Fun with Numbers — Questions and Answers
Welcome to HSLC Guru. This guide covers ASSEB Class 8 General Mathematics Chapter 16, Fun with Numbers, with the summary, the divisibility tests, the number games, numerical patterns, and letter-for-digit puzzles, and full step-by-step worked answers to every question of Exercise 16.1, 16.2 and 16.3.
Summary
This chapter explores the interesting properties and games of numbers. It first develops the tests of divisibility — quick ways to decide whether a number is divisible by another without actually dividing. The rules for divisibility by 2, 4 and 8; by 5, 25 and 125; by 3 and 9; by 6; and by 11 are proved, and $1001 = 7 \times 11 \times 13$ is used to test divisibility by 7 and 13.
Next come several fascinating number games — the four numbers 153, 370, 371 and 407 (each equal to the sum of the cubes of its digits), the magic of 37, tricks with consecutive numbers, the game of 9, the game of 1089 (Kaprekar’s number) and the game of 1001. The chapter closes with numerical patterns of squares and with letters for digits (cryptarithm) puzzles solved step by step.
An important idea introduced here is that a number divisible by two different numbers is divisible by their product only when the two numbers are co-prime (their HCF is 1). For example, since 2 and 3 are co-prime, a number divisible by both is divisible by 6; but a number divisible by 2 and 4 need not be divisible by 8, because the HCF of 2 and 4 is 2.
Summary: ASSEB Class 8 General Mathematics Chapter 16 Fun with Numbers covers divisibility tests (by 2, 4, 8; 5, 25, 125; 3, 9; 6; 11; and 7, 13 using 1001 = 7 × 11 × 13), amazing number games (153, 370, 371, 407; the 37 game; consecutive-number tricks; the games of 9, 1089 and 1001), numerical patterns of squares, and letter-for-digit (cryptarithm) puzzles. This HSLC Guru guide gives full step-by-step worked answers to Exercise 16.1, 16.2 and 16.3.
Textbook Questions and Answers
16.1 Tests of Divisibility — Rules and Examples
Rules of divisibility:
- By 2: the unit’s digit is $0$ or even.
- By 4: the number formed by the ten’s and unit’s digits is $00$ or divisible by $4$.
- By 8: the number formed by the hundred’s, ten’s and unit’s digits is $000$ or divisible by $8$.
- By 5: the unit’s digit is $0$ or $5$.
- By 25: the last two digits form $00$ or a multiple of $25$.
- By 125: the last three digits form $000$ or a multiple of $125$.
- By 3 or 9: the sum of all digits is divisible by $3$ or $9$.
- By 6: the number is divisible by both $2$ and $3$ (since $6 = 2 \times 3$).
- By 11: the difference of the sums of digits in odd and even places is $0$ or a multiple of $11$.
Example 1. Investigate the divisibility of $2359874$ by $2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$ and $13$.
Answer: Unit’s digit $4$ is even, so it is divisible by 2. Last two digits $74$ are not divisible by $4$, so it is not divisible by $4$ or $8$. Digit sum $2+3+5+9+8+7+4 = 38$ is not divisible by $3$ or $9$, so it is not divisible by $3, 6, 9$ or $12$. The unit’s digit is neither $0$ nor $5$, so it is not divisible by $5, 10, 25$ or $125$.
For $7, 11, 13$, group three digits from the right: $874,\ 359,\ 2$. The difference of the alternate-group sums is $(874 + 2) – 359 = 876 – 359 = 517$. Since $517 = 11 \times 47$, it is divisible by 11, but $517$ is divisible by neither $7$ nor $13$. Hence $2359874$ is divisible only by 2 and 11.
Example 2. Show that $151452$ is divisible by $21$.
Answer: $21 = 3 \times 7$. Digit sum $2+5+4+1+5+1 = 18$ is divisible by $3$. Grouping three digits from the right gives $452$ and $151$; $452 – 151 = 301 = 7 \times 43$, divisible by $7$. As $3$ and $7$ are co-prime, $151452$ is divisible by $3 \times 7 = 21$.
