Question 0
For the pictured arrangement of seven children, write the number each child calls out.
Steps only
Move from the front of the line to the back.
At each place, count only earlier people who are taller.
Answer
The code is 0,0,1,0,3,0,3.
CLASS 7 · GANITA PRAKASH PART I · CHAPTER 6
Number Play
Explore parity, height codes, grid puzzles and number patterns with clear reasoning.
45 worked answersAll 2 exercise setsAll 11 end-of-chapter questions
What you will learn
- Recognise how position and comparison create number codes.
- Use even/odd reasoning to prove whether totals are possible.
- Build and analyse magic squares and recursive sequences.
45 worked answers · short steps first, explanations when you need them
6.1 Numbers tell us things
Textbook pages 128–129
Question 1
Each person in a line counts only the people before them who are taller. Give one arrangement for each code: (a) 0,1,1,2,4,1,5; (b) 0,0,0,0,0,0,0; (c) 0,1,2,3,4,5,6; (d) 0,1,0,1,0,1,0; (e) 0,1,1,1,1,1,1; (f) 0,0,0,3,3,3,3.
Steps only
Compare each person only with people earlier in the line.
Arrange the heights so each position has the required count of earlier taller people.
Answer
(a) The cutout order F,C,B,G,A,D,E is one valid answer. If heights are represented by ranks 1=shortest through 7=tallest, valid left-to-right rank orders are (a) 7,3,5,4,1,6,2; (b) 1,2,3,4,5,6,7; (c) 7,6,5,4,3,2,1; (d) 2,1,4,3,6,5,7; (e) 7,1,2,3,4,5,6; (f) 5,6,7,1,2,3,4. Each list reproduces its code. The number counts earlier taller people, not all taller people in the line.
Question 2
Classify: (a) someone saying 0 means tallest; (b) tallest person says 0; (c) first person says 0; (d) an in-between person cannot say 0; (e) the largest number belongs to the shortest; (f) largest possible code in a line of 8.
Steps only
A person’s count includes only taller people before them.
Test whether the statement holds for every arrangement, at least one, or none.
Answer
(a) Sometimes; (b) always; (c) always; (d) sometimes; (e) sometimes; (f) 7.
6.2 Even and odd (parity)
Textbook pages 129–132
Question 1
Find the parity of: (a) two even plus two odd numbers; (b) two odd plus three even; (c) five even numbers; (d) eight odd numbers.
Steps only
Even numbers contribute no leftover after pairing.
An odd summand contributes one leftover; an even number of odd summands pair those leftovers.
Answer
(a) even; (b) even; (c) even; (d) even.
Question 2
Lakpa has an odd number of ₹1 coins, an odd number of ₹5 coins and an even number of ₹10 coins. Can their total be ₹205?
Steps only
Odd×₹1 is odd; odd×₹5 is odd; even×₹10 is even.
Odd+odd+even is even.
Answer
No. The total must be even, so ₹205 is impossible under those conditions.
Question 3
Find the parity: even−even; odd−odd; even−odd; odd−even.
Steps only
Subtracting an integer has the same parity effect as adding it: negative even is even, negative odd is odd.
Answer
even; even; odd; odd.
Question 4
State when an m×n grid contains an even or odd number of unit squares. Find the parity for 27×13, 42×78 and 135×654.
Steps only
The count is mn.
A product is odd only when both factors are odd.
Answer
Odd exactly when both m and n are odd; otherwise even. 27×13: odd; 42×78: even; 135×654: even.
Question 6
Can five odd-number cards add to 30? Can two siblings born exactly one year apart have ages adding to 112?
Steps only
An odd number of odd addends has odd parity.
Two consecutive whole-number ages have one even age and one odd age, so their sum is odd.
Answer
Neither is possible. Five odd numbers have an odd sum, not 30. Consecutive ages add to an odd number, not 112.
Question 8
A story claims that 1,470,369 people got married in one year. Can this be the exact count if every marriage has two partners?
Steps only
Each marriage includes two people.
A total of pairs is even.
Check whether the reported count is even or odd.
Answer
No. 1,470,369 is odd, but counting two people for each marriage must give an even total.
Question 7
Find the parity of sums of 3, 4, 5 and 6 odd numbers.
