Chapter 6 exercise answers: Number Play

Open NCERT p. 128Checked by computerAnswers can differ: one example

(a) 0, 1, 1, 2, 4, 1, 5

1. Rule: each child says how many children in front of them are taller.
2. Build the line from the front. Keep the children placed so far in order from tallest to shortest. A child who says k goes just below the k tallest of them.
3. Child 1 (0): list is C1.
4. Child 2 (1): one taller in front, so below C1: C1, C2.
5. Child 3 (1): below the tallest one: C1, C3, C2.
6. Child 4 (2): below the two tallest: C1, C3, C4, C2.
7. Child 5 (4): below all four: C1, C3, C4, C2, C5.
8. Child 6 (1): below the tallest one: C1, C6, C3, C4, C2, C5.
9. Child 7 (5): below the five tallest: C1, C6, C3, C4, C2, C7, C5.
10. Number the heights from 7 (tallest) down to 1 (shortest): C1 = 7, C6 = 6, C3 = 5, C4 = 4, C2 = 3, C7 = 2, C5 = 1.
11. Check: 7, 3, 5, 4, 1, 6, 2 gives 0, 1, 1, 2, 4, 1, 5.

AnswerHeights front to back (1 = shortest, 7 = tallest): 7, 3, 5, 4, 1, 6, 2. Each step left no choice, so this is the only arrangement.

(b) 0, 0, 0, 0, 0, 0, 0

1. Every child says 0, so each child is taller than everyone in front.
2. So the heights go up from front to back.

AnswerHeights front to back: 1, 2, 3, 4, 5, 6, 7 (shortest at the front, tallest at the back).

(c) 0, 1, 2, 3, 4, 5, 6

1. Child k says k − 1, which is everyone in front of them, so each child is shorter than everyone in front.
2. So the heights go down from front to back.

AnswerHeights front to back: 7, 6, 5, 4, 3, 2, 1 (tallest at the front, shortest at the back).

(d) 0, 1, 0, 1, 0, 1, 0

1. Use the same method as (a), tallest first.
2. C1 (0): C1. C2 (1): C1, C2. C3 (0): C3, C1, C2. C4 (1): C3, C4, C1, C2.
3. C5 (0): C5, C3, C4, C1, C2. C6 (1): C5, C6, C3, C4, C1, C2. C7 (0): C7, C5, C6, C3, C4, C1, C2.
4. Heights from 7 down: C7 = 7, C5 = 6, C6 = 5, C3 = 4, C4 = 3, C1 = 2, C2 = 1.

AnswerHeights front to back: 2, 1, 4, 3, 6, 5, 7.

(e) 0, 1, 1, 1, 1, 1, 1

1. Child 2 says 1, so child 1 is taller than child 2.
2. Child 3 says 1. If child 2 were taller than child 3, child 1 would be too, and child 3 would say 2. So child 3 is taller than child 2, and child 1 is the one taller than child 3.
3. The same happens for every later child: each is taller than every child in front of them except child 1, and shorter than child 1. So children 2 to 7 get taller one after another, and child 1 is the tallest.

AnswerHeights front to back: 7, 1, 2, 3, 4, 5, 6 (the tallest in front, then the rest from shortest to tallest).

(f) 0, 0, 0, 3, 3, 3, 3

1. Children 1, 2, 3 say 0, so they get taller one after another.
2. Children 4 to 7 each say 3, so children 1, 2, 3 are taller than each of them, and nobody else in front is taller.
3. So children 4 to 7 must also get taller one after another, while staying shorter than children 1, 2, 3.

AnswerHeights front to back: 5, 6, 7, 1, 2, 3, 4.

Watch this explained “Rebuilding the row”, 6:44 into Encoding a line-up as a sequence of numbers · हिंदी में देखें

Open NCERT p. 128Matches NCERT’s answer

(a) If a person says '0', then they are the tallest …

1. The tallest child always says 0, so a child who says 0 can be the tallest.
2. But the child at the front also always says 0, however short they are. Line up three children as shortest, tallest, middle: the front child says 0 and is the shortest.

AnswerOnly Sometimes True.

(b) If a person is the tallest, then their number is '0'.

1. Nobody can be taller than the tallest child, wherever they stand.
2. So the count of taller children in front of them is always 0.

AnswerAlways True.

(c) The first person's number is '0'.

1. Nobody stands in front of the first person, so there is nobody to count.

AnswerAlways True.