Example 3. Find the value of $x$ so that $13×2741$ is divisible by $11$.
Answer: The difference of the sums of digits in even and odd places must be $0$ or a multiple of $11$.
$$(1 + 7 + x + 1) – (4 + 2 + 3) = (9 + x) – 9 = x$$
Since $x$ is a single digit, $x = 0$ (a non-zero multiple of $11$ has at least two digits). Hence the number is $1302741$.
Example 4. Find the least number which when added to $2311$ makes it divisible by (i) $3$ (ii) $4$.
Answer: (i) Digit sum $2+3+1+1 = 7$; the next multiple of $3$ is $9$, so add $9 – 7 = 2$. (ii) Last two digits $11$; the next multiple of $4$ is $12$, so add $12 – 11 = 1$ (then $2311 + 1 = 2312$ is divisible by $4$).
Example 5. Put a digit to the right of $2785$ so that the new number is divisible by (i) $9$ (ii) $11$.
Answer: Let the new number be $2785x$. (i) Digit sum $22 + x$; the only multiple of $9$ between $22$ and $31$ is $27$, so $22 + x = 27 \Rightarrow x = 5$. (ii) For $11$, $x – 5 + 8 – 7 + 2 = x – 2$ must be $0$, so $x = 2$.
Exercise 16.1
1. Examine the divisibility of the following by $2, 3, 5, 9$: (i) $4253$ (ii) $18935$ (iii) $12123232$ (iv) $8753973$ (v) $333666$ (vi) $785634$.
Answer:
- (i) $4253$: unit $3$ (odd) → not by $2$; digit sum $14$ → not by $3, 9$; not ending in $0/5$ → not by $5$. Divisible by none.
- (ii) $18935$: ends in $5$ → divisible by 5; digit sum $26$ → not by $3, 9$; odd → not by $2$.
- (iii) $12123232$: ends in $2$ → divisible by 2; digit sum $16$ → not by $3, 9$; not by $5$.
- (iv) $8753973$: odd → not by $2$; digit sum $42$ → divisible by 3 but not by $9$; not by $5$.
- (v) $333666$: ends in $6$ → divisible by 2; digit sum $27$ → divisible by 3 and 9; not by $5$.
- (vi) $785634$: ends in $4$ → divisible by 2; digit sum $33$ → divisible by 3 but not by $9$; not by $5$.
2. Examine the divisibility of the following by $4, 6, 8, 11$: (i) $532740$ (ii) $347435$ (iii) $123456$ (iv) $693011$ (v) $1238932$.
Answer:
- (i) $532740$: last two $40$ → by 4; ends in $0$ (so by $2$) and digit sum $21$ (by $3$) → by 6; last three $740$ not by $8$; $11$: $(0+7+3)-(4+2+5) = -1$ → not by $11$.
- (ii) $347435$: last two $35$ → not by $4$; odd → not by $6, 8$; $11$: $(5+4+4)-(3+7+3) = 0$ → by 11.
- (iii) $123456$: last two $56$ → by 4; ends in $6$ and digit sum $21$ → by 6; last three $456 = 8 \times 57$ → by 8; $11$: $(6+4+2)-(5+3+1) = 3$ → not by $11$.
- (iv) $693011$: last two $11$ → not by $4$; odd → not by $6, 8$; $11$: $(1+0+9)-(1+3+6) = 0$ → by 11.
- (v) $1238932$: last two $32$ → by 4; digit sum $28$ (not by $3$) → not by $6$; last three $932$ not by $8$; $11$: $(2+9+3+1)-(3+8+2) = 2$ → not by $11$.
3. Examine the divisibility of the following by $7$ and $13$: (i) $2561876$ (ii) $864192$ (iii) $1604928$.
Answer: Group three digits from the right, take the difference of the alternate-group sums, and test that by $7$ and $13$.
