Question

Choose the correct option.
Find the units digit in the product of 7,^74 * 6^53+ 3^45?​

Answer 1

Step-by-step explanation:

3^65 * 6^59 * 7^71

Lets see cycle

3^1 = 3 , 7^1 = 7

3^2 = 9, 7^2 = 9

3^3 = 7, 7^3 = 3

3^4 = 1, 7^4 = 1

3^5 = 3, 7^5 = 7

. .

. .

Last digit of 3 & 7 starts to repeat after 4th power in cycle of 4 I.e (3,9,7,1) & (7,9,3,1)

Power / cycle

65/4 = 1 remainder

So 3^65 can be written as 3^1

71/4 = 3 remainder

So 7^71 can be written as 7^3

Now remember

6^n = 6 <--- last digit

So

3^65 * 6^59 * 7^71

= 3^1 * 6 * 7^3

= 3 * 6 * 3

= 4 <--- last/unit digit

Answer 2

To Find :- The units digit of :- 7⁷⁴ × 6⁵³ + 3⁴⁵ ?

Concept used :-

• 3¹ = 3 , 3² = 9, 3³ = 27, 3⁴ = 81 . So, unit digits of power of 3 is 3,9,7 or 1 . Also 3^4n = 1
• 6^n = unit digit 6 always .
• 7¹ = 7 , 7² = 49, 7³ = 343, 7⁴ = 2401 . So, unit digits of power of 3 is 7,9,3 or 1 . Also 7^4n = 1

Solution :-

→ 7⁷⁴ × 6⁵³ + 3⁴⁵

→ 7^(72 + 2) × 6^(52 + 1) + 3^(44 + 1)

→ {7^(4*18) × 7²} × {6^(4*13) × 6} + {3^(4*11) × 3}

since 7^4n and 3^4n is equal to 1,

→ (1 × 7²) × 6 × (1 × 3)

→ 49 × 6 × 3

→ 49 × 18

→ 882

→ Unit digit 2 .

Hence, unit digit will be 2 .

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