Exercise: 4-C
Q1:
A A + B 6 ----- C 2 5
Step 1: Add the units digit: A + 6 = 5 or 15 (with carry)
→ Since 6 + A = 5 is not possible without a carry, we consider 6 + A = 15 (with carry 1)
⇒ So, A = 9 and carry = 1 ✅
Step 2: Now go to tens place: A + B + carry = 2
We already know A = 9 and carry = 1
So, 9 + B + 1 = 2 (or 12 if we consider carry to next digit)
→ 9 + B + 1 = 12 ⇒ B = 2 ✅ (since 9 + 2 + 1 = 12 → 2 at tens place, carry 1)
Step 3: Now go to hundreds: Carry from tens = 1
⇒ 1 + nothing (left of B) = C
So, 1 + nothing = C ⇒ C = 1 ✅
Step 4: Check final values:
A = 9, B = 2, C = 1
→ A A = 99
→ B 6 = 26
→ C25 = 125
Verification:
9 9 + 2 6 ----- 1 2 5
Answer: A = 9, B = 2, C = 1
Q2:
5 C B 8 + D 9 7 A ------------- 8 2 4 3
Step 1: Units place → 8 + A = 3 (or 13)
8 + A must give 3, so we consider 8 + A = 13
⇒ A = 5 (since 8 + 5 = 13), carry = 1 ✅
Step 2: Tens place → B + 7 + carry = 4 (or 14)
Let’s assume carry = 1 from previous step
Then: B + 7 + 1 = 14 ⇒ B = 6 ✅
Step 3: Hundreds place → C + 9 + carry = 2 (or 12)
Carry from previous = 1
So, C + 9 + 1 = 12 ⇒ C = 2 ✅
Step 4: Thousands place → 5 + D + carry = 8
Carry from previous = 1
So, 5 + D + 1 = 8 ⇒ D = 2 ✅
Step 5: Final digits:
C = 2
B = 6
A = 5
D = 2
Verification:
5 2 6 8 + 2 9 7 5 ------------- 8 2 4 3
Answer: A = 5, B = 6, C = 2, D = 2
Q3:
A + A + A ----- B A
Step 1: Add A + A + A = 3A
So, the total is 3 × A = a two-digit number: BA
We need a digit A (1–9) such that 3 × A gives a 2-digit number ending in A
Let’s try values of A from 1 to 9:
A = 1 → 3 × 1 = 3 → Not 2-digit ❌
A = 2 → 3 × 2 = 6 ❌
A = 3 → 3 × 3 = 9 ❌
A = 4 → 3 × 4 = 12 → ends in 2 ❌
A = 5 → 3 × 5 = 15 → ends in 5 ✅
→ So A = 5, BA = 15 ⇒ B = 1 ✅
Verification:
5 + 5 + 5 ----- 1 5
Answer: A = 5, B = 1
Q4:
6 A - A B ----- 3 7
Step 1: We know the subtraction: (6A) – (AB) = 37
Let’s write it in algebraic form:
Let A = a digit, B = b digit
→ (60 + A) – (10 × A + B) = 37
Step 2: Expand the equation:
(60 + A) – (10A + B) = 37
→ 60 + A – 10A – B = 37
→ 60 – 9A – B = 37
Step 3: Solve the equation:
60 – 9A – B = 37
→ 9A + B = 60 – 37 = 23
Now try values of A (1 to 9), and check if B becomes a digit (0–9):
A = 1 → 9(1) = 9 → B = 23 – 9 = 14 ❌
A = 2 → 18 → B = 5 ✅
Step 4: Verify:
6 2 - 2 5 ----- 3 7
Answer: A = 2, B = 5
Q5:
C B 6 - 2 9 A ------- 5 3 9
Step 1: Analyze unit’s place → 6 − A = 9
But 6 − A cannot be 9 directly, so we must have borrowed from tens digit.
