WORD PROBLEMS ON DIGITS AND NUMBERS

Problem 1 :

In a two-digit number, the digit in the ten’s place is twice the digit in the unit’s place. If 27 be subtracted from the number, the digits are reversed. Find the number.

Solution :

In the required two digit number, let x be the digit in one's place. Then the digit in the ten's place is 2x.

Then, the number is

(2x)x

If 27 be subtracted from the number, the digits are reversed.

(2x)x - 27 = x(2x)

Then, we have

2x ⋅ 10 + x ⋅ 1 - 27 = x ⋅ 10 + 2x ⋅ 1

20x + x - 27 = 10x + 2x

21x - 27 = 12x

9x = 27

x = 3

2x = 6

(2x)x = 63

Therefore, the required two-digit number is 63.

Problem 2 :

In a three-digit number, the middle digit is zero  and the sum of the other digits is 9. The number formed by interchanging the first and third digits is more than the original number by 297. Find the number.

Solution :

In the three-digit number, since the middle digit is zero, the number can be assumed as a0b.

Since the sum of the other digits is 9,

a + b = 9

b = 9 - a ----(1)

It is given that the number formed by interchanging the first and third digits is more than the original number by 297.

b0a - a0b = 297

[100⋅b + 10⋅0 + 1⋅a] - [100⋅a + 10⋅0 + 1⋅b] = 297

[100b + 0 + a] - [100a + 0 + b] = 297

[100b + a] - [100a + b] = 297

100b + a - 100a - b = 297

-99a + 99b = 297

Divide both sides by 99.

-a + b = 3 ----(2)

Substitute b = 9 - a.

-a + 9 - a = 3

-2a + 9 = 3

-2a = -6

a = 3

Substitute a = 3 into (1).

b = 9 - 3

a = 6

a0b = 306

Therefore, the three-digit number is 306.

Problem 3 :

A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the digits are reversed find the number.

Solution :

Any number between 10 and 100 is a two-digit number.

So, the required number can be assumed as xy.

It is given that the number is equal to five times the sum of its digits.

xy = 5(x + y)

10⋅x + y⋅1 = 5x + 5y

10x + y = 5x + 5y

5x - 4y = 0 ----(1)

If 9 be added to it the digits are reversed find the number.

xy + 9 = yx

10⋅x + y⋅1 + 9 = 10⋅y + x⋅1

10x + y + 9 = 10y + x

9x - 9y = -9

Divide both sides by 9.

x - y = -1

x = y - 1 ----(2)

Substitute x = y - 1 into (1).

5(y - 1) - 4y = 0

5y - 5 - 4y = 0

y - 5 = 0

y = 5

Substitute y = 5 into (2).

x = 5 - 1

x = 4

xy = 45

Therefore, the two-digit number is 45.

Problem 4 :

The sum of the digits in a three digit number is 12. If the digits are reversed the number is increased by 495 but reversing only of the ten’s and one's digits increases the number by 36. The number is

(A) 327     (B) 372     (C) 237     (D) 273

Solution :

We have three conditions.

(i) Sum of the digits in the three-digit number is 12.

(ii) If the digits are reversed the number is increased by 495.

(iii) Reversing only of the ten’s and one's digits increases the number by 36.

The correct answer choice have to meet all the three conditions above.

All the four answer choices meet the first condition.

Now we have to check the second and third conditions for each answer choice.

Choice A :

In 327, if the digits are reversed, then we have 723.

723 - 327 = 396

In the digits are reversed, the number increases 396, which does not meet the second condition. Since, the second condition is not met, we don't have to verify the third condition.

Choice A is incorrect.

Choice B :

In 372, if the digits are reversed, then we have 273.

In the digits are reversed, the number decreases, which does not meet the second condition.

Choice B is incorrect.

Choice C :

In 237, if the digits are reversed, then we have 732.

732 - 237 = 495

In the digits are reversed, the number increases by 495, the second condition is met.

In 237, reversing only of the ten’s and one's digits, we get 273.

273 - 237 = 36

Reversing only of the ten’s and unit digits increases the number by 36, the third condition is met. 

Choice Cis correct.

Problem 5 :

Find the digit at one's place in the following product.

(2467)¹⁵³ x (341)⁷²

Solution :

7² gives the digit 9 at one's place

7⁴ = 7^{2 + 2} = 7² x 7²

7⁴ gives the digit 1 at one's place, as 9 x 9 = 81.

7¹⁵² = 7^{4 x 38} = (7⁴)³⁸

7¹⁵² gives the digit 1 at one's place

7¹⁵³ gives the digit (1 x 7) = 7 at one's place

1⁷² gives the digit 1 at one's place

Hence, the digit at one's place in the given product is

= 7 x 1

= 7

Problem 6 :

Find the digit at one's place in the following addition.

(264)¹⁰² + (264)¹⁰³

Solution :

The required digit at one's place is the digit at one's place in 4¹⁰² + 4¹⁰³.

4² gives the digit at one's place is 6

4¹⁰² = 4^{2 x 51} = (4²)⁵¹

4¹⁰² gives the digit at one's place is 6

4¹⁰³ = 4^{102 + 1} = 4¹⁰² x 4

4¹⁰³ gives the digit at one's place is 4

Sum of the digits at one's place in 4¹⁰² and 4¹⁰³ :

= 6 + 4

= 10

Hence, the digit at one's place in the given addition is

= 0

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