 to multiply so the third digit from the left would be another 0 rather than 4. Given that the answer is in hex, I'd assume 12 and 10 are too. --- See children, Trumbles are bad, and if you don't believe me ask your dad... <cite>Trumble posted [p:975662744] in 12! + 10! ... [t:80546597]...</cite> <quote>Serious_Cat posted... </cite><quote>I feel like I'm missing something here. 12! would include another 10 and both have a 5 and 2 to multiply so the third digit from the left would be another 0 rather than 4.</quote>Given that the answer is in hex, I'd assume 12 and 10 are too. </quote> ... Copied to Clipboard!
Homeless_Waifu 08/25/23 6:01:40 PM #5:	How does one even get letters? Sorry for my ignorance, i get lost when it comes to math --- I bought the Taito Egret II Mini just to play puchi carat <cite>Homeless_Waifu posted [p:975662746] in 12! + 10! ... [t:80546597]...</cite> <quote>How does one even get letters?Sorry for my ignorance, i get lost when it comes to math </quote> ... Copied to Clipboard!
Serious_Cat 08/25/23 6:03:22 PM #6:	Ah. I thought the letters were the unknowns. --- I are Serious Cat This is serious thread <cite>Serious_Cat posted [p:975662774] in 12! + 10! ... [t:80546597]...</cite> <quote>Ah. I thought the letters were the unknowns. </quote> ... Copied to Clipboard!
AngelsNAirwav3s 08/25/23 6:03:49 PM #7:	Trumble posted... Given that the answer is in hex, I'd assume 12 and 10 are too. (ie. They represent the decimal values 18 and 16) I think the numbers are base 10, and TC just wrote the unknown digits as a/b/c instead of the normal x/y/z --- Hello world! <cite>AngelsNAirwav3s posted [p:975662782] in 12! + 10! ... [t:80546597]...</cite> <quote>Trumble posted... </cite><quote>Given that the answer is in hex, I'd assume 12 and 10 are too. (ie. They represent the decimal values 18 and 16)</quote>I think the numbers are base 10, and TC just wrote the unknown digits as a/b/c instead of the normal x/y/z </quote> ... Copied to Clipboard!
Tyranthraxus 08/25/23 6:06:52 PM #8:	Serious_Cat posted... I feel like I'm missing something here. 12! would include another 10 and both have a 5 and 2 to multiply so the third digit from the left would be another 0 rather than 4. 1052 is 100 so that means c is zero, not the third digit. --- It says right here in Matthew 16:4 "Jesus doth not need a giant Mecha." https://i.imgur.com/dQgC4kv.jpg <cite>Tyranthraxus posted [p:975662829] in 12! + 10! ... [t:80546597]...</cite> <quote>Serious_Cat posted... </cite><quote>I feel like I'm missing something here. 12! would include another 10 and both have a 5 and 2 to multiply so the third digit from the left would be another 0 rather than 4.</quote>1052 is 100 so that means c is zero, not the third digit. </quote> ... Copied to Clipboard!
Giacomo_Hawkins 08/25/23 6:10:52 PM #9:	Serious_Cat posted... I feel like I'm missing something here. 12! would include another 10 and both have a 5 and 2 to multiply so the third digit from the left would be another 0 rather than 4. You're on the right track, but not quite. The 5 and 2 mean that all factorials 5! or higher end in a zero. The 10 adds a second zero digit. 15 would add a third. 20 adds a fourth. Ect. Adding two factorials 10! or higher does not mean that the third digit would become zero, but you can safely conclude that the 'c' is zero. --- Will the little voice in the back of my mind screaming "This is a bad idea" please yield the floor. --Mikey Chivalry be hanged, and so will you. <cite>Giacomo_Hawkins posted [p:975662879] in 12! + 10! ... [t:80546597]...</cite> <quote>Serious_Cat posted... </cite><quote>I feel like I'm missing something here. 12! would include another 10 and both have a 5 and 2 to multiply so the third digit from the left would be another 0 rather than 4.</quote>You're on the right track, but not quite.The 5 and 2 mean that all factorials 5! or higher end in a zero. The 10 adds a second zero digit. 15 would add a third. 20 adds a fourth. Ect.Adding two factorials 10! or higher does not mean that the third digit would become zero, but you can safely conclude that the 'c' is zero. </quote> ... Copied to Clipboard!
Giacomo_Hawkins 08/25/23 6:20:36 PM #10:	Also, echoes of this topic. https://gamefaqs.gamespot.com/boards/400-current-events/80544903 I'm sensing a theme here. --- Will the little voice in the back of my mind screaming "This is a bad idea" please yield the floor. --Mikey Chivalry be hanged, and so will you. <cite>Giacomo_Hawkins posted [p:975663034] in 12! + 10! ... [t:80546597]...</cite> <quote>Also, echoes of this topic.https://gamefaqs.gamespot.com/boards/400-current-events/80544903I'm sensing a theme here. </quote> ... Copied to Clipboard!