(d) If a person is not first or last in line …

1. A child in the middle who is taller than everyone in front says 0. In the book's first picture, the third child says 0.
2. A child in the middle with a taller child in front does not say 0. So it depends on the line.

AnswerOnly Sometimes True.

(e) The person who calls out the largest number is the shortest.

1. True in a line going from tallest to shortest: the numbers are 0, 1, 2, 3, …, and the last child, the shortest, says the largest number.
2. False in the line shortest, tallest, middle: the numbers are 0, 0, 1, and the largest number, 1, is said by the middle-height child, not the shortest.

AnswerOnly Sometimes True.

(f) What is the largest number possible in a group of 8 people?

1. A child can count only the children in front of them.
2. In a line of 8, the last child has 7 in front, so no one can say more than 7.
3. 7 does happen: in a line from tallest to shortest, the last child says 7.

Answer7.

Watch this explained “Always, sometimes, never”, 7:30 into Encoding a line-up as a sequence of numbers · हिंदी में देखें

Open NCERT p. 131Matches NCERT’s answer

(a) Sum of 2 even numbers and 2 odd numbers

1. Even numbers never bring a leftover dot, so they never change the parity of a sum.
2. Only the 2 odd numbers matter: 2 odd numbers pair up their leftover dots, leaving none stranded.

AnswerEven.

(b) Sum of 2 odd numbers and 3 even numbers

1. Ignore the 3 even numbers — they cannot change the parity.
2. The 2 odd numbers pair up their leftover dots completely.

AnswerEven.

(c) Sum of 5 even numbers

1. All 5 numbers are even, so there is no leftover dot anywhere.

AnswerEven.

(d) Sum of 8 odd numbers

1. 8 odd numbers give 8 leftover dots.
2. 8 is even, so the leftover dots pair up completely, leaving nothing stranded.

AnswerEven.

Watch this explained “A long sum”, 1:40 into The parity of a sum or product is fixed before you compute it · हिंदी में देखें

Open NCERT p. 131Checked by computerAnswers can differ: one example

1. The ₹1 coins: an odd count of them contributes an odd amount (odd number of ₹1's is odd).
2. The ₹5 coins: an odd count times an odd value (₹5) also contributes an odd amount.
3. The ₹10 coins: any count of them contributes an even amount, since 10 itself is even.
4. Odd + odd + even = even, so the total must always be even, no matter what the actual counts are.
5. ₹205 is odd, so no combination of an odd number of ₹1's, an odd number of ₹5's and an even number of ₹10's can ever total ₹205.

AnswerYes, Lakpa made a mistake — the total must always come out even, but ₹205 is odd, so no valid combination of coins gives ₹205.

Watch this explained “A piggy bank”, 2:29 into The parity of a sum or product is fixed before you compute it · हिंदी में देखें

Open NCERT p. 131Matches NCERT’s answer

(d)

1. An even number is made of whole pairs.
2. Taking away whole pairs leaves whole pairs. Example: 10 − 4 = 6.

Answereven − even = even.

(e)

1. Each odd number is whole pairs plus one extra dot.
2. Take away the smaller number's pairs from the bigger number's pairs, and its extra dot from the bigger number's extra dot. Only whole pairs are left. Example: 9 − 5 = 4.

Answerodd − odd = even.

(f)

1. The even number is whole pairs. The odd number is whole pairs plus one extra dot.
2. Taking away the odd number's pairs leaves whole pairs. Taking away its one extra dot then breaks one pair and leaves its partner alone. Example: 10 − 3 = 7.

Answereven − odd = odd.

(g)

1. The odd number has one extra dot.
2. Taking away an even number removes only whole pairs, so the extra dot is still there. Example: 9 − 4 = 5.

Answerodd − even = odd.

Watch this explained “Taking away pairs”, 0:48 into The parity of a sum or product is fixed before you compute it · हिंदी में देखें

Open NCERT p. 136Checked by computerReads two ways: both answers shown

1. From the chapter: the magic sum is 15, 5 must be in the centre, and 1 and 9 must be in middle-of-an-edge positions, opposite each other.
2. 1 can go in any of the 4 edge-middle positions, and then 9 goes opposite it.
3. Say 1 is in the middle of the top row. The other two numbers in the top row must add up to 15 − 1 = 14. Of the numbers still free (2, 3, 4, 6, 7, 8), the only pair that adds up to 14 is 6 and 8, and they can go in either order: 2 ways.
4. After that, every other square is fixed by the sums of 15. For example: 8 1 6 / 3 5 7 / 4 9 2.
5. So there are 4 × 2 = 8 ways to fill the grid.
6. But all 8 are the same square turned around or flipped over. If turned or flipped squares are not counted as new, there is just 1 magic square. NCERT's answer key uses this reading.