- (i) $2561876$: groups $876,\ 561,\ 2$; difference $(876+2) – 561 = 317$. $317$ is divisible by neither $7$ nor $13$ → divisible by neither.
- (ii) $864192$: groups $192,\ 864$; difference $864 – 192 = 672 = 7 \times 96$ → divisible by 7; $672$ not divisible by $13$ → not by $13$. ($864192 = 7 \times 123456$.)
- (iii) $1604928$: groups $928,\ 604,\ 1$; difference $(928+1) – 604 = 325 = 13 \times 25$ → divisible by 13; $325$ not divisible by $7$ → not by $7$. ($1604928 = 13 \times 123456$.)
4. Find the value of $x$ in $25372x$ so that the number becomes divisible by (i) $3$ (ii) $9$.
Answer: Digit sum $= 2+5+3+7+2+x = 19 + x$.
(i) For $3$: $19 + x$ must be a multiple of $3$ → $19+x = 21, 24, 27$, so $x = 2, 5$ or $8$.
(ii) For $9$: $19 + x$ must be a multiple of $9$ → $19 + x = 27$, so $x = 8$.
5. Find the value of $x$ in $25×043$ so that it becomes divisible by $11$.
Answer: The difference of the sums of digits in odd and even places must be $0$.
$$(3 + 0 + 5) – (4 + x + 2) = 8 – (6 + x) = 2 – x$$
$2 – x = 0 \Rightarrow x = 2$. Hence the number is $252043$ ($= 11 \times 22913$). $x = 2$.
Simple Games of Numbers (16.2)
Sum of cubes of digits: each of $153, 370, 371$ and $407$ equals the sum of the cubes of its digits —
$$153 = 1^3 + 5^3 + 3^3,\quad 407 = 4^3 + 0^3 + 7^3$$
Fun with 37: multiply $37$ by $3$ or multiples of $3$ and fill the blanks.
Answer: $37 \times 18 = 666$, $37 \times 21 = 777$, $37 \times 24 = 888$, $37 \times 27 = 999$. Then a new order: $37 \times 30 = 1110$, $37 \times 33 = 1221$, $37 \times 36 = 1332$, $37 \times 39 = 1443$, $37 \times 42 = 1554$, $37 \times 45 = 1665$. Also $37037 \times 3 = 111111$ and $37037037 \times 3 = 111111111$.
Three consecutive numbers — find the sum: (i) $24+25+26$ (ii) $28+29+30$ (iii) $69+70+71$.
Answer: Sum of three consecutive numbers $=$ middle number $\times 3$. (i) $25 \times 3 = 75$ (ii) $29 \times 3 = 87$ (iii) $70 \times 3 = 210$.
Four consecutive numbers — find the sum: (i) $7+8+9+10$ (ii) $12+13+14+15$ (iii) $20+21+22+23$.
Answer: Sum $=$ (sum of the two middle numbers) $\times 2$. (i) $(8+9)\times 2 = 34$ (ii) $(13+14)\times 2 = 54$ (iii) $(21+22)\times 2 = 86$.
Game of 9 — reverse the two-digit number, subtract the smaller from the larger, divide by $9$: (i) $16$ (ii) $45$ (iii) $62$.
Answer: For $ab$ and its reverse $ba$, the difference is always $9(a-b)$, a multiple of $9$. (i) $61-16 = 45$, $45 \div 9 = 5$ (ii) $54-45 = 9$, $9 \div 9 = 1$ (iii) $62-26 = 36$, $36 \div 9 = 4$.
Game of 1089 — from a three-digit number (all digits different) make the largest and smallest numbers, subtract, then add the reverse of the result: (i) $327$ (ii) $291$ (iii) $456$ (iv) $786$.