Justin2Krelian 08/25/23 6:24:44 PM #11:	Reminds me of when someone used a number that was something like 111!!!!!!!!!! in a "biggest numbers" contest. --- -J2K Currently Streaming: The Witcher, Homeland, Travelers, The Bear <cite>Justin2Krelian posted [p:975663100] in 12! + 10! ... [t:80546597]...</cite> <quote>Reminds me of when someone used a number that was something like 111!!!!!!!!!! in a "biggest numbers" contest. </quote> ... Copied to Clipboard!
Giacomo_Hawkins 08/25/23 7:45:16 PM #12:	Justin2Krelian posted... Reminds me of when someone used a number that was something like 111!!!!!!!!!! in a "biggest numbers" contest. Not as large a number as they may have intended: https://mathworld.wolfram.com/Multifactorial.html Every additional exclamation point makes the result smaller. --- Will the little voice in the back of my mind screaming "This is a bad idea" please yield the floor. --Mikey Chivalry be hanged, and so will you. <cite>Giacomo_Hawkins posted [p:975664582] in 12! + 10! ... [t:80546597]...</cite> <quote>Justin2Krelian posted... </cite><quote>Reminds me of when someone used a number that was something like 111!!!!!!!!!! in a "biggest numbers" contest.</quote>Not as large a number as they may have intended:https://mathworld.wolfram.com/Multifactorial.htmlEvery additional exclamation point makes the result smaller. </quote> ... Copied to Clipboard!
BewmHedshot 08/25/23 7:55:18 PM #13:	Are we like pretending we don't have a calculator? 482 million isn't remotely a computationally intractable number where you need modular arithmetic. <cite>BewmHedshot posted [p:975664807] in 12! + 10! ... [t:80546597]...</cite> <quote>Are we like pretending we don't have a calculator? 482 million isn't remotely a computationally intractable number where you need modular arithmetic.</quote> ... Copied to Clipboard!
Anteaterking 08/25/23 8:05:49 PM #14:	You can use modular arithmetic for problems of this sort (reducing things modulo 100 will give you the bottom two digits and you only need to track those two values throughout which makes it easier to do things by hand). e.g. I can tell you that 3^(5000000) ends 01 because 3^20 = 1 mod 100. This method doesn't really help that much with factorials though because the digits are all slowly annihilated by the 5s and finding digits closer to the front would be basically the same amount of memory and work as just calculating the factorial. --- http://i18.photobucket.com/albums/b136/Anteaterking/scan00021.jpg http://i18.photobucket.com/albums/b136/Anteaterking/scan00021.jpg <cite>Anteaterking posted [p:975665039] in 12! + 10! ... [t:80546597]...</cite> <quote>You can use modular arithmetic for problems of this sort (reducing things modulo 100 will give you the bottom two digits and you only need to track those two values throughout which makes it easier to do things by hand).e.g. I can tell you that 3^(5000000) ends 01 because 3^20 = 1 mod 100.This method doesn't really help that much with factorials though because the digits are all slowly annihilated by the 5s and finding digits closer to the front would be basically the same amount of memory and work as just calculating the factorial. </quote> ... Copied to Clipboard!
Justin2Krelian 08/25/23 8:58:45 PM #15:	Giacomo_Hawkins posted... Not as large a number as they may have intended: https://mathworld.wolfram.com/Multifactorial.html Every additional exclamation point makes the result smaller. Interesting, though they must have intended the factorial of the factorial of the factorial... and so on. These were smart mathematicians so there must be some detail I'm missing. --- -J2K Currently Streaming: The Witcher, Homeland, Travelers, The Bear <cite>Justin2Krelian posted [p:975665968] in 12! + 10! ... [t:80546597]...</cite> <quote>Giacomo_Hawkins posted... </cite><quote>Not as large a number as they may have intended:https://mathworld.wolfram.com/Multifactorial.htmlEvery additional exclamation point makes the result smaller.</quote>Interesting, though they must have intended the factorial of the factorial of the factorial... and so on. These were smart mathematicians so there must be some detail I'm missing. </quote> ... Copied to Clipboard!