AnswerNot counting turned or flipped squares as new (NCERT's answer key): 1 magic square. Counting every different filling of the grid: 8.

Watch this explained “One line, and the rest falls out”, 5:16 into Constructing magic squares, and where they came from · हिंदी में देखें

Open NCERT p. 136One way to think about it

1. The numbers 2 to 10 are the numbers 1 to 9 with 1 added to each.
2. Start from a 1–9 magic square, for example 2, 7, 6 / 9, 5, 1 / 4, 3, 8 (magic sum 15).
3. Add 1 to every number: 3, 8, 7 / 10, 6, 2 / 5, 4, 9.
4. Each line has 3 numbers, so each line total goes up by 3. The magic sum is 15 + 3 = 18. Check the top row: 3 + 8 + 7 = 18.
5. You can also find it directly: 2 + 3 + … + 10 = 54, and the 3 rows share this, so the magic sum is 54 ÷ 3 = 18. The centre must be the middle number 6.
6. Compared with the 1–9 squares: the same pattern, every entry 1 more, centre 6 instead of 5, magic sum 18 instead of 15. The smallest and largest numbers (2 and 10) sit in the middle of edges, just like 1 and 9.

In shortOne magic square using 2–10 (rows, top to bottom): 3, 8, 7; 10, 6, 2; 5, 4, 9, with magic sum 18. Strategy: add 1 to every entry of a 1–9 magic square. Turning or flipping this square gives other correct answers (8 in all).

Watch this explained “Add one, and the sum goes up by three”, 6:09 into Constructing magic squares, and where they came from · हिंदी में देखें

Open NCERT p. 136Checked by computer

(a) increase each number by 1

1. Take the magic square 8 1 6 / 3 5 7 / 4 9 2 (magic sum 15). Adding 1 to each number gives 9 2 7 / 4 6 8 / 5 10 3.
2. Every row, column and diagonal has 3 numbers, so every line sum goes up by 3 × 1 = 3.
3. All the line sums were equal before, so they are still equal: the new grid is a magic square.
4. This works for any magic square: the magic sum always increases by 3. For our square, 15 becomes 15 + 3 = 18.

AnswerYes, it is still a magic square. The magic sum increases by 3 (for the 1–9 square, 15 becomes 18).

(b) double each number

1. Doubling each number gives 16 2 12 / 6 10 14 / 8 18 4.
2. In each line, 2 × a + 2 × b + 2 × c = 2 × (a + b + c), so every line sum doubles.
3. All the line sums were equal before, so they are still equal: the new grid is a magic square.
4. This works for any magic square: the magic sum is multiplied by 2 (NCERT's answer key says the same). For our square, 15 becomes 30. The increase, 15, is the old magic sum, so it depends on which square you start with.

AnswerYes, it is still a magic square. The magic sum is multiplied by 2 (for the 1–9 square, 15 becomes 30).

Watch this explained “Add one, and the sum goes up by three”, 6:09 into Constructing magic squares, and where they came from · हिंदी में देखें

Open NCERT p. 136Checked by computerAnswers can differ: one example

1. Turn the whole square a quarter turn (90°). For example, 2, 7, 6 / 9, 5, 1 / 4, 3, 8 becomes 4, 9, 2 / 3, 5, 7 / 8, 1, 6.
2. After the turn, each old row has become a column, each old column has become a row, and the two diagonals have swapped places.
3. So the same 8 line totals are still there, just in new places. Every line still adds up to 15, and the square is still magic.
4. Other operations also work: reflecting the square like a mirror image, adding the same number to every entry, multiplying every entry by the same number, or subtracting every entry from the same number (for 1–9, replacing each number n by 10 − n).

AnswerRotating the whole square by 90° gives another magic square with the same magic sum. The question has many correct answers; others include reflecting the square, adding the same number to every entry, multiplying every entry by the same number, and subtracting every entry from the same number.

