Table of Contents
Puzzle From A Math Teacher – If AAA + BBB + CCC = BAAC, What Are A, B, C = ? | ข่าวทั่วไปรายวัน
[penci_button link = “#” icon = “” icon_position = “left”] เห็นแล้ว [/penci_button]
คุณสามารถดู baac ข่าวหรือข่าวอื่น ๆ ที่เกี่ยวข้องมากขึ้น เราด้วย แบ่งปัน Puzzle From A Math Teacher – If AAA + BBB + CCC = BAAC, What Are A, B, C = ? และรูปภาพที่เกี่ยวข้อง baac
baac และข้อมูลที่เกี่ยวข้อง
Luuk ทำให้ครูคณิตศาสตร์ของเขาสะดุดกับหนึ่งในปริศนาของฉัน จากนั้นครูคณิตศาสตร์ของเขาก็ให้ปัญหาที่ลูกแนะนำแก่ฉัน ถ้า AAA + BBB + CCC = ธ.ก.ส. แต่ละหลักมีค่าเท่าไร? ตัวเลขมีความชัดเจนและเป็นจำนวนเต็มตั้งแต่ 1 ถึง 9 ดูวิดีโอเพื่อดูวิธีแก้ปัญหา บล็อกโพสต์ของฉันสำหรับวิดีโอนี้ หากคุณชอบวิดีโอของฉัน คุณสามารถสนับสนุนฉันได้ที่ Patreon: Connect บนโซเชียลมีเดีย ฉันอัปเดตแต่ละไซต์เมื่อมีวิดีโอหรือบล็อกโพสต์ใหม่ ดังนั้นคุณสามารถติดตามฉันได้ด้วยวิธีใดก็ได้ที่สะดวกที่สุดสำหรับคุณ บล็อกของฉัน: Twitter:
Facebook: Google+: Pinterest: Tumblr: Instagram: Patreon: จดหมายข่าว (ส่งประมาณปีละ 2 ครั้ง): หากคุณซื้อจากลิงก์ด้านล่าง ฉันอาจได้รับค่าคอมมิชชั่นสำหรับการขาย สิ่งนี้ไม่มีผลกระทบต่อราคาสำหรับคุณ หนังสือของฉัน “The Joy of Game Theory” แสดงให้เห็นว่าคุณสามารถใช้คณิตศาสตร์เพื่อเอาชนะคู่แข่งได้อย่างไร (ได้คะแนน 3.8/5 ดาวจาก 31 บทวิจารณ์) “ภาพลวงตาที่ไร้เหตุผล: วิธีการตัดสินใจอย่างชาญฉลาดและเอาชนะอคติ” เป็นคู่มือที่อธิบายวิธีการต่างๆ ที่เราลำเอียงเกี่ยวกับการตัดสินใจและเทคนิคในการตัดสินใจอย่างชาญฉลาด (ได้คะแนน 5/5 ดาวจาก 2 รีวิว) “Math Puzzles Volume 1” นำเสนอของเล่นพัฒนาสมองและปริศนาคลาสสิกพร้อมวิธีแก้ปัญหาที่สมบูรณ์สำหรับปัญหาในการนับ เรขาคณิต ความน่าจะเป็น และทฤษฎีเกม เล่มที่ 1 ได้รับการจัดอันดับ 4.4/5 ดาวจาก 13 บทวิจารณ์ “Math Puzzles Volume 2” เป็นหนังสือภาคต่อที่มีปัญหามากกว่า (ให้คะแนน 5/5 ดาวจากบทวิจารณ์ 3) “Math Puzzles Volume 3” เป็นชุดที่สามในซีรีส์ (ได้คะแนน 3.8/5 ดาวจาก 4 บทวิจารณ์) “40 Paradoxes ในตรรกะ ความน่าจะเป็น และทฤษฎีเกม” มีผลลัพธ์ที่กระตุ้นความคิดและตอบโต้กับสัญชาตญาณ (ได้คะแนน 4.3/5 stars จาก 12 รีวิว) “The Best Mental Math Tricks” สอนวิธีที่คุณสามารถดูเหมือนอัจฉริยะทางคณิตศาสตร์โดยการแก้ปัญหาในหัวของคุณ (ได้คะแนน 4.7/5 stars จาก 4 รีวิว) “Multiply Numbers By Drawing Lines” This หนังสือเป็นคู่มืออ้างอิงสำหรับวิดีโอของฉันซึ่งมีการดูมากกว่า 1 ล้านครั้งในวิธีเรขาคณิตเพื่อคูณตัวเลข (ให้คะแนน 5/5 ดาวจาก 3 รีวิว) .