thisworld 08/25/23 9:23:34 PM #16:	So basically I have to find out a, b, and c? Fine, let's do this by hand. 12! + 10! = 4a26b04c0 (12 * 11 * 10!) + 10! = 4a26b04c0 (132 * 10!) + 10! = 4a26b04c0 (132 + 1) * 10! = 4a26b04c0 133 * 10! = 4a26b04c0 c = 0 because the left side of the equation has 10 * 5 * 2 = 100 as factor (comes from 10!). Divide the whole equation by (10 * 5 * 2) 133 * (9 * 8 * 7 * 6 * 4 * 3) = 4a26b04 Find the prime factorization for the left side of the equation (7 * 19) * (3^2 * 2^3 * 7 * (23) 2^2 * 3) = 4a26b04 2^6 * 3^4 * 7^2 * 19 = 4a26b04 Left side has 2^6 = 64 as a factor. This includes 32, 16, 8, 4, and 2 as factors too obviously. Observe the divisibility rule with 2^x as divisor: -------------------------------- 2 as factor = the last digit is divisable by 2 4 as factor = the last 2 digits are divisable by 4 ... 64 as factor = the last 6 digits are divisable by 64 -------------------------------- 2^6 * 3^4 * 7^2 * 19 = 4a26b04 4a26b04 has 2^5 = 32 as a factor which means the last 5 digits, 26b04, are divisable by 64. Calculate 26b04/64 by hand with all the possible combination (b = 0, b = 1 ... b = 9). Only one possible value of b makes 26b04 divisable by 64 and that is b = 3 2^6 * 3^4 * 7^2 * 19 = 4a26b04 2^6 * 3^4 * 7^2 * 19 = 4a26304 Left side has 3^4 = 81 as a factor which means 27, 9, and 3 as factors too. Observe the divisibility rule with 9 as divisor: -------------------------------- Sum the digits. The result must be divisible by 9 -------------------------------- 4a26304 has 9 as a factor. 4+a+2+6+3+0+4 = 19+a must be divisable by 9. a = 8 c=0, b=3, a=8 12! + 10! = 4a26b04c0 12! + 10! = 482630400 QED ...wtf did i just do? doing someone's homework on gamefaqs lol <cite>thisworld posted [p:975666399] in 12! + 10! ... [t:80546597]...</cite> <quote>So basically I have to find out a, b, and c? Fine, let's do this by hand.12! + 10! = 4a26b04c0(12 * 11 * 10!) + 10! = 4a26b04c0(132 * 10!) + 10! = 4a26b04c0(132 + 1) * 10! = 4a26b04c0133 * 10! = 4a26b04c0c = 0 because the left side of the equation has 10 * 5 * 2 = 100 as factor (comes from 10!). Divide the whole equation by (10 * 5 * 2)133 * (9 * 8 * 7 * 6 * 4 * 3) = 4a26b04Find the prime factorization for the left side of the equation(7 * 19) * (3^2 * 2^3 * 7 * (23) 2^2 * 3) = 4a26b042^6 * 3^4 * 7^2 * 19 = 4a26b04Left side has 2^6 = 64 as a factor. This includes 32, 16, 8, 4, and 2 as factors too obviously. Observe the divisibility rule with 2^x as divisor:-------------------------------- 2 as factor = the last digit is divisable by 2 4 as factor = the last 2 digits are divisable by 4 ... 64 as factor = the last 6 digits are divisable by 64--------------------------------2^6 * 3^4 * 7^2 * 19 = 4a26b044a26b04 has 2^5 = 32 as a factor which means the last 5 digits, 26b04, are divisable by 64. Calculate 26b04/64 by hand with all the possible combination (b = 0, b = 1 ... b = 9). Only one possible value of b makes 26b04 divisable by 64 and that is b = 32^6 * 3^4 * 7^2 * 19 = 4a26b042^6 * 3^4 * 7^2 * 19 = 4a26304Left side has 3^4 = 81 as a factor which means 27, 9, and 3 as factors too. Observe the divisibility rule with 9 as divisor:--------------------------------Sum the digits. The result must be divisible by 9--------------------------------4a26304 has 9 as a factor. 4+a+2+6+3+0+4 = 19+a must be divisable by 9. a = 8c=0, b=3, a=812! + 10! = 4a26b04c012! + 10! = 482630400QED...wtf did i just do? doing someone's homework on gamefaqs lol </quote> ... Copied to Clipboard!
Anteaterking 08/25/23 9:32:07 PM #17:	thisworld posted... So basically I have to find out a, b, c without doing a single multiplication operation? thisworld posted... c = 0 because the left side of the equation has *10 5 * 2 = 100** as factor --- http://i18.photobucket.com/albums/b136/Anteaterking/scan00021.jpg http://i18.photobucket.com/albums/b136/Anteaterking/scan00021.jpg <cite>Anteaterking posted [p:975666531] in 12! + 10! ... [t:80546597]...</cite> <quote>thisworld posted... </cite><quote>So basically I have to find out a, b, c without doing a single multiplication operation?</quote>thisworld posted... </cite><quote>c = 0 because the left side of the equation has 10 * 5 * 2 = 100 as factor</quote> </quote> ... Copied to Clipboard!