Watch this explained “Add one, and the sum goes up by three”, 6:09 into Constructing magic squares, and where they came from · हिंदी में देखें

Open NCERT p. 136One way to think about it

1. Any 9 consecutive numbers are 1–9 with the same number added to each. For example, 9–17 is 1–9 with 8 added.
2. So take a 1–9 magic square, such as 2, 7, 6 / 9, 5, 1 / 4, 3, 8, and add that number to every entry. Each line gains 3 times that number, so all the lines stay equal.
3. For 9–17, add 8: 10, 15, 14 / 17, 13, 9 / 12, 11, 16. The magic sum is 15 + 3 × 8 = 39.
4. Another way is to reason directly, as for 1–9: the magic sum is the total of the 9 numbers divided by 3, the centre must be the middle (5th) number, and the smallest and largest numbers go in the middle of edges, not in corners. For 9–17: total 117, magic sum 39, centre 13.
5. Either way, the magic sum is 3 times the middle number.

In shortAdd the same number to every entry of a 1–9 magic square; for 9–17, adding 8 gives 10, 15, 14; 17, 13, 9; 12, 11, 16 with magic sum 39. Or build it directly with the middle number in the centre. The magic sum is always 3 times the middle number.

Watch this explained “Every cell as an offset”, 6:55 into Constructing magic squares, and where they came from · हिंदी में देखें

Open NCERT p. 137Checked by computerAnswers can differ: one example

1. In the generalised form, every cell is the centre number m plus a fixed offset (the offsets used are −3, 2, 1, 4, 0, −4, −1, −2, 3, arranged the same way as in the 1–9 square).
2. Set m = 25 and add each offset to it to get the 9 cell values.
3. Check: every row, column and diagonal adds up to 3 × 25 = 75.

AnswerOne valid square (rows, top to bottom): 22, 27, 26; 29, 25, 21; 24, 23, 28 — every row, column and diagonal adds up to 75.

Watch this explained “Every cell as an offset”, 6:55 into Constructing magic squares, and where they came from · हिंदी में देखें

Open NCERT p. 137Matches NCERT’s answer

1. Every cell is written as m plus an offset.
2. Adding the 3 cells of any row, column or diagonal adds the 3 m's together, plus the 3 offsets on that line.
3. On every single line, the 3 offsets happen to add up to 0 (checked directly for all 8 lines), so only the 3 m's remain.

Answer3m.

Watch this explained “Every cell as an offset”, 6:55 into Constructing magic squares, and where they came from · हिंदी में देखें

Open NCERT p. 137Checked by computer

(a) adding 1 to every term in the generalised form.

1. The original sum of any line is 3m.
2. Adding 1 to each of the 3 terms on a line adds 1 + 1 + 1 = 3 to that line's total.

Answer3m + 3.

(b) doubling every term in the generalised form

1. The original sum of any line is 3m.
2. Doubling each of the 3 terms on a line doubles the whole line total.

Answer6m.

Watch this explained “Add one, and the sum goes up by three”, 6:09 into Constructing magic squares, and where they came from · हिंदी में देखें

Open NCERT p. 137Checked by computerAnswers can differ: one example

1. The magic sum of the generalised form is 3m, so 3m = 60 gives m = 20.
2. Add the same fixed offsets (−4 to 4) used in the 1–9 square to 20 to get the 9 cell values, which run from 16 to 24.

AnswerOne valid square (rows, top to bottom): 17, 22, 21; 24, 20, 16; 19, 18, 23 — using the numbers 16 to 24, with magic sum 60.

Watch this explained “Every cell as an offset”, 6:55 into Constructing magic squares, and where they came from · हिंदी में देखें

Open NCERT p. 137Matches NCERT’s answer

1. Doubling every number in the standard 1–9 magic square keeps it magic (shown earlier in this set), with magic sum 30.
2. The doubled numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18 — these go up in steps of 2, so they are not consecutive.
3. This is a magic square made entirely of non-consecutive numbers.

AnswerYes — for example, doubling every number of the 1–9 magic square gives a magic square using 2, 4, 6, …, 18, which are non-consecutive (they differ by 2, not 1), with magic sum 30.

Watch this explained “Add one, and the sum goes up by three”, 6:09 into Constructing magic squares, and where they came from · हिंदी में देखें

Open NCERT p. 143Matches NCERT’s answer

1. Every toggle flips the bulb's state: on becomes off, and off becomes on.
2. Two toggles in a row bring the bulb back to how it started, so only whether the number of toggles is odd or even matters.
3. 77 is odd, so after 77 toggles the bulb ends up in the opposite state to where it started.
4. It started ON, so it ends OFF.

AnswerOff, because 77 is odd, and an odd number of toggles always leaves the bulb in the opposite state to how it started.

