Скачать или смотреть Class 7 Math | Ch-6 Number Play | 6.5 Digits in Disguise | Page 142 - 143

Are you ready to solve the most interesting puzzles in your math book? 🕵️‍♂️ In this video, we dive into "Digits in Disguise," also known as Cryptarithms or Alphametics, from Chapter 6, 'Number Play' of the Class 7 Maths curriculum. These puzzles replace digits with letters, and your mission is to crack the code!

We'll teach you the fundamental rules and logic to solve any cryptarithm problem with ease. Forget memorizing, and start understanding the simple tricks to find which digit each letter stands for. This is a perfect brain teaser and a fun way to improve your logical reasoning and problem-solving skills!

Join us as we break down every single example and question from page 142-143, step-by-step!

⏰ TIMESTAMPS:
0:00 - Introduction to Cryptarithms, also known as Digits in Disguise
1:15 - The Basic Rules of Alphametics
2:30 - Puzzle 1: T + T + T = UT - The Logic Explained!
4:45 - Puzzle 2: K2 + K2 = HMM - Finding H and M
6:50 - Practice Problem 1: YY + Z = ZOO
8:30 - Practice Problem 2: B5 + 3D = ED5
10:12 - Practice Problem 3: KP + KP = PRR
12:05 - Practice Problem 4: C1 + C = 1FF
14:00 - Summary of Tricks and Tips

🔢 Problems Solved in this Video:

T + T + T = UT

Logic: The result of 3 x T is a number where the units digit is T itself. This only works for T=0 or T=5. Since the sum UT is a 2-digit number, T can't be 0.

Solution: T = 5. So, 5 + 5 + 5 = 15. This means U = 1.

K2 + K2 = HMM

Logic: In the units place, 2 + 2 = 4. So, M = 4. In the tens place, K + K = HM. Since M is 4, the sum is H44. This means K+K results in a number ending with 4. Possibilities for K are 2 or 7. K can't be 2 as it's already used. Let's try K=7. 72 + 72 = 144.

Solution: K = 7, H = 1, M = 4.

YY + Z = ZOO

Logic: Adding a 2-digit number and a 1-digit number gives a 3-digit number. This is only possible if the sum just crosses over 100. This makes Z in the hundreds place 1. So, YY + 1 = 1OO. Now, Y+1 must end in O. Looking at the tens place, Y plus a carry results in 1O. This implies Y=9.

Solution: 99 + 1 = 100. So Y = 9, Z = 1, O = 0.

B5 + 3D = ED5

Logic: In the units place, 5 + D must end in 5. This means D must be 0. The sum is now B5 + 30 = E05. This means a carry occurred. So, in the tens place, B + 3 = 10.

Solution: B = 7. The sum is 75 + 30 = 105. So E = 1, D = 0.

KP + KP = PRR

Logic: This is 2 times KP equals PRR. The result is a 3-digit number, so the carry from the tens column is P. This means P must be 1. Now, in the units column, 1 + 1 = R, so R=2. In the tens column, K + K must result in a number ending in R=2 with a carry of P=1. So 2K = 12.

Solution: K = 6. The sum is 61 + 61 = 122. So K = 6, P = 1, R = 2.

C1 + C = 1FF

Logic: In the units place, 1 + C results in F with a carry. In the tens place, C plus that carry results in 1F. The only single digits where C - F = 9 are C=9 and F=0.

Solution: C = 9, F = 0. The sum is 91 + 9 = 100. So 1FF is 100.

Did you solve them differently? Let us know your method in the comments below! 👇

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