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PLAYING WITH NUMBERS - Q&A

EXERCISE 5(A)

1. Multiple Choice Type: Choose the correct answer from the options given below.

(i) Generalized form of a three-digit number xyz is:
(a) x + y + z
(b) 100x + 10y + z
(c) 100z + 10y + x
(d) 100y + 10x + z
Answer: (b) 100x + 10y + z
Explanation: In the number xyz, x is at the hundreds place, y is at the tens place, and z is at the units place.
Value = (x × 100) + (y × 10) + (z × 1) = 100x + 10y + z.

(ii) The usual form of 100a + b + 10c is:
(a) abc
(b) cab
(c) bac
(d) acb
Answer: (d) acb
Explanation:
100a → 'a' is at the hundreds place.
10c → 'c' is at the tens place.
b → 'b' is at the units place.
So, the number is acb.

(iii) If 5 × A = CA, then the values of A and C are:
(a) A = 5, C = 1
(b) A = 4, C = 2
(c) A = 5, C = 2
(d) A = 2, C = 5
Answer: (c) A = 5, C = 2
Steps:
The equation is 5 × A = CA.
If A = 5, then 5 × 5 = 25.
Here, the unit digit is 5 (which is A) and the tens digit is 2 (which is C).
This satisfies the condition CA = 25. Thus, A=5, C=2.

(iv) If 5A + 25 is equal to B2, then the value of A + B is:
(a) 15
(b) 10
(c) 8
(d) 7
Answer: (a) 15
Steps:
Arranging in column addition:
  5 A
+ 2 5
-------
  B 2
For the units column: A + 5 = number ending in 2.
A + 5 = 12 implies A = 7.
Carry over is 1.
For the tens column: 1 (carry) + 5 + 2 = B.
1 + 5 + 2 = 8. So, B = 8.
Value of A + B = 7 + 8 = 15.

2. Write the following numbers in generalized form:

(i) 85
Answer: 10 × 8 + 5

(ii) 502
Answer: 100 × 5 + 10 × 0 + 2

(iii) 347
Answer: 100 × 3 + 10 × 4 + 7

3. Write the usual form of the following:

(i) 10 × 4 + 9
Answer: 49

(ii) 100 × 5 + 10 × 2 + 8
Answer: 528

(iii) 100 × 3 + 7
Answer: 307
Explanation: There is no tens term, so the tens digit is 0. Number = 307.

4. The sum of the digits of a two-digit number is 9. The number formed by reversing the digits is greater than the original number by 27. Find the original number.
Answer: 36
Steps:
Let the number be 10x + y.
Given: x + y = 9 ... (i)
Number reversed = 10y + x.
Condition: (10y + x) - (10x + y) = 27
9y - 9x = 27 ⇒ 9(y - x) = 27 ⇒ y - x = 3 ... (ii)
Adding (i) and (ii): (x + y) + (y - x) = 9 + 3 ⇒ 2y = 12 ⇒ y = 6.
From (i), x + 6 = 9 ⇒ x = 3.
Original number = 10(3) + 6 = 36.

5. In a two-digit number, the unit's digit is 3 more than the ten's digit. The sum of the number and the number formed by reversing the digits is 143. Find the number.
Answer: 58
Steps:
Let tens digit = x, units digit = y. Number = 10x + y.
Given: y = x + 3 ... (i)
Sum of number and reverse: (10x + y) + (10y + x) = 143
11x + 11y = 143 ⇒ 11(x + y) = 143 ⇒ x + y = 13 ... (ii)
Substitute (i) into (ii): x + (x + 3) = 13 ⇒ 2x = 10 ⇒ x = 5.
y = 5 + 3 = 8.
The number is 58.

6. Without actual calculation, find the quotient when the sum of 25 and 52 is divided by:
(i) 11 Answer: 7
Reason: The sum of a two-digit number ab and its reverse ba is 11(a+b). Here a=2, b=5. Sum = 11(2+5) = 11 × 7. Quotient when divided by 11 is (a+b) = 7.
(ii) 7 Answer: 11
Reason: As shown above, Sum = 11 × 7. Quotient when divided by 7 is 11.

7. Without actual calculation, find the quotient when the difference of 62 and 26 is divided by:
(i) 9 Answer: 4
Reason: The difference of ab and ba (where a > b) is 9(a - b). Here a=6, b=2. Difference = 9(6 - 2) = 9 × 4. Quotient when divided by 9 is 4.
(ii) 4 Answer: 9
Reason: As shown above, Difference = 9 × 4. Quotient when divided by 4 is 9.