thisworld 08/25/23 9:55:15 PM #18:	Anteaterking posted... thisworld posted... So basically I have to find out a, b, c without doing a single multiplication operation? thisworld posted... c = 0 because the left side of the equation has *10 5 * 2 = 100** as factor lmao hahahaha! okay okay... Goddammit I had a brain fart there. You're absolutely right though. It is a multiplication operation, no matter how simple. I erased that 'no multiplication' stuff. Don't tell anyone... lol <cite>thisworld posted [p:975666945] in 12! + 10! ... [t:80546597]...</cite> <quote>Anteaterking posted...</cite><quote>thisworld posted...</cite><quote>So basically I have to find out a, b, c without doing a single multiplication operation? </quote>thisworld posted...</cite><quote>c = 0 because the left side of the equation has 10 * 5 * 2 = 100 as factor</quote></quote>lmao hahahaha! okay okay... Goddammit I had a brain fart there. You're absolutely right though. It is a multiplication operation, no matter how simple.I erased that 'no multiplication' stuff. Don't tell anyone... lol</quote> ... Copied to Clipboard!
thisworld 08/26/23 12:21:18 AM #19:	Jesus I just realized that I made a typo up there! thisworld posted... 4a26b04 has 2^5 = 32 as a factor which means the last 5 digits, 26b04, are divisable by 64. Calculate 26b04/64 by hand with all the possible combination (b = 0, b = 1 ... b = 9). Only one possible value of b makes 26b04 divisable by 64 and that is b = 3 It is 32 NOT 64. It should've been: 4a26b04 has 2^5 = 32 as a factor which means the last 5 digits, 26b04, are divisable by 32. Calculate 26b04/32 by hand with all the possible combination (b = 0, b = 1 ... b = 9). Only one possible value of b makes 26b04 divisable by 32 and that is b = 3 I actually did the proper calculation using 32 as divisor. Bloody hell I have no idea how that '64' managed to slip in there >.< I'm sorry everyone. <cite>thisworld posted [p:975669016] in 12! + 10! ... [t:80546597]...</cite> <quote>Jesus I just realized that I made a typo up there!thisworld posted...</cite><quote>4a26b04 has 2^5 = 32 as a factor which means the last 5 digits, 26b04, are divisable by 64. Calculate 26b04/64 by hand with all the possible combination (b = 0, b = 1 ... b = 9). Only one possible value of b makes 26b04 divisable by 64 and that is b = 3</quote>It is 32 NOT 64. It should've been:<quote>4a26b04 has 2^5 = 32 as a factor which means the last 5 digits, 26b04, are divisable by 32. Calculate 26b04/32 by hand with all the possible combination (b = 0, b = 1 ... b = 9). Only one possible value of b makes 26b04 divisable by 32 and that is b = 3</quote>I actually did the proper calculation using 32 as divisor. Bloody hell I have no idea how that '64' managed to slip in there >.<I'm sorry everyone.</quote> ... Copied to Clipboard!
ChocoboMogALT 08/26/23 12:36:56 AM #20:	thisworld posted... I actually did the proper calculation using 32 as divisor. Bloody hell I have no idea how that '64' managed to slip in there >.< I'm sorry everyone. That's what you get for throwing in the QED. --- "We live in a country Hasire.." ~ yosouf06 REVOLVER STAKE! http://img.photobucket.com/albums/v717/ChocoboMog123/AltEisenRChocoboMog.png <cite>ChocoboMogALT posted [p:975669221] in 12! + 10! ... [t:80546597]...</cite> <quote>thisworld posted... </cite><quote>I actually did the proper calculation using 32 as divisor. Bloody hell I have no idea how that '64' managed to slip in there >.<I'm sorry everyone.</quote>That's what you get for throwing in the QED. </quote> ... Copied to Clipboard!
thisworld 08/26/23 12:55:09 AM #21:	ChocoboMogALT posted... That's what you get for throwing in the QED. Haha yeah. Apparently my brain was out of sync with my hand or something during that particular part. Still can't believe I missed that!! Well lesson learned, gamefaqs and math don't mix <cite>thisworld posted [p:975669412] in 12! + 10! ... [t:80546597]...</cite> <quote>ChocoboMogALT posted...</cite><quote>That's what you get for throwing in the QED.</quote>Haha yeah. Apparently my brain was out of sync with my hand or something during that particular part. Still can't believe I missed that!! Well lesson learned, gamefaqs and math don't mix </quote> ... Copied to Clipboard!
Robot2600 08/26/23 1:29:53 AM #22:	okay now explain how in the fuck that was easier than computing on scratch paper --- --- <cite>Robot2600 posted [p:975669758] in 12! + 10! ... [t:80546597]...</cite> <quote>okay now explain how in the fuck that was easier than computing on scratch paper </quote> ... Copied to Clipboard!
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Current Events > 12! + 10! = 4a26b04c0 (+) Yeah! (+)

Current Events > 12! + 10! = 4a26b04c0