Cryptarithms: recovering digits from constraints alone

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What the video covers, what to say before it, where a class usually goes wrong, and what to set afterwards. An account is free, and it opens every chapter of every book.

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What to assume they know

Column addition with carrying, and knowing which column feeds which
Place value in two- and three-digit numbers: what the tens digit contributes
Predicting how many digits a product will have, before multiplying: reasoning about how many digits an answer will have before computing it
The parity of a sum or product is fixed before you compute it: settling a digit or a parity from a constraint rather than from a calculation
Multiples of a one-digit number, and reading off the units digit of a product

What they should be able to do

Read a letter-for-digit addition and say what each column asserts
Identify which column of a given puzzle carries the most information, and start there
Use the units column to pin a digit, giving the reason rather than the answer
Bound a carry, and say why the bound differs when three numbers are added instead of two
Argue that the leading digit of a sum which has gained a place must be small, and say how small
Solve a two-line letter addition completely and check the solution by substituting back
Decide whether a puzzle has one solution or several
Explain a solution to somebody else as a sequence of forced steps

Where it usually goes wrong

"Letters keep their values from one puzzle to the next." They do not. T is 5 in the first puzzle on p.143 and 1 in the last puzzle on p.144. Each puzzle is a fresh assignment, and a student who carries values across will get contradictions and blame themselves.
"You have to try all ten digits for each letter." Three letters would be a thousand combinations. The chapter's own questions — could it be 2, could it be 3 — look like trial but are a demonstration of elimination.
"A carry is always 1." With two numbers being added, yes. With three, a column can pass 2. Puzzle 1 adds three numbers, so the bound there is different, and stating the rule flatly would be wrong on the chapter's very first example.
"If the sum has one more digit than the numbers being added, that extra digit could be anything." Two two-digit numbers reach at most 198, so an extra leading digit can only be 1. That single observation cracks two of the printed puzzles.
"Different letters must stand for different digits." That is a common convention elsewhere but this chapter does not impose it. Nothing here needs it.
"Start wherever the puzzle looks easiest." Start where information flows from. Columns feed leftwards, so the rightmost column is the only one that can be read without knowing anything else. Any other opening move is a gamble.
"Once I find digits that work, I am done." Not until you have shown nothing else works. Some puzzles admit several answers, and the difference between finding one and showing it is the only one is the same distinction the grid section made.

Questions to check understanding

Solve a letter-for-digit addition and state the value of every letter
Say which column you started from and why that column was forced
Justify a claimed digit by the constraint that produced it, not by substitution after the fact
Decide whether a given puzzle has exactly one solution
Construct a puzzle of this kind with a known unique answer, which is much harder than solving one and makes an excellent extension task
Explain why the leading digit of a lengthened sum must be 1
Chapter-end items in this book mix these with the parity questions on the same page, so expect a cryptarithm and a parity argument in the same set

Examples worth working on the board

Puzzle 1 (Part I, §6.5, p.143). Checked against the printed page: three copies of T written one under another with a plus sign, ruled off, and UT beneath. The chapter asks whether T could be 2 or 3, and then states the outcome: T is 5 and the two-digit sum is 15. The reasoning behind it is added here to supply — the units digit of three copies of a digit has to come back as that same digit, and only 0 and 5 manage that, and 0 gives no two-digit sum at all.
Puzzle 2 (Part I, §6.5, p.143). Checked against the printed page: K2 written twice with a plus, ruled off, and HMM beneath. The chapter spells out that K2 denotes a two-digit number whose tens digit is K and whose units digit is 2, asks what M must be, and then asks whether H could be 2 or 3. It stops there — no values are printed. Inputs: the puzzle, the place-value reading, and the observation that the last two digits of the answer match. The answer is 72 + 72 = 144; I verified it and it is the only one. Present the chain, not the answer: the units column fixes M, the fact that two two-digit numbers cannot reach 200 fixes H, and K then follows.
The four side puzzles (Part I, §6.5, p.143). Checked against the printed page, printed side by side in one row: YY + Z = ZOO; B5 + 3D = ED5; KP + KP = PRR; and C1 + C = 1FF. No answers are printed anywhere in the chapter. I worked each one and each has exactly one solution: 99 + 1 = 100; 75 + 30 = 105; 61 + 61 = 122; and 91 + 9 = 100.
The parting puzzle (Part I, §6.5 Figure it Out, p.144, question 11). UT + TA = TAT. This one is worth building the closing section around, because the units column alone does not crack it and the reasoning has to look at the whole sum at once. Its solution is 91 + 10 = 101, which the solutions appendix bound after the chapter also gives.
Carry sizes (not in the book). Not printed. Adding two digits and an incoming carry reaches at most 19, so a column passes at most 1 leftwards. Adding three digits reaches at most 27, so puzzle 1's column could in principle pass 2. This distinction is worth a beat: it is why the first puzzle and the second puzzle are not the same shape.
The closing comic (Part I, p.145, below the SUMMARY). Checked against the printed page. One child reports a marriage figure of 14,70,369 for the year and another asks whether it should not have been even; an owl in a mortarboard reacts. It is a parity joke, and it makes a neat sign-off tying the last section of the chapter back to §6.2.

Figures to have open

A column-addition frame in which letters can be swapped for digits one cell at a time, with a carry box above each column. This is the only figure the topic really needs, and it must be able to grey out columns that are not in play.
A pair of maximum-total panels, one for two addends and one for three, for the carry bound in section 8.
A small legend panel mapping letters to digit boxes, refreshed per puzzle, so the reset between puzzles is visible.
The printed puzzles are typeset arithmetic, not artwork; redraw them freely. No photograph or dataset from the textbook is required.

Where this sits in the book

NCERT Ganita Prakash, Class 7, Part I, printed Chapter 6 "Number Play", §6.5 "Digits in Disguise", pp.142–144 — the opening question about doing arithmetic with letters (p.142), the two worked puzzles, the four side puzzles and the naming of the puzzle type (p.143), and the Figure it Out set (pp.143–144) — its heading and question 1 are at the foot of p.143, below the four side puzzles, and questions 2 to 11 are on p.144
Same part, SUMMARY, p.145, last bullet, for the chapter's own one-line account of this section
Same part, the three-panel comic below the SUMMARY, p.145, which closes the chapter on a parity joke
Solutions appendix bound after p.145 in the cached PDF (not part of the printed book), which supplies the answer to question 11 only