Exercise 15.5

1. Find the missing digits in the following additions:

i. 111 + A + 77 = 197

Solution:

Let's break down the addition column by column, from right to left:

Units Column: 1 + A + 7 = a number ending in 7. This means 1 + A + 7 must equal (since it can't be 7, 27, etc., as we're dealing with single digits).

So, + A = , which means A = .

Tens Column: 1 (carried over from the units column) + 1 + 7 = . This matches the tens digit in the answer (9), so no carry-over to the hundreds column.

Hundreds Column: 1 + 0 + 0 = . This matches the hundreds digit.

Therefore, A = .

ii. 22 + 8 + BB = 285

Solution:

Let's analyze the addition column by column, from right to left:

Units Column: 2 + 8 + B = a number ending in 5. This means 10 + B must equal which means B = .

Tens Column: 2 + 0 + 5 = . However, the tens digit in the answer is .

Hundreds Column: 2 + 0 + 0 + (1 carried over) = .

However, if we ignore the hundreds column, then B = .

Therefore, B = 5.

iii. AAA + 7 + AA = 373

Solution:

Let's analyze the addition column by column, from right to left:

Units Column: A + 7 + A = a number ending in 3. This means 2A + 7 must equal either 13 or 23

If 2A + 7 = 13, then 2A = , and A = .

If 2A + 7 = 23, then 2A = , and A = .

Tens Column: Let's consider both possibilities for A:

If A = 3: 3 + 0 + 3 = .

If A = 8: 8 + 0 + 8 = .

Hundreds Column:

If A = 3: 3 + 0 + 0 + (1 carried over) = .

Therefore, A = .

iv. 2222 + 8 + AAA = 299A

Solution:

Units Column: 2 + 8 + 4 = .

Tens Column: 1 (carry-over) + 2 + 0 + 4 = .

Hundreds Column: 2 + 0 + 4 = .

Thousands Column: 2 + 0 + 0 = .

Therefore, A = , and the sum is

v. BB + 6 + AAA = 285

Solution:

Units Column: 1 + 6 + 4 = .

Tens Column: 1 (carry-over) + 1 + 4 + 4 = .

Hundreds Column: 1 (carry-over) + 0 + 0 + 4 = .

Therefore, while A = and B = .

2. Find the value of A in the following

(a) 7A – 16 = A9

Solution:

7A – 16 = A9 can be rewritten as 70 + A – 16 = 10A + 9 which gives: 54 + A = 10A + 9: = 9A

A =

(b) 107 – A9 = 1A

Solution:

107 – A9 = 1A can be rewritten as 107 - (10A + 9) = 10 + A

107 - 10A - 9 = 10 + A

- 10A = 10 + A

= A

A =

(c) A36 – 1A4 = 742

Solution:

A36 – 1A4 = 742 can be rewritten as: A + 36 – ( + 10A + ) = 742

100A + 36 - 100 - 10A - 4 = 742

A - = 742

90A =

A =

4. Replace the letters with appropriate digits

(a) 73K ÷ 8 = 9L

Solution:

73K = 8 × 9L. So, 73K must be a of 8.

Trying values for K:

If K = 0, 7308 = (not an integer)

If K = 1, 7318 = (not an integer)

If K = 2, 7328 = (not an integer)

If K = 3, 7338 = (not an integer)

If K = 4, 7348 = (not an integer)

If K = 5, 7358 = (not an integer)

If K = 6, 7368 = . So, K = and L = .

Therefore, K = 6 and L = 2.

(b) 1MN ÷ 3 = MN

Solution:

1MN = 3 × MN. This means + 10M + N = 3 × (10M + N)

100 + 10M + N = 30M + 3N

100 = M + N

= M + N

Since M and N are digits (0-9), M must be and N must be .

Therefore, M = 5 and N = 0.

5. If ABB × 999 = ABC123 (where A, B, C are digits) find the values of A, B, C.

Solution:

ABB × 999 = ABB × (1000 - 1) = ABB000 - ABB = ABC123

This means that when ABB is subtracted from ABB000, the result is ABC123.

Since the last three digits of the result are 123, B must be since, 1000 - =

So, we have A11 × 999 = A11 × (1000 - 1) = A11000 -

A11000 - A11 = A10889.

Therefore, A = , B = , and C = .