So, effectively: 16 − A = 9
⇒ A = 7 ✅
Step 2: Tens place → B − 9
We borrowed 1 earlier, so B − 9 − 1 = 3 ⇒ B − 10 = 3
⇒ B = 13 ❌ (Not valid, since B must be a digit)
Try borrowing again:
→ B + 10 − 9 − 1 = 3 ⇒ B + 0 = 3 ⇒ B = 3 ✅
Step 3: Hundreds place → C − 2
No borrow here
C − 2 = 5 ⇒ C = 7 ✅
Step 4: Final values:
C = 7, B = 3, A = 7
Now substitute:
Minuend: CB6 = 736
Subtrahend: 29A = 297
Check: 736 − 297 = 439 ✅
But we want result = 539 ❌
Something is off. Let’s retry…
Try A = 7 ⇒ 6 − 7 not possible
→ Borrow: 16 − 7 = 9 ✅
Now units place is valid.
Tens place: B − 9 (after borrow = B − 10)
We need result = 3
So, B − 10 = 3 ⇒ B = 13 ❌
→ But if we borrow again from hundreds digit:
Then B + 10 − 9 − 1 = 3
→ B = 3 ✅
Now hundreds: C − 2 − 1 (since borrowed for tens) = 5
⇒ C − 3 = 5 ⇒ C = 8 ✅
Step 5: Verify
8 3 6 - 2 9 7 ------- 5 3 9
Answer: C = 8, B = 3, A = 7
Q6:
A B × 3 ----- C A B
Step 1: Let the two-digit number be \( 10A + B \)
→ Multiply it by 3: \( 3 × (10A + B) = 100C + 10A + B \)
So we want:
\( 3 × (10A + B) = 100C + 10A + B \)
Step 2: Rearranging the equation:
Left side = \( 30A + 3B \)
Right side = \( 100C + 10A + B \)
So,
\( 30A + 3B = 100C + 10A + B \)
Subtract \(10A + B\) from both sides:
\( (30A – 10A) + (3B – B) = 100C \)
⇒ \( 20A + 2B = 100C \)
Divide by 2:
⇒ \( 10A + B = 50C \)
Step 3: Try values of C (since C is a digit, 1–9):
If C = 1 → 50 × 1 = 50 → Try 10A + B = 50
→ A = 5, B = 0 ✅
Now check full multiplication:
AB = 50
50 × 3 = 150
→ CAB = 150 ⇒ C = 1, A = 5, B = 0 ✅
Step 4: Verification
5 0 × 3 ----- 1 5 0
Answer: A = 5, B = 0, C = 1
Q7:
A B
× B A
---------
(B+1) C B
We are given a multiplication puzzle:
Two-digit number AB is multiplied by BA to give a three-digit number (B+1)CB.
Step 1: Assume the two-digit numbers:
→ AB = 10A + B
→ BA = 10B + A
Their product is:
(10A + B)(10B + A) = 100(B + 1) + 10C + B
Let’s test possible digit values (A, B ∈ 1–9, C ∈ 0–9). Try A = 1, B = 3:
AB = 13
BA = 31
→ 13 × 31 = ?
→ 13 × 31 = 403 ✅
Now break 403 into the form (B+1)CB:
→ B = 3 → B + 1 = 4
→ So (B+1)CB = 403 ⇒ 4 0 3 ✅
Thus, C = 0, A = 1, B = 3
1 3
× 3 1
--------
4 0 3
Answer: A = 1, B = 3, C = 0
Q8:
______
6 | 5 A B | 9 C
-5 4
------
3 B
- 3 6
--------
0
We are given a long division setup with missing digits A, B, C, and asked to find them such that the division is correct.
Let’s decode step-by-step:
Step 1: From the divisor line:
→ Divisor = 6
→ Dividend starts with: 5 A B
→ Quotient starts with 9 C
Also, we see:
→ First subtraction: 54 from 5AB → so the first partial dividend is 54 ✅
So 6 × 9 = 54 ⇒ Quotient starts with 9 ✅
Step 2: Next digit in dividend after subtracting 54 is: 3B
→ Now 6 × C = something ending in 36
Try C = 6 → 6 × 6 = 36 ✅
So B = 6 ✅
Step 3: What is A?
We now know:
→ Quotient is 96
→ So 6 × 96 = 576
Let’s test this:
→ 6 × 96 = 576
So the dividend = 576 ✅
Now match with digits 5 A B → A = 7, B = 6 ✅
And remainder part works out:
576 ÷ 6 = 96, remainder = 0
Step 4: Final values:
A = 7
B = 6
C = 6
______
6 | 576 | 96
-54
------
36
-36
-----
0
Answer: A = 7, B = 6, C = 6
Q9: Find the quotient when you divide the number of four-digit numbers by the number of three-digit numbers.
