NCERT Solutions Ganita Prakash (Part 1) Chapter 6 –144Section 6.5 Digits in Disguise — Figure it Out

Book page 143 Updated on2026-09-05

Q1.

A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why?

Answer

The bulb will be OFF.

Each toggle changes the state: ON → OFF → ON → OFF …
After an even number of toggles the bulb is back to ON.
After an odd number of toggles the bulb is OFF.
77 is odd → the bulb is OFF.

Toggles	0	1	2	3	4	…	77
State	ON	OFF	ON	OFF	ON	…	OFF

Why it happens: the two toggles “ON → OFF” and “OFF → ON” cancel each other, exactly like a pair of dots. 77 toggles make 38 such pairs with one toggle left over, and that last toggle switches the bulb off.

Q2.

Liswini has a large old encyclopaedia. When she opened it, several loose pages fell out of it. She counted 50 sheets in total, each printed on both sides. Can the sum of the page numbers of the loose sheets be 6000? Why or why not?

Answer

Parity says yes, but a closer check says no — the sum can never be exactly 6000.

Step 1 — the parity test. Each sheet carries two page numbers, one odd and one even.

One sheet: odd + even = odd
50 sheets: 50 odd numbers added together
50 is an even count of odd numbers → the total is even
6000 is even, so parity does not rule it out.

Step 2 — look closer. If a sheet is the k^th leaf, its two pages are 2k – 1 and 2k.

Sum for one sheet = (2k – 1) + 2k = 4k – 1
Sum for 50 sheets = 4 × (k₁ + k₂ + … + k₅₀) – 50
If this equals 6000: 4 × (sum of k's) = 6050
sum of k's = 6050 ÷ 4 = 1512.5 — not a whole number!

So no set of 50 sheets can total 6000. The total always leaves remainder 2 when divided by 4, while 6000 leaves remainder 0.

Why it happens: the parity test is a good first filter, but it only checks divisibility by 2. Checking divisibility by 4 is a sharper test, and it settles this question.

Check it yourself: the closest reachable totals near 6000 are 5998 and 6002 — both leave remainder 2 on division by 4.

Q3.

Here is a 2 × 3 grid. For each row and column, the parity of the sum is written in the circle; ‘e’ for even and ‘o’ for odd. Fill the 6 boxes with 3 odd numbers (‘o’) and 3 even numbers (‘e’) to satisfy the parity of the row and column sums.

Answer

The circles read: row 1 → o, row 2 → e, column 1 → e, column 2 → e, column 3 → o.

One arrangement that works:

	C1	C2	C3	Row sum	Required
R1	1	2	4	7	o ✓
R2	3	6	5	14	e ✓
Column sum	4	8	9	—	—
Required	e ✓	e ✓	o ✓	—	—

Odd numbers used: 1, 3, 5 (three of them). Even numbers used: 2, 4, 6 (three of them). ✓

In terms of parity the pattern is:

Row 1: o e e → o + e + e = odd ✓
Row 2: o e o → o + e + o = even ✓
Column 1: o + o = even ✓ Column 2: e + e = even ✓ Column 3: e + o = odd ✓

Try This: another pattern that works is e o e in row 1 and e o o in row 2 — for example 2, 1, 4 over 6, 3, 5.

Q4.

Make a 3 × 3 magic square with 0 as the magic sum. All numbers can not be zero. Use negative numbers, as needed.

Answer

Magic sum = 3m, so take m = 0 in the generalised form.

3	– 4	1
– 2	0	2
– 1	4	– 3

Rows: 3 – 4 + 1 = 0, –2 + 0 + 2 = 0, –1 + 4 – 3 = 0 ✓
Columns: 3 – 2 – 1 = 0, –4 + 0 + 4 = 0, 1 + 2 – 3 = 0 ✓
Diagonals: 3 + 0 – 3 = 0, 1 + 0 – 1 = 0 ✓

Why it happens: the nine entries are – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4 — nine consecutive numbers with 0 in the middle. Every number is matched by its opposite, so all lines cancel to zero.

Tip: a quick way to get it — take 8 1 6 / 3 5 7 / 4 9 2 and subtract 5 from every entry.

Q5.

Fill in the following blanks with ‘odd’ or ‘even’: (a) Sum of an odd number of even numbers is ______ (b) Sum of an even number of odd numbers is ______ (c) Sum of an even number of even numbers is ______ (d) Sum of an odd number of odd numbers is ______

Answer

	Sum of …	Answer	Example
(a)	an odd number of even numbers	even	2 + 4 + 6 = 12
(b)	an even number of odd numbers	even	1 + 3 + 5 + 7 = 16
(c)	an even number of even numbers	even	2 + 4 = 6
(d)	an odd number of odd numbers	odd	1 + 3 + 5 = 9

Why it happens: even numbers never leave a stray dot, so however many you add, the sum stays even — that settles (a) and (c). Odd numbers each leave one stray dot; those strays cancel two at a time, so an even count gives even (b) and an odd count leaves one dot over, giving odd (d).

Q6.

What is the parity of the sum of the numbers from 1 to 100?

Answer

The sum is even.