Steps only
Pair the odd numbers.
Each pair has an even sum; one unpaired odd number remains only when the count is odd.
Answer
3 odd numbers: odd; 4: even; 5: odd; 6: even.
Question 5
Give a rule that generates all positive even numbers and one that generates all positive odd numbers. Give an expression whose parity may vary with n.
Steps only
Even numbers are twice a counting number.
Odd numbers are one less than the next even number.
For 3n+4, check odd and even n.
Answer
2n lists the positive even numbers; 2n−1 lists the positive odd numbers. 3n+4 is odd when n is odd and even when n is even.
6.3 Grid sums and magic squares
Textbook pages 133–137
Question 1
Why do row sums and column sums in a 3×3 grid containing 1–9 each total 45?
Steps only
Add 1+2+…+9.
Every entry appears in exactly one row and exactly one column.
Answer
Both the three row totals together and the three column totals together equal 45.
Question 2
A 3×3 grid uses 1–9 once each. Can a row or column sum be 5 or 26?
Steps only
The least possible sum of three distinct entries is 1+2+3=6.
The greatest is 9+8+7=24.
Answer
No. Every row/column sum must be between 6 and 24, inclusive.
Question 3
Find the magic sum of a 3×3 magic square containing 1 through 9 once each.
Steps only
The three row sums are equal.
They add to the total 45.
Divide 45 by 3.
Answer
Magic sum=15.
Question 4
Give a 3×3 magic square using 1–9 and explain its uniqueness up to rotations/reflections.
Steps only
Place 5 in the centre.
Pair numbers summing to 10 on opposite sides.
Arrange to make each row, column and diagonal sum 15.
Answer
One is 8 1 6 / 3 5 7 / 4 9 2. It is unique up to turning or reflecting the square.
Question 5
If every entry in a 3×3 magic square is increased by 1, or doubled, does it remain magic? What happens to the magic sum?
Steps only
Every row, column and diagonal has three entries.
Adding 1 adds 3 to each line total; doubling doubles each line total.
Answer
Yes. Starting from magic sum 15, increasing every entry by 1 gives sum 18; doubling gives sum 30.
Question 6
Create a magic square from 2–10. How can one be made for any nine consecutive integers?
Steps only
Add 1 to each entry of the 1–9 square.
For a sequence starting at k, add k−1 to each entry.
Answer
For 2–10: 9 2 7 / 4 6 8 / 5 10 3. For k through k+8, add k−1 to each entry of the base square; magic sum=3(k+4).
Question 7
If the centre entry is m, express the other entries in one magic-square arrangement and its line sum.
Steps only
Add or subtract 1 through 4 from m in the corresponding cells.
Each row/column/diagonal contains offsets whose sum is zero.
Answer
m+3, m−4, m+1 / m−2, m, m+2 / m−1, m+4, m−3. Every row, column and diagonal sums to 3m.
Question 8
Use the generalised pattern to make a magic square with centre 25.
Steps only
Substitute m=25 in each position.
Answer
28 21 26 / 23 25 27 / 24 29 22. Magic sum=75.
Question 9
Give a 3×3 magic square with magic sum 60.
Steps only
For this pattern the magic sum is three times the centre.
Choose centre 20 and substitute.
Answer
23 16 21 / 18 20 22 / 19 24 17. Each row, column and diagonal sums to 60.
Question 11
Fill each 3×3 grid with 1–9 once, matching its row and column totals. Grid A has row totals 13,14,18; column totals 24,9,12; top-left cell 9 and bottom-right cell 5. Grid B has row totals 24,15,6; column totals 12,16,17; middle-left cell 4 and bottom-right cell 3.
Steps only
Place the given numbers first.
Use each row total to find a missing row entry, then check each column total.
Check that 1–9 appear once in each grid.
Answer
Grid A: 9 1 3 / 8 2 4 / 7 6 5. Grid B: 7 8 9 / 4 6 5 / 1 2 3 is one solution. Another valid Grid B is 7 9 8 / 4 5 6 / 1 2 3.
Question 12
Give a 3×3 grid using 1–9 once and list row and column totals that another student could use to reconstruct it.
Steps only
Choose a known 1–9 grid.
Add across each row and down each column.
