Numbers Exercise 3

Let the number be x.

Then, x² – (74)² = 5340

⇒ x² = 5340 + 5476 = 10816

⇒ x =

Let the number be = x

According to the question

x² – (74)² = 3740

⇒ x² = 3740 + 5476 = 9216

∴ x = = 96

Let the number be x

∴ x² – (46)² = 485

⇒ x² = 485 + (46)² = 2601

∴ x = = 51

Let the number be = x

According to the question,

x² + 57² = 8010

⇒ x² + 3249 = 8010

⇒ x² = 8010 – 3249 = 4761

⇒ x = = 69

Let the required number be x

∴ x² – (9)³ = 567

x² = 567 + 729 = 1296

∴ x = = 36

Let the number be x.

According to the question,

x² – 78² = 6460

⇒ x² = 6460 + 6084

⇒ x² = 12544

= 66.33

∴ Required number = 67² – 4400

= 4489 – 4400 = 89

⇒ x = = 112

∴ required number = 91 × 91 – 8115 = 166

∴ (66)² – 4321 = 4356 – 4321 = 35

69 × 69 = 4761

68 × 68 = 4624

Clearly, 4624 < 4700 < 4761

∴ Hence, 61 should be added to make = 4761 – 4700 = 61

= 63.13

∴ Here, 63² = (64)² – 3986

= 4096 – 3986 = 110

38² = 1444

39² = 1521

∴ Required number

= 1521 – 1500 = 21

Let the larger and smaller numbers be x and y respectively.

Then, x – y = 3 ...(i)

and, x² – y² = 63

⇒ (x + y) (x – y) = 63

⇒ ...(ii)

From equation (i) and (ii),

x = 12

Let the number be x.

x² - (12)3 = 976

∴ x² = 976 + 1728 = 2704

∴ x =

Let the number be x.

x² + (56)² = 4985

⇒ x² = 4985 – 3136 = 1849

∴ x = = 43

Let the two-digit number be

= 10 x + y, where x < y.

Number obtained after interchanging the digits = 10 y + x

According to the question,

10 y + x – 10 x – y = 9

⇒ 9y – 9x = 9

⇒ 9(y –x) = 9

⇒ y – x = 1 ...(i)

and x + y = 15 ...(ii)

From equations (i) and (ii),

y = 8 and x = 7

∴ Required product = 8 × 7 = 56

Let the number be (10x + y)

Then, (10x + y) – (10y + x) = 18

⇒ 9x – 9y = 18

⇒ x – y = 2 ...(i)

and, x + y = 16 ...(ii)

∴ x = 9, y = 7

From equations (i) and (ii),

So, the number is (10 × 9 + 7) = 97

The numbers from 1 to 100 which are exactly divisible by

4 are 4, 8, 12, 16, 20,24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100.

But each number should have 4 as its digit.

The required numbers are 4, 24, 40, 44, 48, 64, 84.

Clearly, there are 7 such numbers.

Sum of the digits of the ‘super’ number

= 1 + 2 + 3 +.... + 29

=

Now, sum of digits in the number 435 = 4 + 3 + 5 = 12 which gives a remainder of 3 when divided by 9.

Here 44 = 11 × 4

∴ the number must be divisible by 4 and 11 respectively. Test of 4 says that 9y must be divisible by 4 and since y > 5, so y = 6

Again , x 9596 is divisible by 11,

so x + 5 + 6 = 9 + 9

⇒ x = 7

Thus x = 7, y = 6

24162 = 89x + 43

⇒ x = (24162 – 43) ÷ 89 = 271

Clearly, unit’s digit in the given product = unit’s digit in 7¹⁵³ × 1⁷².

Now, 7⁴ gives unit digit 1.

∴ 7¹⁵³ gives unit digit (1 × 7) = 7. Also 1⁷² gives unit digit 1.

Hence, unit’s digit in the product

= (7 × 1) = 7.

Unit digit in 7⁴ is 1.

Unit digit in 7⁶⁸ is 1.

∴ Unit digit in 7⁷¹ = 1 × 7³ = 3

Again, every power of 6 will give unit digit 6.

∴ Unit digit in 6⁵⁹ is 6.

Unit digit in 3⁴ is 1.

∴ Unit digit in 3⁶⁴ is 1.

Unit digit in 3⁶⁵ is 3.

∴ Unit digit in (7⁷¹ × 6⁵⁹ × 3⁶⁵)

= Unit digit in (3 × 6 × 3) = 4.

Let x be the number of times, then

79x + 43759 = 50,000

⇒ x = (50000 – 43759) ÷ 79 = 79

Unit digit in 7⁴ is 1. So, unit digit in 7⁹² is 1.

∴ Unit digit in 7⁹⁵ is 3.

Unit digit in 3⁴ is 1.

∴ Unit digit in 3⁵⁶ is 1.

∴ Unit digit in 3⁵⁸ is 9.

∴ Unit digit in (7⁹⁵ – 3⁵⁸) = (13 – 9) = 4.

The digit in the unit’s place of 2⁵¹ is equal to the remainder when 2⁵¹ is divided by 10. 2⁵ = 32 leaves the remainder 2 when divided by 10. Then 2⁵⁰ = (2⁵)¹⁰ leaves the remainder 2¹⁰ = (2⁵)² which in turn leaves the remainder 2² = 4. Then
2⁵¹ = 2⁵⁰×2, when divided by 10, leaves the remainder 4×2 = 8.