Step 1: Count of three-digit numbers:
→ Smallest 3-digit number = 100
→ Largest 3-digit number = 999
→ So, total = 999 − 100 + 1 = 900 ✅
Step 2: Count of four-digit numbers:
→ Smallest 4-digit number = 1000
→ Largest 4-digit number = 9999
→ So, total = 9999 − 1000 + 1 = 9000 ✅
Step 3: Now divide:
→ Quotient = (Number of 4-digit numbers) ÷ (Number of 3-digit numbers)
→ \( \frac{9000}{900} = 10 \) ✅
Answer: The quotient is 10.
Q10: Find the largest natural number by which the difference between a three-digit number and the number obtained by reversing its digits is always divisible.
Step 1: Let the three-digit number be \( 100a + 10b + c \)
→ Its reverse is \( 100c + 10b + a \)
Step 2: Take the difference:
\[
(100a + 10b + c) – (100c + 10b + a)
\]
Cancel out common terms:
\[
= 100a – a + c – 100c = 99a – 99c = 99(a – c)
\]Step 3: So the difference is always a multiple of 99 ✅
Hence, the largest natural number that always divides such a difference is:
Answer: 99
Q11: On multiplying a number by 7, the product is a number each of whose digits is 3. Find the smallest such number.
Step 1: Let the required number be \( x \), such that:
\[
7 \times x = 333\ldots3 \ (\text{all digits are 3})
\]We will check small numbers made of all 3s and divide by 7:
→ \( \frac{3}{7} = \) not a whole number ❌
→ \( \frac{33}{7} = 4.714 \) ❌
→ \( \frac{333}{7} = 47.571 \) ❌
→ \( \frac{3333}{7} = 476.14 \) ❌
→ \( \frac{33333}{7} = 4761.857 \) ❌
→ \( \frac{333333}{7} = 47619.0 \) ✅
Step 2: So, 333333 is divisible by 7
⇒ \( x = \frac{333333}{7} = 47619 \) ✅
Step 3: Verification
→ 47619 × 7 = 333333 ✅
47619
× 7
---------
333333
Answer: The required number is 47619
Q12: The letters A, B, C, D, E, F, G, H and I are assigned values from 1 to 9 but not in that order. 4 is assigned to E. The difference between E and I is 5. The difference between C and I is 3. What is the value assigned to C?
Step 1: Given:
→ E = 4 ✅
→ |E − I| = 5 ⇒ |4 − I| = 5
→ So, I = 4 + 5 = 9 or I = 4 − 5 = -1 (not valid)
⇒ I = 9 ✅
Step 2: Now use:
→ |C − I| = 3 ⇒ |C − 9| = 3
→ So, C = 9 + 3 = 12 ❌ or C = 9 − 3 = 6 ✅
Step 3: Final values:
E = 4
I = 9
C = 6 ✅
Answer: The value assigned to C is 6.
Q13: Find the values of A and B in the division sum shown below:
_______
AB | 2 5 2 | B A
-2 4
------
1 2
- 1 2
-------
0
Step 1: According to the setup:
→ AB × B = 24
→ AB × A = 12
Step 2: Use the division identity:
\[
\frac{AB \times B}{AB \times A} = \frac{24}{12}\\
\frac{B}{A} = \frac{2}{1}
\]Hence
\[
A = 1 \\
B = 2
\]
Step 3: Verification
_______
12 | 2 5 2 | 21
-2 4
------
1 2
-1 2
------
0
Answer: A = 1, B = 2
Q14: If X and Y represent digits, what is the maximum possible value of Y in the following?
5 X 9
3 2 7
+ 2 Y 8
----------
1 1 1 4
We are adding three 3-digit numbers to get a 4-digit number.
Step 1: Analyze the unit (ones) column:
→ 9 + 7 + 8 = 24 ⇒ Unit digit = 4 (matches the last digit of 1114) ✅
→ Carry over = 2
Step 2: Analyze the tens column (X + 2 + Y + carry 2 = ?):
Let’s compute:
X + 2 + Y + 2 = X + Y + 4 = 1 (tens digit of 1114) with possible carryover 1
So:
→ (X + Y + 4) ends in 1 ⇒ Possible totals: 11, 21, etc.