From 1 to 100 there are 50 odd numbers (1, 3, 5, …, 99) and 50 even numbers.
The 50 even numbers add to an even total.
50 odd numbers = an even count of odd numbers → even total.
even + even = even

Checking with the actual value:

1 + 2 + 3 + … + 100 = (1 + 100) × 100 ÷ 2 = 101 × 50 = 5050 — even ✓

Tip: the parity was found without computing 5050 at all — just by counting how many odd numbers there are.

Q7.

Two consecutive numbers in the Virahāṅka sequence are 987 and 1597. What are the next 2 numbers in the sequence? What are the previous 2 numbers in the sequence?

Answer

Next two: 2584 and 4181. Previous two: 610 and 377.

Going forward (add the two before):
987 + 1597 = 2584
1597 + 2584 = 4181

Going backward (subtract):
1597 – 987 = 610
987 – 610 = 377

So the stretch of the sequence is:

…, 377, 610, 987, 1597, 2584, 4181, …

Why it happens: the rule a + b = c can be read backwards as c – b = a, so the same rule runs in both directions.

Check it yourself: 377 + 610 = 987 ✓ and 610 + 987 = 1597 ✓.

Q8.

Angaan wants to climb an 8-step staircase. His playful rule is that he can take either 1 step or 2 steps at a time. For example, one of his paths is 1, 2, 2, 1, 2. In how many different ways can he reach the top?

Answer

Angaan can reach the top in 34 different ways.

Staircase of n steps	1	2	3	4	5	6	7	8
Number of ways	1	2	3	5	8	13	21	34

ways(3) = ways(2) + ways(1) = 2 + 1 = 3
ways(4) = 3 + 2 = 5
ways(5) = 5 + 3 = 8
ways(6) = 8 + 5 = 13
ways(7) = 13 + 8 = 21
ways(8) = 21 + 13 = 34

Why it happens: Angaan's last move is either a 1-step (so he was on step 7) or a 2-step (so he was on step 6). These two cases never overlap, so the counts add — exactly the Virahāṅka rule. It is the very same problem as writing 8 as a sum of 1s and 2s.

Did you know? this is why 8 beats of short and long syllables also give 34 rhythms — the poem and the staircase are the same puzzle in different clothes.

Q9.

What is the parity of the 20th term of the Virahāṅka sequence?

Answer

The 20th term is even.

Parity pattern: O, E, O, O, E, O, O, E, O, O, E, …
After the first term the block odd, odd, even repeats.
Even terms sit at positions 2, 5, 8, 11, 14, 17, 20, …
Each of these is 3 more than the previous, and 20 = 2 + 3 × 6 → the 20th term is even.

Confirming with the actual sequence:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946
The 20th term is 10946, which is even ✓

Why it happens: parity depends only on the two previous parities, so it must fall into a repeating cycle. Here the cycle has length 3, so you only need to know the position number's remainder on division by 3.

Q10.

Identify the statements that are true. (a) The expression 4m – 1 always gives odd numbers. (b) All even numbers can be expressed as 6j – 4. (c) Both expressions 2p + 1 and 2q – 1 describe all odd numbers. (d) The expression 2f + 3 gives both even and odd numbers.

Answer

Only statement (a) is true.

	Statement	True / False	Reason
(a)	4m – 1 always gives odd numbers	True	4m is always even, and even – 1 = odd. Values: 3, 7, 11, 15, …
(b)	All even numbers can be written as 6j – 4	False	6j – 4 gives 2, 8, 14, 20, … It skips 4, 6, 10, 12, 16, …
(c)	Both 2p + 1 and 2q – 1 describe all odd numbers	False	With letter-numbers 1, 2, 3, …, 2q – 1 gives 1, 3, 5, … but 2p + 1 gives 3, 5, 7, … and misses 1.
(d)	2f + 3 gives both even and odd numbers	False	2f is even and 3 is odd, so 2f + 3 is always odd: 5, 7, 9, 11, …

Why it happens: for (b), 6j – 4 = 2(3j – 2), so it only produces even numbers whose half is of the form 3j – 2 — it jumps in steps of 6 and misses most even numbers. For (d) the parity is locked: an even part plus an odd part is always odd.

Tip: in (c), if the letter-number were allowed to be 0 as well, then 2p + 1 with p = 0, 1, 2, … would also give 1, 3, 5, … and the statement would become true. Here we follow the book, where a letter-number stands for a counting number 1, 2, 3, …

Q11.

Solve this cryptarithm: UT + TA = TAT

Answer

U = 9, T = 1, A = 0, that is, 91 + 10 = 101.

Write the numbers using place value:
UT = 10U + T • TA = 10T + A • TAT = 100T + 10A + T = 101T + 10A

(10U + T) + (10T + A) = 101T + 10A
10U + 11T + A = 101T + 10A
10U = 90T + 9A
10U = 9 × (10T + A)

The right side is a multiple of 9, so 10U must be a multiple of 9 → U = 0 or U = 9.
U = 0 is impossible (UT would not be a 2-digit number), so U = 9.
90 = 9 × (10T + A) → 10T + A = 10 → T = 1, A = 0

Check the addition:

9 1
+ 1 0
——
1 0 1 ✓

Why it happens: TAT is a 3-digit answer from adding two 2-digit numbers, so its leading digit T can only be a carry of 1. Once T = 1, the units column gives T + A ending in T, that is 1 + A ending in 1, so A = 0 — and then U falls out at once.

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