Answer
For 4 7 5 / 6 1 2 / 3 9 8, row totals are 16,9,20 and column totals are 13,17,15.
Question 13
Why must the centre be 5, and why can 1 and 9 not be in corners?
Steps only
Every line through the centre has two opposite cells whose numbers add to 15−m.
The four opposite pairs contain all numbers except m, whose total is 45−m. So 4(15−m)=45−m, which gives m=5.
A corner lies on a row, a column and a diagonal. If 1 were there, each line needs a pair summing 14; only 5+9 and 6+8 work, not three distinct pairs. For 9, only 1+5 and 2+4 sum to 6.
Answer
The centre must be 5. The 1 and 9 must be in middle-edge cells, not corners.
Question 14
What operations, applied equally to every entry, preserve a magic square?
Steps only
Every line has the same number of cells.
Adding or subtracting the same constant changes every line total equally; multiplying or dividing by the same nonzero constant also changes every line total equally.
Answer
Add or subtract a fixed number, multiply by a fixed number, or divide by the same nonzero number. Each operation keeps the row, column and diagonal totals equal.
Question 15
Starting with the centre-m square m+3,m−4,m+1 / m−2,m,m+2 / m−1,m+4,m−3, write the grids formed by adding 1 to every entry and by doubling every entry.
Steps only
For the first grid, add 1 to each constant offset.
For the second, double both m and each offset.
Answer
Add 1: m+4,m−3,m+2 / m−1,m+1,m+3 / m,m+5,m−2. Double: 2m+6,2m−8,2m+2 / 2m−4,2m,2m+4 / 2m−2,2m+8,2m−6.
Question 16
The historical 4×4 square has rows, columns and diagonals summing to 34. Why is it called Chautīsā, and give two other groups of four entries that also total 34.
Steps only
Add a row or column to check the common total.
Try the four corners and the central four cells.
Answer
It is called Chautīsā because chautīs means 34. The four corners 7+14+9+4=34; the central block 13+8+3+10=34.
Question 10
Can a magic square use nine non-consecutive integers? Give an example.
Steps only
Apply a common multiplier to each entry of a known magic square.
Answer
Yes. Example: 24 3 18 / 9 15 21 / 12 27 6 has every line sum 45.
6.5 Digits in disguise
Textbook pages 142–143
Question 1
A digit T is added to itself three times to make the two-digit number UT. Find T and U.
Steps only
Three times T must end in T.
Test the column equation 3T=10U+T.
The remaining amount 2T is the tens digit.
Answer
T=5 and U=1, because 5+5+5=15.
Question 2
K2 is added to itself to make HMM. Find K, H and M.
Steps only
The units column gives 2+2=4, so M=4.
The tens column is K+K=10H+M.
Find the digit K that makes the tens sum 14.
Answer
K=7, H=1 and M=4: 72+72=144.
Question 3
Each different letter stands for a different digit. Find YY+Z=ZOO.
Steps only
The hundreds digit equals Z, so the carry from the tens column must be 1.
A repeated-digit number YY is 11Y; test the column addition with distinct digits.
Answer
Y=9, Z=1 and O=0: 99+1=100.
Question 4
Different letters stand for different digits. Find B5+3D=ED5.
Steps only
Units: 5+D ends in 5, so D=0.
Tens: B+3 must end in 0 and carry E into the hundreds.
Use the nonzero leading digit.
Answer
B=7, E=1 and D=0: 75+30=105.
Question 5
Different letters stand for different digits. Find KP+KP=PRR.
Steps only
The hundreds digit P equals the carry from the tens column.
Check the units and tens columns with a repeated addend.
Answer
K=6, P=1 and R=2: 61+61=122.
Question 6
Different letters stand for different digits. Find C1+C=1FF.
Steps only
The units column is 1+C and ends in F.
The tens column adds C and the carry, giving the hundreds digit 1.
Answer
C=9 and F=0: 91+9=100.
Virahanka sequence and parity challenges
Textbook pages 142–144
Question 1
Continue 1,2,3,5,8,13,21,34,55,89 by three terms. What is the next term’s parity?
Steps only
Add the previous two terms each time.
Track odd/even pattern: odd, even, odd repeats every three terms.
Answer
144, 233, 377; the following term is even.