→ But 11 is the only realistic one (sum of two digits + 4 = 11)
→ So X + Y + 4 = 11 ⇒ X + Y = 7 ✅
Step 3: We need to maximize Y, subject to:
X + Y = 7, where X and Y are digits (0–9)
→ Maximum Y occurs when X is minimum
→ Try X = 0 ⇒ Y = 7 ✅
→ Try X = 1 ⇒ Y = 6 ✅
→ Try X = 2 ⇒ Y = 5 ✅
…
→ But to maximize Y, we choose X = 0 ⇒ Y = 7 ✅
Answer: The maximum possible value of Y is 7.
Q15: What would be the maximum value of Q in the following addition sum?
5 P 7
8 Q 9
+ R 3 2
----------
1 9 2 8
Step 1: Analyze the unit (ones) column:
→ 7 + 9 + 2 = 18 ⇒ Unit digit = 8 (matches the last digit of 1928) ✅
→ Carry over = 1
Step 2: Analyze the tens column (P + Q + 3 + carry 1 = ?)
→ (P + Q + 3 + 1) = P + Q + 4
→ Middle digit in sum is 2 ⇒ So (P + Q + 4) should end in 2
→ Possible total: 12 (only realistic one, since two digits max to 18)
⇒ P + Q + 4 = 12 ⇒ P + Q = 8 ✅
Step 3: To maximize Q, minimize P
→ Try P = 0 ⇒ Q = 8 ✅
→ Try P = 1 ⇒ Q = 7 ✅
…
→ But maximum value of Q is when P = 0 ⇒ Q = 8 ✅
Step 4: Let’s verify with assumed digits:
P = 0, Q = 8, now check hundreds place:
→ 5 + 8 + R + any carry from tens
→ From previous step: P + Q + 3 + 1 = 12 ⇒ carry = 1 again
→ Now: 5 + 8 + R + 1 = 9 (hundreds digit of 1928 is 9)
→ 5 + 8 + R + 1 = 14 ⇒ R = 0 ❌
→ Try other values (assume P = 1, Q = 7 ⇒ P + Q = 8 ✅)
→ 5 + 8 + 0 + 1 = 14 ❌
→ Try P = 2, Q = 6 ⇒ P + Q = 8
→ 5 + 8 + R + 1 = ?
But to answer the question, only maximum value of Q matters ✅
Answer: The maximum possible value of Q is 8.
Q16: Find two numbers whose product is a 1-digit number and the sum is a 2-digit number.
We are looking for two numbers:
→ Product is a 1-digit number (i.e., from 1 to 9) ✅
→ Sum is a 2-digit number (i.e., 10 or more) ✅
Step 1: Try small integers for product
Let’s take:
→ Product = 6 ⇒ Possible factor pairs: (1,6), (2,3), (3,2), (6,1)
Try with these:
→ 1 + 6 = 7 ❌ (not 2-digit)
→ 2 + 3 = 5 ❌
→ Try 1 × 8 = 8 ⇒ 1 + 8 = 9 ❌
→ Try 1 × 9 = 9 ⇒ 1 + 9 = 10 ✅
So, (1, 9) satisfies both conditions:
→ Product = 1 × 9 = 9 ✅
→ Sum = 1 + 9 = 10 ✅
Let the numbers be 1 and 9
→ Product = 1 × 9 = 9
→ Sum = 1 + 9 = 10
Answer: The numbers are 1 and 9.
Q17: Find three natural numbers whose sum and product are equal.
We need to find three natural numbers, say \( a, b, c \), such that:
→ \( a + b + c = a \times b \times c \)
Step 1: Try small natural numbers using trial-and-error
Try \( a = 1, b = 2, c = 3 \)
→ Sum = 1 + 2 + 3 = 6
→ Product = 1 × 2 × 3 = 6 ✅
This satisfies the condition.
Let the numbers be: 1, 2, 3
→ Sum = 1 + 2 + 3 = 6
→ Product = 1 × 2 × 3 = 6
Answer: The numbers are 1, 2, and 3.