Answer: Each gives $1089$ (Kaprekar’s number). (i) $732-237 = 495$, $495+594 = 1089$ (ii) $921-129 = 792$, $792+297 = 1089$ (iii) $654-456 = 198$, $198+891 = 1089$ (iv) $876-678 = 198$, $198+891 = 1089$.
Game of 1001 — write a three-digit number twice: (i) $234$ (ii) $175$ (iii) $432$.
Answer: Writing a three-digit number twice gives $1001$ times the number, and $1001 = 7 \times 11 \times 13$. (i) $234234 = 234 \times 1001$; $\div 7 = 33462$, $\div 11 = 3042$, $\div 13 = 234$ (ii) $175175 = 175 \times 1001$ (iii) $432432 = 432 \times 1001$. Dividing successively by $7, 11, 13$ returns the original number.
Some Numerical Patterns (16.3)
Pattern 1 — product and sum (fill the blanks):
Answer: Reversing the digits of the product gives the sum. $9 \times 9 = 81,\ 9+9 = 18$; $24 \times 3 = 72,\ 24+3 = 27$; $47 \times 2 = 94,\ 47+2 = 49$; $497 \times 2 = 994,\ 497+2 = 499$.
Pattern 2 — the eights (fill the blanks):
Answer: $9 \times 9 + 7 = 88$; $98 \times 9 + 8 = 888$; $987 \times 9 + 5 = 8888$; $\dots$; $98765432 \times 9 + 0 = 88888888$; $987654321 \times 9 + (-1) = 8888888888$; $9876543210 \times 9 + (-2) = 88888888888$.
Pattern 3 — difference of squares:
Answer: $6^2 – 5^2 = 11 = 11 \times 1$; $56^2 – 45^2 = 1111 = 101 \times 11$; $556^2 – 445^2 = 111111 = 1001 \times 111$.
Pattern 4 — what is special about $3816547290$?
Answer: It uses every digit $0$ to $9$ once, and each left-hand group of digits is divisible by the number of digits in that group — $3$ by $1$, $38$ by $2$, $381$ by $3$, $3816$ by $4$, and so on.
Pattern 5 — what is the significance of $\dfrac{148}{296} + \dfrac{35}{70}$?
Answer: Both fractions equal $\dfrac{1}{2}$, so
$$\frac{148}{296} + \frac{35}{70} = \frac{1}{2} + \frac{1}{2} = 1$$
Patterns of squares: $4^2 = 16,\ 34^2 = 1156,\ 334^2 = 111556,\ 3334^2 = 11115556, \dots$; $1^2 = 1,\ 11^2 = 121,\ 111^2 = 12321,\ 1111^2 = 1234321, \dots$ (palindrome patterns); $9^2 = 81,\ 99^2 = 9801,\ 999^2 = 998001,\ 9999^2 = 99980001, \dots$.
Exercise 16.2
Each triangle has a number at the apex, one at each base corner, and one in the middle. In each problem, find the rule (technique) from parts (a), (b), (c) and fill in the blank (?).
1. Middle number $=$ apex $\times$ (sum of the two base corners). ($2\times(3+5)=16$, $5\times(1+7)=40$, $3\times(9+2)=33$ …) Fill in the blanks.
Answer:
- (a) apex $8$, corners $3$ and $6$ → $? = 8 \times (3+6) = 8 \times 9 = \mathbf{72}$.
- (b) middle $56$, corners $5$ and $3$ → apex $? \times (5+3) = 56 \Rightarrow ? = 56 \div 8 = \mathbf{7}$.
- (c) apex $6$, middle $42$, corners $3$ and $?$ → $6 \times (3 + ?) = 42 \Rightarrow 3 + ? = 7 \Rightarrow ? = \mathbf{4}$.
2. Middle number $=$ (apex $-$ left corner) $\times$ right corner. ($(7-2)\times2=10$, $(14-9)\times4=20$ …) Fill in the blanks.
Answer:
- (a) apex $10$, corners $3$ and $2$ → $? = (10-3)\times 2 = 7 \times 2 = \mathbf{14}$.