Watch this explained “Odd, drawn”, 2:45 into Odd and even as "can this be arranged in pairs" · हिंदी में देखें

Open NCERT p. 144Checked by computer

1. In a book, the two page numbers on one sheet follow each other, and the odd one comes first: 1 and 2, 3 and 4, 5 and 6, and so on.
2. So one sheet's two page numbers add up to 3, 7, 11, 15, … If the front page is 2k + 1, the back page is 2k + 2, and together they make 4k + 3: a multiple of 4, plus 3.
3. Adding 50 sheets gives (a multiple of 4) + 50 × 3 = (a multiple of 4) + 150.
4. For the total to be 6000, 6000 − 150 = 5850 would have to be a multiple of 4. But 5850 ÷ 4 = 1462 with remainder 2.
5. Note that parity alone does not decide it: each sheet's total is odd, and 50 odd numbers add up to an even number, just like 6000. Counting in fours is what rules it out.
6. The answer key at the back of the book prints 'yes' because 6000 is even; but each sheet has pages like 1 and 2, 3 and 4 (sums 3, 7, 11, … each one less than a multiple of 4), so 50 sheets always add up to 50 less than a multiple of 4, and 6000 + 50 = 6050 is not a multiple of 4, so the answer is no.

AnswerNo. Whichever 50 sheets fell out, their page numbers add up to a multiple of 4 plus 150, and 6000 is not of that form (6000 − 150 = 5850 is not a multiple of 4).

Watch this explained “Two ages, one year apart”, 6:25 into Odd and even as "can this be arranged in pairs" · हिंदी में देखें

Open NCERT p. 144Checked by computerAnswers can differ: one example

1. Call the top row a, b, c and the bottom row d, e, f. The circles ask for: top row odd, bottom row even; columns even, even, odd.
2. Two numbers add to an even number when they have the same parity, and to an odd number when they differ. So a and d match, b and e match, and c and f differ.
3. The pairs (a, d) and (b, e) are each two odds or two evens, and (c, f) is one odd and one even. To get exactly 3 odd numbers, one matching pair must be odd and the other even.
4. Choice 1: a and d odd, b and e even. The top row a, b, c needs an odd number of odds, so c is even, and then f is odd. Bottom row: odd, even, odd has two odds, so its sum is even, as needed.
5. Choice 2: b and e odd, a and d even. The same steps give c even and f odd: top row even, odd, even; bottom row even, odd, odd.

AnswerOne filling (rows, top to bottom): odd, even, even; odd, even, odd. For example, 1, 2, 4 on top and 3, 6, 5 below. The only other filling is even, odd, even; even, odd, odd.

Watch this explained “A long sum”, 1:40 into The parity of a sum or product is fixed before you compute it · हिंदी में देखें

Open NCERT p. 144Checked by computerAnswers can differ: one example

1. The magic sum of the generalised form is 3m, so setting m = 0 gives a magic sum of 3 × 0 = 0.
2. Keep the same fixed offsets used in the 1–9 square (−3, 2, 1, 4, 0, −4, −1, −2, 3) but add them to 0 instead of 5.
3. The offsets themselves become the 9 cell values, some of them negative.

Answer−3, 2, 1; 4, 0, −4; −1, −2, 3 (rows, top to bottom) — every row, column and diagonal adds up to 0, and not every number is 0.

Watch this explained “One square, nine times over”, 8:50 into Constructing magic squares, and where they came from · हिंदी में देखें

Open NCERT p. 144Matches NCERT’s answer

(a) Sum of an odd number of even numbers is

1. Even numbers never contribute a leftover dot, no matter how many are added.

AnswerEven.

(b) Sum of an even number of odd numbers is

1. Each odd number contributes one leftover dot.
2. An even count of leftover dots pairs up completely.

AnswerEven.

(c) Sum of an even number of even numbers is

1. Even numbers never contribute a leftover dot, no matter how many are added.

AnswerEven.

(d) Sum of an odd number of odd numbers is

1. Each odd number contributes one leftover dot.
2. An odd count of leftover dots always leaves exactly one stranded.

AnswerOdd.

Watch this explained “A long sum”, 1:40 into The parity of a sum or product is fixed before you compute it · हिंदी में देखें

Open NCERT p. 144Matches NCERT’s answer

1. From 1 to 100 there are 50 odd numbers and 50 even numbers.
2. The even numbers never contribute a leftover dot.
3. The 50 odd numbers contribute 50 leftover dots, and 50 is even, so they all pair up completely.
4. So the total sum is even. (It is exactly 5050.)

AnswerEven.