>>> คุณยังสามารถดูข้อมูลที่น่าสนใจเพิ่มเติมได้ที่นี่ us also
แบ่งปันที่นี่
#Puzzle #Math #Teacher #AAA #BBB #CCC #BAAC.
mathematics,math,maths,riddle,brain teaser,puzzle,math puzzle,algebra,logic puzzle,addition puzzle.
Puzzle From A Math Teacher – If AAA + BBB + CCC = BAAC, What Are A, B, C = ?.
[คีย์เวิร์ด].
เราหวังว่าคุณจะพบข้อมูลเกี่ยวกับ baac ที่นี่
ขอบคุณที่รับชมเนื้อหานี้
it took me a while but I could do it. Great channel!!
That's the second easiest problem you've put on your channel!
guessing? there are clues right there – solved it right away
So many solutions. I came up with another path to victory.
AAA
+BBB
+CCC
—–
BAAC
1. Highest possible solution: 9+8+7=24= <30
That means: B=2 OR B=1
looking at the right column:
9+B+8=19 OR 18, wich means reminder can max be 1
We've got
B=1
AAA
+111
+CCC
—–
1AAC
Now taking the right column doing some maths (Or something related)
A+1+C = 10+C | -1 (10 comes from the reminder on the left, which carries over to the most left column)
A+C = 9+C | -C
A=9
We've got
B=1
A=9
999
+111
+CCC
—–
199C
Now looking at the center column
9+1+C+1=10+9 (second +1 on the left comes from the reminder from the right, which is carried over to the left)
11+C=19 | -11
C=8
We've got
B=1
A=9
C=8
999
+111
+888
—–
1998
Pretty sure some or most of you did that faster and different than me, but that's okay. I solved it myself. I'm proud of myself.
Did it without guessing. First observed as you did that A+B=10, or the first (ones) column wouldn’t work. Then, for the tens column to work, 10+1+C=A+10. If A+B=10, then A+B+C<20. Therefore B=1. Therefore, A=9 because A+B=10. Therefore, C=8 because 10+1+C=A+10, or 1+C=A. Done.
A-9( B_1 C-8 using my head
You don't need to give values to B in fact :
With the first column, you determine that A + B = 10 (first eqution) you get C as unite and A and B are different so they can't be both equal to 0 so A +B +C = C +10 in the first column which is correct with the assumption.
Then comes the second column, re-written as A+B+C+1=C+10 which gives a second equation : A = C+1
Then for the third column, A+B+C+1=10B+A which leads to B=1 with the two prvious equations. There goes A=9 and C=8.
I believe he found out the same digits, I haven't watched the full video.
Too simple and requires NO guessing or checking. C in the units sum means A+B is 10, and the tens carry means A is one more than C. B can only be 1 or 2 since the largest A+B+C is 24, but that would require B greater than 2, thus B is 1. A+1=10 means A=9, and A-1=C means C=8. QED.
I had a different method, because A+B+C = 1C, to get C in the units place and A in the 10s and 100s place, that means that A+B = 10. and that A+B+C+1 = 1A. Because these numbers each represent singular digits, they can't sum to more than 19, and B cannot equal zero. So, B must be 1. From that, we can deduce that A+1 = 10 by substituting B. So A = 9. So then we just plug A = 9 and B = 1 into A+B+C+1 = 1A. to get 9+1+1+C = 19. –> 11+c = 19. –> C = 8. Just plug those values for A B and C into the original equation, and you get 999 + 111 + 888 = 1998, which is true. So A = 9, B = 1, and C =8.
I finally did it
Okay, question. In this video, it seems like you make an assumption that A + 1 + C = C + 10, or that A + 2 + C = C + 10. Why can't they be equal to say, C + 20, or C + 30, etc?
A+B+C =X+C
So A+B=X.
X has to be 10 as there is no way that with the numbers 1 through 9 that 2 of them can equal 20
In the second column there is effectively 10A+10B+10c +10 =10X+10A so dividing by 10 you have A+B+C+1=X+A
Since A+B=X we can drop that to get C+1=A
Knowing that A+B+C can not be above 20 due to A+B being 10, B has to be the number 1.
So 1998
Different way to the answer but still works.