- (b) second triangle: $(8-4)\times 3 = \mathbf{12}$; fourth triangle: $(7-3)\times ? = 24 \Rightarrow 4 \times ? = 24 \Rightarrow ? = \mathbf{6}$.
- (c) second triangle: $(4-2)\times ? = 20 \Rightarrow 2 \times ? = 20 \Rightarrow ? = \mathbf{10}$; fourth triangle: $(5-?)\times 4 = 12 \Rightarrow 5 – ? = 3 \Rightarrow ? = \mathbf{2}$.
3. Middle number $=$ (left corner $\times$ right corner $\times$ base number) $\div\ 10$. ($5\times6\times4\div10=12$, $6\times7\times5\div10=21$ …) Fill in the blanks.
Answer:
- (a) corners $3, 2$ and base $5$ → $? = \dfrac{3 \times 2 \times 5}{10} = \dfrac{30}{10} = \mathbf{3}$.
- (b) second: $\dfrac{? \times 10 \times 6}{10} = 42 \Rightarrow 6? = 42 \Rightarrow ? = \mathbf{7}$; fourth: $\dfrac{7 \times ? \times 2}{10} = 7 \Rightarrow 14? = 70 \Rightarrow ? = \mathbf{5}$.
- (c) second: $\dfrac{2 \times 2 \times ?}{10} = 8 \Rightarrow 4? = 80 \Rightarrow ? = \mathbf{20}$; third: $\dfrac{6 \times ? \times 4}{10} = 12 \Rightarrow 24? = 120 \Rightarrow ? = \mathbf{5}$; fourth: $\dfrac{? \times 8 \times 5}{10} = 24 \Rightarrow 40? = 240 \Rightarrow ? = \mathbf{6}$.
Game with Letters for Digits (16.4) — Examples
Rules: one letter stands for one digit only; the first digit of a number cannot be zero; each problem has one solution.
Example 1. In $A4 + 1B = 49$, find $A$ and $B$.
Answer: Unit’s place $4 + B = 9 \Rightarrow B = 5$; ten’s place $A + 1 = 4 \Rightarrow A = 3$. So $A = 3, B = 5$ ($34 + 15 = 49$).
Example 2. In $5AB + AB2 = B78$, find $A$ and $B$.
Answer: Unit’s place $B + 2 = 8 \Rightarrow B = 6$; ten’s place $A + B = 7 \Rightarrow A = 1$; hundred’s place $5 + A = B \Rightarrow 5 + 1 = 6$ ✓. So $A = 1, B = 6$ ($516 + 162 = 678$).
Example 3. In $ABB + ABB + ABB = 19AB$, find $A$ and $B$.
Answer: The unit’s digit of $B+B+B$ must be $B$, which forces $B = 5$ ($5+5+5 = 15$); carrying $1$, the hundreds give $A = 6$ ($6+6+6+1 = 19$). So $A = 6, B = 5$ ($655 \times 3 = 1965$).
Example 4. In $AB \times B = CAB$, find $A, B, C$.
Answer: For the unit’s digit of $B \times B$ to be $B$, $B = 5$ (or $6$). With $B = 5$: $(10A+5)\times 5 = 100C + 10A + 5 \Rightarrow 40A + 20 = 100C \Rightarrow 2A + 1 = 5C$, giving $A = 2, C = 1$. So $A = 2, B = 5, C = 1$ ($25 \times 5 = 125$).
Exercise 16.3
1. Find the value of the letters for digits (step wise) in each of the following:
(i) $6A + 87 = BA2$
Answer: For the unit’s digit of $A + 7$ to be $2$, $A = 5$ ($5+7 = 12$, carry $1$); $60 + 5 + 87 = 152 = BA2 \Rightarrow B = 1$. $A = 5, B = 1$ ($65 + 87 = 152$).