EXERCISE 5(B)

1. Find the values of the letters in each of the following and give reasons for the steps involved:

(i)
  3 A
+ 2 5
-------
  B 2
Answer: A = 7, B = 6
Steps:
1. Units column: A + 5 = 2. This is impossible for digits unless A + 5 = 12. So, A = 12 - 5 = 7.
2. A carry of 1 goes to the tens column.
3. Tens column: 1 (carry) + 3 + 2 = B ⇒ B = 6.

(ii)
  4 A
+ 9 8
-------
C B 3
Answer: A = 5, B = 4, C = 1
Steps:
1. Units column: A + 8 ends in 3. So, A + 8 = 13 ⇒ A = 5. Carry = 1.
2. Tens column: 1 (carry) + 4 + 9 = 14.
3. The result is CB, so C = 1 and B = 4.

(iii)
  1 A
× A
-------
  9 A
Answer: A = 6
Steps:
1. Units digit: A × A ends in A. Possible values for A: 0, 1, 5, 6.
2. Tens digit check:
If A=0: 10 × 0 = 0 (Not 90)
If A=1: 11 × 1 = 11 (Not 91)
If A=5: 15 × 5 = 75 (Not 95)
If A=6: 16 × 6 = 96 (Matches 9A). So, A = 6.

(iv)
  A B
+ 3 7
-------
  6 A
Answer: A = 2, B = 5
Steps:
1. Units column: B + 7 = A (ends in A).
2. Tens column: A + 3 + (carry?) = 6.
Let's test A. Since A + 3 is close to 6, A is likely 2 or 3.
If carry is 0: A + 3 = 6 ⇒ A = 3. Then B + 7 ends in 3 ⇒ B = 6. Check: 36+37 = 73 (Not 63). Fail.
If carry is 1: A + 3 + 1 = 6 ⇒ A = 2. Then B + 7 ends in 2 (12) ⇒ B = 5. Check: 25+37 = 62 (Matches 6A). So A=2, B=5.

2. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Answer: 1
Steps:
Divisibility rule for 9: Sum of digits must be a multiple of 9.
Sum = 2 + 1 + y + 5 = 8 + y.
Possible multiples of 9 are 9, 18, etc.
8 + y = 9 ⇒ y = 1.
8 + y = 18 ⇒ y = 10 (Not a digit).
So, y = 1.

3. If 24x is a multiple of 3, where x is a digit, what is the value of x?
Answer: 0, 3, 6, or 9
Steps:
Divisibility rule for 3: Sum of digits must be a multiple of 3.
Sum = 2 + 4 + x = 6 + x.
Since 6 is already divisible by 3, x must be a multiple of 3.
x = 0, 3, 6, 9.

Test Yourself

1. Multiple Choice Type: Choose the correct answer from the options given below.

(i) If the sum of digits of a number is divisible by 3, then the number is always divisible by:
(a) 2
(b) 3
(c) 6
(d) 9
Answer: (b) 3
Reason: This is the standard divisibility test for 3.

(ii) If the division N ÷ 5 leaves a remainder of 3, what might be the one's digit of N?
(a) 1 or 6
(b) 2 or 7
(c) 3 or 8
(d) 4 or 9
Answer: (c) 3 or 8
Steps: Numbers divisible by 5 end in 0 or 5.
Remainder 3 means the unit digit is 0+3=3 or 5+3=8.

(iii) The value of abc - cba is always divisible by:
(a) 9
(b) 11
(c) 99
(d) none of the above
Answer: (c) 99
Explanation: abc = 100a + 10b + c. cba = 100c + 10b + a.
Difference = (100a + 10b + c) - (100c + 10b + a) = 99a - 99c = 99(a - c).
Since 99 is a factor, it is divisible by 99 (and also 9 and 11). 99 is the strongest answer.

(iv) If abc is a three-digit number, then abc + bca + cab is always divisible by:
(a) 3
(b) 37
(c) 9
(d) 11
Answer: (b) 37
Explanation:
abc = 100a + 10b + c
bca = 100b + 10c + a
cab = 100c + 10a + b
Sum = 111a + 111b + 111c = 111(a + b + c) = 3 × 37 × (a + b + c).
So it is divisible by 37 (and 3, 111). Option (b) is the distinct property.

(v) Statement 1: If a number is divisible by 3, it must be divisible by 9.
Statement 2: If a number is divisible by 9, it must be divisible by 3.
(a) Both statements are true.
(b) Both statements are false.
(c) Statement 1 is true, and Statement 2 is false.
(d) Statement 1 is false, and Statement 2 is true.
Answer: (d) Statement 1 is false, and Statement 2 is true.
Reason: 6 is divisible by 3 but not 9 (S1 False). 9 is a multiple of 3, so any number divisible by 9 is divisible by 3 (S2 True).