Question 2
A bulb is on. Dorjee toggles it 77 times. What is its final state?
Steps only
Each pair of toggles returns the bulb to its earlier state.
77=2×38+1, so one toggle remains.
Answer
Off. An odd number of toggles changes the initial state.
Question 3
A loose set has 50 sheets, each printed on both sides. Could their page-number sum be 6,000?
Steps only
Each physical sheet has one odd-numbered side and one even-numbered side.
Their sum is odd.
Fifty odd sheet-sums make an even total.
Answer
Yes, ₹6,000 is possible by the parity check: the total of 50 odd sheet-sums is even, and ₹6,000 is even. The given information does not list the actual page pairs, so parity does not determine the exact total; it shows that 6,000 is not ruled out.
Question 4
A 2×3 grid has even/odd row and column-sum labels. How should the six cells be filled with three odd and three even values?
Steps only
Use O for an odd cell and E for an even cell.
Put O,E,E in the first row and O,E,O in the second.
The rows then have odd/even parity; the columns have even/even/odd parity, and there are three O cells.
Answer
One valid pattern is O E E / O E O. The row sums are odd and even; the three column sums are even, even and odd. It contains exactly three odd cells.
Question 5
Give a nonzero 3×3 magic square with magic sum zero.
Steps only
Use opposite entries that cancel and make the centre zero.
Answer
3, −4, 1 / −2, 0, 2 / −1, 4, −3. Every row, column and diagonal sums to zero.
Question 6
State: sum of odd number of evens; even number of odds; even number of evens; odd number of odds. Find the parity of 1+2+…+100.
Steps only
Even summands contribute no odd remainder.
Pair odd summands; an odd count leaves one odd.
Pair 1 with 100, 2 with 99, etc.
Answer
Even; even; even; odd. The sum 1+…+100=5,050 is even.
Question 7
Two consecutive Virahanka terms are 987 and 1,597. Find the next two and previous two.
Steps only
Forward: add the latest pair.
Backward: subtract the earlier term from the later term.
Answer
Next: 2,584 and 4,181. Previous: 377 and 610. Sequence: …,377,610,987,1,597,2,584,4,181,…
Question 8
Angaan climbs 8 steps, taking either 1 or 2 at a time. How many distinct step sequences reach the top?
Steps only
Let W(n) be the number of ways to reach step n.
The last move is either a 1-step from n−1 or a 2-step from n−2, so W(n)=W(n−1)+W(n−2).
Start W(1)=1,W(2)=2.
Answer
W(8)=34 ways.
Question 9
Is the 20th term of 1,2,3,5,8,… odd or even?
Steps only
The odd/even cycle is O,E,O and repeats every three terms.
20 leaves remainder 2 when divided by 3.
Answer
Even.
Question 10
Decide whether these are true: (a) 4m−1 always odd; (b) all even numbers are 6j−4; (c) 2p+1 and 2q−1 each describe all odd numbers; (d) 2f+3 gives both parities.
Steps only
Check parity for arbitrary integer inputs.
A counterexample disproves an “all” claim.
Answer
(a) true; (b) false; (c) false as phrased—each form gives odd numbers, but with positive integer inputs they do not generate every odd number; (d) false: 2f+3 is always odd.
Question 12
A short syllable takes 1 beat and a long syllable takes 2. How many rhythms make 5 beats and 8 beats? Continue the count sequence and state how many petals the three pictured daisies have.
Steps only
A rhythm starts with either a 1-beat or a 2-beat syllable.
So the count for n beats equals the counts for n−1 and n−2 beats.
Start with counts 1,2 for one and two beats.
Answer
Counts are 1,2,3,5,8,13,21,34,… . There are 8 rhythms for 5 beats and 34 for 8 beats. The daisies shown have 13, 21 and 34 petals.
Question 11
Solve the printed letter-digit puzzle.
Steps only
Read the addition by columns: UT + TA = TAT.
The units column gives T+A=T, so A=0 and there is no carry.
The tens column then gives U+T=10T; since T is the nonzero hundreds digit, T=1 and U=9.
Answer
U=9, T=1 and A=0, so 91+10=101. In the units column, T+0 ends in T. In the tens column, U+T must make 10, carrying 1 to the hundreds column. Therefore T=1 and U=9.