(ii) $21A + 1A3 = 368$
Answer: $(210 + A) + (100 + 10A + 3) = 313 + 11A = 368 \Rightarrow 11A = 55 \Rightarrow A = 5$. $A = 5$ ($215 + 153 = 368$).
(iii) $1AB + AB1 = B07$
Answer: $(100 + 10A + B) + (100A + 10B + 1) = 100B + 7$, which simplifies to $89B = 110A + 94$; with $A = 4$, $89B = 534 \Rightarrow B = 6$. $A = 4, B = 6$ ($146 + 461 = 607$).
(iv) $B2A + 3AB = A00$
Answer: $(100B + 20 + A) + (300 + 10A + B) = 100A$, which simplifies to $89A = 101B + 320$; with $B = 3$, $89A = 623 \Rightarrow A = 7$. $A = 7, B = 3$ ($327 + 373 = 700$).
(v) $30A6 + 424B + A3B6 = C6A7$
Answer: In the unit’s column, $6 + B + 6$ ends in $7$, so $B = 5$ (carry $1$). Hundreds: $0+2+3+1 = 6$ ✓. Thousands: $3 + 4 + A = C$; since $A$ is the leading digit of $A3B6$ (not $0$) and $C$ is a single digit, $A = 1$, giving $C = 8$. $A = 1, B = 5, C = 8$ ($3016 + 4245 + 1356 = 8617$).
(vi) $AAA + AAA + AAA = CBBA$
Answer: $3 \times AAA = 333A$; for the unit’s digit of $3A$ to be $A$, $A = 5$. $333 \times 5 = 1665 = CBBA \Rightarrow C = 1, B = 6$. $A = 5, B = 6, C = 1$ ($555 \times 3 = 1665$).
(vii) $AB \times 3 = CAB$
Answer: $3(10A+B) = 100C + 10A + B \Rightarrow 20A + 2B = 100C \Rightarrow 10A + B = 50C$; with $C = 1$, $AB = 50$. $A = 5, B = 0, C = 1$ ($50 \times 3 = 150$).
(viii) $BA \times B3 = 57A$
Answer: The product is in the $500$s, so $B = 2$; $(20+A)\times 23 = 570 + A \Rightarrow 460 + 23A = 570 + A \Rightarrow 22A = 110 \Rightarrow A = 5$. $A = 5, B = 2$ ($25 \times 23 = 575$).
(ix) $AB \times 6 = CBB$
Answer: $6(10A+B) = 100C + 11B \Rightarrow 12A – B = 20C$; for the unit’s digit of $6B$ to be $B$, $B$ must be even. Taking $A = 2, B = 4, C = 1$ gives $24 \times 6 = 144$ ✓ (one solution; $48\times6=288$, $50\times6=300$, $98\times6=588$ also satisfy the pattern). $A = 2, B = 4, C = 1$.
(x) $AB \times 6 = BBB$
Answer: $6(10A+B) = 111B \Rightarrow 60A = 105B \Rightarrow 4A = 7B$, so $A = 7, B = 4$. $A = 7, B = 4$ ($74 \times 6 = 444$).
(xi) $AB \times 5 = CAB$
Answer: $5(10A+B) = 100C + 10A + B \Rightarrow 40A + 4B = 100C \Rightarrow 10A + B = 25C$; with $C = 1$, $AB = 25$. $A = 2, B = 5, C = 1$ ($25 \times 5 = 125$; $50\times5=250$, $75\times5=375$ also satisfy the pattern).
(xii) $AB \times AB = CAB$
Answer: The last two digits of the product must be $AB$, i.e. $AB^2$ ends in $AB$; the only such two-digit number giving a three-digit square is $25$ ($25^2 = 625$). $A = 2, B = 5, C = 6$ ($25 \times 25 = 625$).
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. For divisibility by 8, the number formed by how many end digits must be $000$ or a multiple of $8$? (a) one (b) two (c) three (d) four
Answer: (c) three (unit’s, ten’s, hundred’s places).
