NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 5 — Figure it Out (Pages 132–134)#

The sum of the digits must be a multiple of 9:

Sum of digits $= 3 + 1 + z + 5$
Sum of digits $= 9 + z$

$9 + z$ is a multiple of 9 when $z = 0$ (sum 9) or $z = 9$ (sum 18). The next one, 27, would need $z = 18$, which is not a digit.

∴ z = 0 or z = 9, giving 3105 and 3195. There are two answers because 9 more than a multiple of 9 is again a multiple of 9, and both $0$ and $9$ are single digits.

The numbers are $12n + 8$ and $12m - 4$. Their sum is:

Writing $k = n + m$, the sum is $12k + 4$:

• $k = 1$: $16$, a multiple of 8 ✓ (for example, $8 + 8$)
• $k = 2$: $28$, not a multiple of 8 ✗ (for example, $20 + 8$)

∴ Snehal’s claim is false. The sum is a multiple of 8 only sometimes, when $12k + 4$ happens to be (that is, when $k$ is odd).

Let the two multiples of 3 be $3m$ and $3n$. Their sum is $3(m + n)$, which is a multiple of 6 only when $m + n$ is even.

Generalisation: the sum of two multiples of 3 is a multiple of 6 exactly when the two numbers are both even or both odd.

True for every multiple of 9. Reversing the digits does not change their sum, and divisibility by 9 depends only on the sum of the digits. For example, $918 = 9 \times 102$ and $819 = 9 \times 91$.

Yes, every shuffle works. Any rearrangement of the digits keeps the same digit sum, so the new number is still a multiple of 9. For example, the shuffles of 918 include 189, 198, 891 and 981, all multiples of 9.

$18 = 2 \times 9$, so the number must be divisible by both 2 and 9.

• By 2: $b$ is even, so $b = 0, 2, 4, 6$ or $8$.
• By 9: the digit sum $4 + 8 + a + 2 + 3 + b = 17 + a + b$ must be a multiple of 9. Since $a + b$ is at most 17, $17 + a + b$ is between 17 and 34, so it must be 18 or 27. That is, $a + b = 1$ or $a + b = 10$.

For each even $b$:

∴ The possible pairs are (a, b) = (1, 0), (8, 2), (6, 4), (4, 6) and (2, 8), giving 481230, 488232, 486234, 484236 and 482238.

$44 = 4 \times 11$, so the number must be divisible by both 4 and 11.

• By 4: the last two digits $q8$ must be a multiple of 4. This is true for $q = 0, 2, 4, 6, 8$ (08, 28, 48, 68, 88).
• By 11: the digits in the odd places (from the left: 3, 7, 8) add up to 18, and those in the even places ($p$, $q$) add up to $p + q$. So $18 - (p + q)$ must be 0 or a multiple of 11.
  • $18 - (p + q) = 0$ gives $p + q = 18$, so $p = q = 9$, but $q = 9$ is not allowed (not even). ✗
  • $18 - (p + q) = 11$ gives $p + q = 7$.

With $q$ even and $p + q = 7$:

∴ The possible pairs are (p, q) = (7, 0), (5, 2), (3, 4) and (1, 6), giving 37708, 35728, 33748 and 31768.

Let the numbers be $n$, $n + 1$ and $n + 2$. We need $n$ to be even, $n + 1$ a multiple of 3, and $n + 2$ a multiple of 4. Trying small even numbers:

2, 3, 4 works: 2 is a multiple of 2, 3 of 3, and 4 of 4.

Adding the LCM of 2, 3 and 4 keeps every condition true:

$LCM(2, 3, 4) = 12$

So the next sets are 14, 15, 16 and 26, 27, 28, and so on.

∴ Yes, there are more: such sets occur once in every 12 numbers, starting at $12k + 2$.

Approach: $36 = 4 \times 9$, so a multiple of 36 must be divisible by 4 (last two digits a multiple of 4) and by 9 (digit sum a multiple of 9).

The first one after 45,000: $45000 = 36 \times 1250$ exactly, so the next multiples are found by adding 36 each time:

$36 \times 1251 = 45036$
$36 \times 1252 = 45072$
$36 \times 1253 = 45108$
$36 \times 1254 = 45144$
$36 \times 1255 = 45180$

Check 45036: its last two digits 36 are a multiple of 4, and $4 + 5 + 0 + 3 + 6 = 18$ is a multiple of 9. ✓

∴ Five such multiples are 45036, 45072, 45108, 45144 and 45180 (there are 55 in all between 45,000 and 47,000).

Consecutive even numbers differ by 2. Two numbers come before $5p$ and two after:

∴ The five numbers are $5p - 4$, $5p - 2$, $5p$, $5p + 2$ and $5p + 4$.