Watch this explained “A long sum”, 1:40 into The parity of a sum or product is fixed before you compute it · हिंदी में देखें

Open NCERT p. 144Matches NCERT’s answer

1. Each number in the sequence is the sum of the two numbers just before it.
2. Going forward: 987 + 1597 = 2584, and then 1597 + 2584 = 4181.
3. Going backward: since 1597 = 987 + (the term before 987), the term before 987 is 1597 − 987 = 610.
4. And since 987 = 610 + (the term before 610), that term is 987 − 610 = 377.

AnswerNext 2 numbers: 2584, 4181. Previous 2 numbers: 377, 610.

Watch this explained “Six beats, free”, 5:28 into Virahāṅka–Fibonacci numbers, and the poetry-counting problem that produced them · हिंदी में देखें

Open NCERT p. 144Matches NCERT’s answer

1. Climbing in strides of 1 or 2 steps is exactly the same counting problem as writing 8 as an ordered sum of 1's and 2's.
2. That is exactly the Virahāṅka-sequence problem: the number of ways to make n beats using 1-beat and 2-beat syllables.
3. The Virahāṅka sequence is 1, 2, 3, 5, 8, 13, 21, 34, … and its 8th term counts the 8-beat case.
4. So the number of ways is the 8th term of the sequence, which is 34.

Answer34.

Watch this explained “The same count, in other clothes”, 9:02 into Virahāṅka–Fibonacci numbers, and the poetry-counting problem that produced them · हिंदी में देखें

Open NCERT p. 144Matches NCERT’s answer

1. Each term is the sum of the two before it, and odd + even = odd, odd + odd = even, even + odd = odd.
2. The sequence starts 1, 2, 3, 5, 8, …, so the parities start odd, even, odd, odd, even, …
3. Terms 4 and 5 are odd, even, just like terms 1 and 2. Each parity depends only on the two before it, so the pattern odd, even, odd repeats every 3 terms.
4. So the even terms are at positions 2, 5, 8, 11, 14, 17, 20, … (2 more than a multiple of 3).
5. 20 = 18 + 2, so the 20th term is even. (It is 10946.)

AnswerEven.

Watch this explained “Odd or even, without adding”, 8:03 into Virahāṅka–Fibonacci numbers, and the poetry-counting problem that produced them · हिंदी में देखें

Open NCERT p. 144Checked by computerReads two ways: both answers shown

1. (a) 4m = 2 × 2m is always even, and 1 less than an even number is odd. So 4m − 1 is always odd: true.
2. (b) 6j − 4 gives 2, 8, 14, 20, … It never gives 4, because 6j − 4 = 4 would need 6j = 8. So not all even numbers can be written this way: false.
3. (d) 2f is always even, and even + 3 is odd. So 2f + 3 always gives odd numbers, never even ones: false.
4. (c) depends on which values p and q may take, and the question does not say. In this chapter letter-numbers take the values 1, 2, 3, … (for example, '6k + 2 for k = 1, 2, 3, …'). Then 2q − 1 gives 1, 3, 5, 7, … (every odd number), but 2p + 1 gives 3, 5, 7, … and never gives 1. So (c) is false.
5. If p is also allowed to be 0, then 2p + 1 gives 1, 3, 5, … too, and (c) is true.

AnswerWith letter-numbers taking the values 1, 2, 3, … (as in this chapter): only (a) is true. If p may also be 0: (a) and (c) are true.

Watch this explained “A promise about every value”, 8:41 into The parity of a sum or product is fixed before you compute it · हिंदी में देखें

Open NCERT p. 144Matches NCERT’s answer

1. Look at the units column first, since it is the only column with nothing arriving from the right: T + A must end in T, so A must be 0 (adding 0 changes nothing).
2. The answer TAT has one more digit than UT and TA, so the hundreds digit of the answer, T, must be the carry out of the tens column.
3. A carry out of a column adding two digits can only ever be 0 or 1, and it cannot be 0 here (T is a leading digit, so T ≠ 0), so T = 1.
4. In the tens column, U + T must end in the tens digit of the answer (which is A = 0) and also produce that carry of 1: U + 1 ends in 0 and carries 1, so U = 9.
5. Check: UT = 91, TA = 10, and 91 + 10 = 101 = TAT.

AnswerU = 9, T = 1, A = 0 (91 + 10 = 101).

Watch this explained “The one the units will not crack”, 8:17 into Cryptarithms: recovering digits from constraints alone · हिंदी में देखें