2. $1001$ is the product of which prime numbers? (a) $7 \times 11 \times 13$ (b) $3 \times 7 \times 11$ (c) $7 \times 11 \times 17$ (d) $11 \times 13 \times 17$
Answer: (a) $7 \times 11 \times 13$.
3. Which number equals the sum of the cubes of its digits? (a) $123$ (b) $153$ (c) $250$ (d) $220$
Answer: (b) $153$ ($1^3 + 5^3 + 3^3 = 153$).
4. $12348$ is divisible by which of the following? (a) $5$ (b) $6$ (c) $11$ (d) $25$
Answer: (b) $6$ (divisible by both $2$ and $3$).
5. The number $1089$ is known as— (a) a prime number (b) Kaprekar’s number (c) a perfect square (d) a cube number
Answer: (b) Kaprekar’s number.
6. A number divisible by both $2$ and $4$ may not be divisible by $8$ ($=2\times4$) because the HCF of $2$ and $4$ is— (a) $1$ (b) $2$ (c) $4$ (d) $8$
Answer: (b) $2$ (they are not co-prime).
7. The value of $37 \times 24$ is— (a) $666$ (b) $777$ (c) $888$ (d) $999$
Answer: (c) $888$.
8. The value of $999^2$ is— (a) $9801$ (b) $99801$ (c) $998001$ (d) $980001$
Answer: (c) $998001$.
9. A number is divisible by $11$ if the difference of the sums of digits in odd and even places is— (a) always $1$ (b) $0$ or a multiple of $11$ (c) even (d) odd
Answer: (b) $0$ or a multiple of $11$.
10. The value of $111111^2$ is— (a) $123456$ (b) $12345654321$ (c) $1234321$ (d) $123454321$
Answer: (b) $12345654321$.
Fill in the Blanks
- A number is divisible by $5$ if its unit’s digit is $0$ or ______ . — $5$
- $6 = 2 \times$ ______ . — $3$
- The sum of three consecutive numbers $=$ the middle number $\times$ ______ . — $3$
- $37037 \times 3 =$ ______ . — $111111$
- $25 \times 23 =$ ______ . — $575$
True or False
- A number divisible by $3$ is always divisible by $9$. — False
- $864192$ is divisible by $7$. — True
- $407 = 4^3 + 0^3 + 7^3$. — True
- The game of 1089, done with any three-digit number of unequal digits, always ends in $1089$. — True
- $123456$ is divisible by $11$. — False
Short Answer Questions
1. How does $1001$ help in testing divisibility by $7$ and $13$?
Answer: $1001 = 7 \times 11 \times 13$. Group three digits at a time from the right, take the difference of the alternate-group sums; if that difference is divisible by $7$ or $13$, the original number is divisible by $7$ or $13$ respectively.
2. When is a number that is separately divisible by two different numbers also divisible by their product?
Answer: When the two numbers are co-prime, i.e. their HCF is $1$. For example, since HCF of $2$ and $3$ is $1$, a number divisible by both $2$ and $3$ is divisible by $6$.
3. State the three rules to follow while solving letters-for-digits problems.
Answer: (a) One letter stands for one digit only; (b) the first digit of a number cannot be zero; (c) each problem has one solution.
Key Terms
| Term | Meaning |
|---|---|
| Divisibility | The property of one number dividing another completely |
| Unit’s place | The first digit position from the right of a number |
| Sum of digits | The total of all the digits of a number |
| Co-prime | Two numbers whose HCF is $1$ |
| HCF | Highest Common Factor |
| Consecutive numbers | Numbers following one another (e.g. $9, 10, 11$) |
| Kaprekar’s number | $1089$ — a number with a special property |
| Numerical pattern | A regular structure hidden within numbers |
| Palindrome | Reads the same forwards and backwards |
| Letters for digits | A cryptarithm using letters to represent digits |