I limited B the same way you did, but then I repeated the same process and found that 999+222+888 was smaller than 2000, so that meant I had only one valid value for B. Next I noticed that A+2+C = 10+A and A+1+C= 10+C, so C+1=A. Finally I substituted that in the first digits A+B+C = 2*C + 2 = 10 + C. That ment C was even and greater than 5. 6 didn't work, but 8 did.
This is the type of equation on a test where you answer "The first 3 letters of the alphabet" and move on
Can you solve by trying to get simultaneous equations ?
I assumed that a plus b plus c is more than 10 but not more than 20. And it worked out
So logically
This is the first one I actually got right the first try
111A+111B+111C=1000B + 110A + C
i.e. A + 110C = 889B
Instantly we can think of only one combination i.e. A = 9, B = 1 & C = 8
But that's not the correct way I know. So loved the solution you gave. 👍👍
2:55 sir can you please tell how did you figure out that A+1+C=C+10?
Probably overthinking here but it does seem a missed opportunity to motivate some algebra. 111(A+B+C)=1000 B+110 A+C ==> A+110 C = 889 B. B can't be 2 or more, must be 1 so A + 110C = 889. Clearly A=9 leaving C=8.
Reminds me of a problem we were given at school" SAVE + MORE = MONEY. Solved in a similar way to this one
I used some logic from a similar question posted here earlier and voila! I was able to solve it in 2 minutes!
I figured it out before watching the video, but I approached it slightly differently. I didn't worry about figuring out what the largest possible sum is. Instead, I just considered the units and the tens columns to start. A+B+C=C will only work if A+B=10. Right off the bat, we can exclude 5 as a possibility for either A or B, as these are different digits and 10-5=5. This also means that we have a carry of 1. With a carry of 1, the tens column becomes A+B+C+1=A. Since A+B+C=C, we can substitute C for A+B+C, making the equation C+1=A. This means that A and C are consecutive integers, with A being the next digit after C. From here, there's only 7 possible solutions. Since A is 1 more than C, A cannot be 1(as that would mean C is 0, and we only have the digits from 1 to 9) and we have already established that it cannot be 5 either(as that would mean that B has to be 5 too, and A, B, and C are all different). This means that A can only be 2, 3, 4, 6, 7, 8, or 9. Once I had these facts established, I decided to try the largest possible values for A and C, 9 and 8, first. Keeping in mind that A+B=10, that makes B=1. Checking this, it turns out to be the answer.
PUZZLE dt 9.5.2021 –
If A, B and C are distinctive and positive integers, decipher their values in the equation –
A B C
A B C
+ A B C
————–
B B A
————–
The guess and check step isn't necessary. Here's a solution in five steps:
(1) Looking at the units column, it's clear that A + B = 10, so 1 must carry over to the 10s column.
(2) The sum of the 10s column, A, has to be C + 1 (the carry over), so we note that A = C + 1.
(3) In the 100s column, there's a final carry over of 1 to the 1000s, so we see that B = 1.
(4) From step 1, A + 1 = 10, so se see that A = 9.
(5) From step 2, we note that C = A – 1 = 9 – 1, therefore C = 8.
This one is easy
We have
A+B+C=C
That means
A+B=10 with this one we know that A+B+C<20 and that means B = 1.
But we also have
A+B(1)+C=A
That means
A=C+1
Now we just have to replace the letters
C+1+1+C-10=C (-10 cause A+B+C≥10)
C=8
A=8+1
A=9
A=9
B=1
C=8
999
111
+888
1998
As long u can get y A+B =10 ….it's easy therafter.
At 3:06, How did u write that step – "A+1+C = C+1"?
Plzz explain!!
Normally I like your solutions as they are generally very simple and elegant. This one however fell flat as, shown by Lee Holzer, didn't require any guessing and was possible to solve completely analytically.
You cancel put the c at the last digit imma leave it for unknown right now
So AAA+BBB+CC0 = BAA0
A+B = 10 so the B at the end is 1 so if you know algebra A+1=10 so A=9
Because at the last digit the C is unknown we know that the second last digit if we leave out the A+B you get C+1=A because A is 9 if you do more algebra C is 8
So you have A=9 B=1 C=8 it really isn't difficult if your not overthinking
You cancel put the c at the last digit imma leave it for unknown right now
So AAA+BBB+CC0 = BAA0
A+B = 10 so the B at the end is 1 so if you know algebra A+1=10 so A=9
Because at the last digit the C is unknown we know that the second last digit if we leave out the A+B you get C+1=A because A is 9 if you do more algebra C is 8
So you have A=9 B=1 C=8 it really isn't difficult if your not overthinking
I solved it