2. Find the value of A and B in the following:
  A B
×   3
-------
C A B
Answer: A = 5, B = 0, C = 1
Steps:
1. Units digit: B × 3 ends in B. This happens if B = 0 (0×3=0) or B = 5 (5×3=15).
Case 1: If B = 0.
A 0
× 3
-----
C A 0
Tens digit: 3 × A ends in A. Possible for A=0 or A=5.
If A=0, Number is 00, not valid. If A=5: 50 × 3 = 150. Here C=1, A=5, B=0. Matches pattern CAB.
Case 2: If B = 5.
A 5
× 3
-----
C A 5
Units: 5×3=15 (Carry 1).
Tens: (3 × A) + 1 ends in A.
3A + 1 = 10k + A ⇒ 2A + 1 = 10k. 2A+1 is odd, 10k is even. No solution.
Therefore, A=5, B=0, C=1.

3. Check the divisibility of 2146587 by 3.
Answer: Divisible
Steps: Sum of digits = 2 + 1 + 4 + 6 + 5 + 8 + 7 = 33.
33 is divisible by 3. Therefore, the number is divisible by 3.

4. Check the divisibility of 15287 by 9.
Answer: Not Divisible
Steps: Sum of digits = 1 + 5 + 2 + 8 + 7 = 23.
23 is not divisible by 9. Therefore, the number is not divisible by 9.

5. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Answer: 0, 3, 6, 9
Steps: Sum = 3 + 1 + z + 5 = 9 + z.
9 is divisible by 3. So z must be divisible by 3.
z = 0, 3, 6, 9.

6. Find the values of the letters in:
  1 2 A
+ 6 A B
-------
  A 0 9
Answer: A = 8, B = 1
Steps:
1. Units column: A + B = 9 (No carry) or 19 (Carry 1).
2. Tens column: 2 + A + (carry) ends in 0.
If carry from units is 0: 2 + A = 10 ⇒ A = 8.
If A=8, then from units A + B = 9 ⇒ 8 + B = 9 ⇒ B = 1. Valid.
Check Hundreds: 1 + 6 + (carry from tens=1) = 8. Matches A=8.
If carry from units is 1: 2 + A + 1 = 10 ⇒ A = 7.
If A=7, then from units A + B = 19 ⇒ 7 + B = 19 ⇒ B = 12 (Not a digit). Invalid.
So, A = 8, B = 1.