Let the number be $abcdef$.

• Divisible by 15 means divisible by 3 and 5. By 5: $f = 0$ or $5$. But the reversed number $fedcba$ must also be a 6-digit number, so $f \ne 0$. Hence $f = 5$.
• Reversed, divisible by 6 means divisible by 2 and 3. Its last digit is $a$, so $a$ is even (and not 0).
• Divisible by 3: the digit sum is the same for both numbers, so it must be a multiple of 3.

Take $a = 2$ and $f = 5$; then $2 + 5 = 7$, so the middle four digits must add to 2, 5, 8, … Choosing $b = c = d = 0$ and $e = 2$:

Number $= 200025$

Check: $200025 = 15 \times 13335$ ✓, and the reversed number $520002 = 6 \times 86667$ ✓.

∴ One such number is 200025. Others include 200055, 200085 and 202005.

Let $n$ be any multiple of 11, so $n = 11k$. Doubling it:

This is always a multiple of 11.

∴ Deepak’s conjecture is false: every multiple of 11 stays a multiple of 11 when doubled (in fact, when multiplied by any whole number).

Always True. The product is $6m \times 3n = 18mn = 9 \times 2mn$, a multiple of 9.

Always True. Three consecutive even numbers are $2n - 2$, $2n$ and $2n + 2$:

which is a multiple of 6. For example, $4 + 6 + 8 = 18$.

Always True. A multiple of 6 must be even and have a digit sum that is a multiple of 3.

• badcef has the same last digit $f$, so it is still even.
• It has the same digits, so the same digit sum, still a multiple of 3.

So badcef is also a multiple of 6.

Never True. We simplify the expression:

It is always 8 more than a multiple of 12, so it is never a multiple of 12.

Every number leaves a remainder of 0, 1 or 2 when divided by 3. The sum is divisible by 3 exactly when the three remainders add up to a multiple of 3 (0, 3 or 6). Going through all the cases:

Generalisation: the sum of three numbers is divisible by 3 when they all leave the same remainder on division by 3, or all different remainders.

Two consecutive integers: yes, always a multiple of 2. Of any two consecutive integers, one is even, so the product is even. Example: $7 \times 8 = 56$.

Three consecutive integers: yes, always a multiple of 6. Among any three consecutive integers, at least one is even, and exactly one is a multiple of 3, so the product is divisible by 2 and by 3. Example: $4 \times 5 \times 6 = 120$.

Four consecutive integers: always a multiple of 24 ($= 1 \times 2 \times 3 \times 4$). Among four consecutive integers, one is a multiple of 4, another is a different even number, and one is a multiple of 3, so the product is divisible by $4 \times 2 \times 3$. Example: $1 \times 2 \times 3 \times 4 = 24$.

Five consecutive integers: always a multiple of 120 ($= 1 \times 2 \times 3 \times 4 \times 5$). On top of the reasons for four integers, one of the five is a multiple of 5. Example: $2 \times 3 \times 4 \times 5 \times 6 = 720 = 120 \times 6$.

$GGG = G \times 111$, and $111 = 3 \times 37$. So the product contains the prime 37. Since $E$ is a single digit, the 37 must come from the 2-digit number $EF$: EF is 37 or 74.

• $EF = 37$, so $E = 3$: $37 \times 3 = 111$ ✓ — this is $GGG$ with $G = 1$.
• $EF = 74$, so $E = 7$: $74 \times 7 = 518$ ✗ — not of the form $GGG$.

∴ E = 3, F = 7, G = 1: $37 \times 3 = 111$.

Finding W. The units digit of $W \times 5$ is $W$ itself. Multiples of 5 end in 0 or 5, and W is a first digit, so it cannot be 0: W = 5.

Finding M. $5O5 \times 5$ lies between $505 \times 5 = 2525$ and $595 \times 5 = 2975$, so M = 2.

Finding O. The product $5O5 \times 5$ must end in “O5”. We try each O:

($O = 5$ is not allowed, since $W = 5$.) Only $O = 7$ works, giving $575 \times 5 = 2875$, so E = 8.

∴ W = 5, O = 7, M = 2, E = 8: $575 \times 5 = 2875$.

• Every multiple of 32 is a multiple of 8 (since $32 = 8 \times 4$), so the multiples of 32 lie inside the multiples of 8.
• Every multiple of 8 is a multiple of 4, so the multiples of 8 lie inside the multiples of 4.
• But not the other way round: 12 is a multiple of 4 and not of 8, and 16 is a multiple of 8 and not of 32.

So the circles must be nested: multiples of 32 innermost, then multiples of 8, with multiples of 4 outermost.

∴ Venn diagram (iv) shows the relationship correctly.