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Quick Review Flashcards - Click to flip and test your knowledge!
Question
What does it mean for a number to be in 'generalized form'?
Answer
It is expressed as the sum of the products of its digits with their respective place values.
Question
How is a general two-digit number with tens digit 'a' and units digit 'b' written in generalized form?
Answer
It is written as $10a + b$.
Question
In the context of generalized forms, what is the crucial difference between the notation 'ab' and '$a \times b$?'
Answer
The notation 'ab' represents a two-digit number, not the product of 'a' and 'b'.
Question
How is a general three-digit number with digits 'a', 'b', and 'c' (from left to right) written in generalized form?
Answer
It is written as $100a + 10b + c$.
Question
For a two-digit number represented as $10a + b$, what is the range of whole numbers for the digit 'a'?
Answer
The digit 'a' can be any whole number from 1 to 9.
Question
For a three-digit number represented as $100a + 10b + c$, what is the range of whole numbers for the digits 'b' and 'c'?
Answer
The digits 'b' and 'c' can be any whole number from 0 to 9.
Question
The sum of a two-digit number `ab` and the number formed by reversing its digits (`ba`) is always completely divisible by what two quantities?
Answer
The sum is always divisible by 11 and by $(a+b)$.
Question
The expression for the sum of a two-digit number and its reverse, $ab + ba$, simplifies to what algebraic expression?
Answer
The expression simplifies to $11(a+b)$.
Question
When the sum of a two-digit number and its reverse, $ab + ba$, is divided by 11, what is the quotient?
Answer
The quotient is $(a+b)$.
Question
When the sum of a two-digit number and its reverse, $ab + ba$, is divided by $(a+b)$, what is the quotient?
Answer
The quotient is 11.
Question
The difference between a two-digit number `ab` and the number formed by reversing its digits (`ba`) is always completely divisible by what two quantities?
Answer
The difference is always divisible by 9 and by $(a-b)$ or $(b-a)$.
Question
If $a > b$, the expression for the difference of a two-digit number and its reverse, $ab - ba$, simplifies to what algebraic expression?
Answer
The expression simplifies to $9(a-b)$.
Question
If $a > b$, what is the quotient when the difference $ab - ba$ is divided by 9?
Answer
The quotient is $(a-b)$.
Question
If $b > a$, what is the quotient when the difference $ba - ab$ is divided by 9?
Answer
The quotient is $(b-a)$.
Question
The sum of a three-digit number `abc` and the numbers formed by its cyclic shifts, `bca` and `cab`, is always completely divisible by what two quantities?
Answer
The sum is always divisible by 111 and by $(a+b+c)$.
Question
The expression for the sum $abc + bca + cab$ simplifies to what algebraic expression?
Answer
The expression simplifies to $111(a+b+c)$.
Question
When the sum $abc + bca + cab$ is divided by 111, what is the quotient?
Answer
The quotient is $(a+b+c)$.
Question
When the sum $abc + bca + cab$ is divided by $(a+b+c)$, what is the quotient?
Answer
The quotient is 111.
Question
What is cryptarithmetics?
Answer
It is a type of mathematical puzzle where letters represent digits in an arithmetic problem, and one must find the value of each letter.
Question
What is the rule for testing divisibility by 10?
Answer
A number is divisible by 10 if its unit digit is 0.
Question
What is the rule for testing divisibility by 5?
Answer
A number is divisible by 5 if its unit digit is 0 or 5.
Question
What is the rule for testing divisibility by 2?
Answer
A number is divisible by 2 if its unit digit is zero or an even number (0, 2, 4, 6, 8).
Question
What is the rule for testing divisibility by 9?
Answer
A number is divisible by 9 if the sum of its digits is divisible by 9.
Question
What is the rule for testing divisibility by 3?
Answer
A number is divisible by 3 if the sum of its digits is divisible by 3.
Question
What is the rule for testing divisibility by 6?
Answer
A number is divisible by 6 if it is divisible by both 2 and 3.
Question
What is the rule for testing divisibility by 11?
Answer
A number is divisible by 11 if the difference between the sum of its digits in odd places and the sum of its digits in even places is either 0 or a multiple of 11.
Question
What is the rule for testing divisibility by 4?
Answer
A number is divisible by 4 if the number formed by its ten's and unit's digits is divisible by 4.
Question
Describe the procedure for testing divisibility by 7.
Answer
Double the last digit and subtract it from the rest of the number; if the result is divisible by 7, the original number is also divisible by 7.
Question
A number is divisible by 6 if it is divisible by both _____ and _____.
Answer
2 and 3
Question
To be divisible by 11, the difference between the sum of digits in odd places and the sum of digits in even places must be _____ or a multiple of _____.
Answer
0 or 11
Question
A number is divisible by 9 if the _____ of its digits is divisible by 9.
Answer
sum
Question
Without performing the full addition, what is the quotient when $62 + 26$ is divided by 11?
Answer
The quotient is $6+2=8$.
Question
Without performing the subtraction, what is the quotient when $73 - 37$ is divided by 9?
Answer
The quotient is $7-3=4$.
Question
Without calculating the full sum, what is the quotient when $374 + 743 + 437$ is divided by $(3+7+4)$?
Answer
The quotient is 111.
Question
If the number $42x$ is divisible by 9, what must be the value of the digit $x$?
Answer
The value of x must be 3, since $4+2+3=9$.
Question
If the number $24a5$ is a multiple of 3, what are the possible values for the digit 'a'?
Answer
The possible values for 'a' are 1, 4, or 7.
Question
If the number $21y8$ is a multiple of 6, what are the possible values for the digit 'y'?
Answer
The possible values for 'y' are 1, 4, or 7.
Question
If the number $2y5$ is divisible by 11, what must be the value of the digit 'y'?
Answer
The value of y must be 7, since $(2+5)-y=0$ implies $y=7$.
Question
If the number $12z4$ is divisible by 4, what are the possible values of the digit 'z'?
Answer
The possible values for 'z' are 0, 2, 4, 6, or 8, as the number 'z4' must be divisible by 4.
Question
In the cryptarithmetic problem $31A + 1A3 = 501$, what is the value of A?
Answer
The value of A is 8.
Question
In the cryptarithmetic problem $A + A + A = BA$, what must be the value of A?
Answer
The value of A must be 5.
Question
In the cryptarithmetic problem $A + A + A = BA$, what must be the value of B?
Answer
The value of B must be 1.
Question
In the subtraction problem $46A - CB9 = 275$, what is the value of A?
Answer
The value of A is 4.
Question
In the subtraction problem $46A - CB9 = 275$, what is the value of B?
Answer
The value of B is 8.
Question
In the subtraction problem $46A - CB9 = 275$, what is the value of C?
Answer
The value of C is 1.
Question
In the multiplication problem $B A \times 6 = C 8 8$, what are the two possible values for A?
Answer
A can be 3 or 8, since $6 \times 3=18$ and $6 \times 8=48$ both end in 8.
Question
In the multiplication problem $B A \times 6 = C 8 8$, if A=8, what is the value of B?
Answer
The value of B is 4.