Find the units digit in the product of $$(127)^{153}×(341)^{89}$$​.

Correct option is D

Given:

Find the units digit of $$(127)^{153} \times (341)^{89}$$​​

Concept Used:
Units digit of $$7^n$$​ follows a 4-term cycle: 7, 9, 3, 1

Units digit of $$1^n = 1$$​ always

Solution:

Units digit of $$7^{153}$$​​

The units digit of powers of 7 cycles every 4:

$$7^1 = 7 \quad (\text{units digit } 7) \\7^2 = 49 \quad (\text{units digit } 9) \\7^3 = 343 \quad (\text{units digit } 3) \\7^4 = 2401 \quad (\text{units digit } 1) \\7^5 = 16807 \quad (\text{units digit } 7) \quad \text{(cycle repeats)}$$​

The cycle is 7, 9, 3, 1.

​$$153 \div 4 = 38 \text{ cycles with remainder } 1 \quad (\text{since } 153 = 4 \times 38 + 1)$$​​

The remainder 1 corresponds to the first position in the cycle, which is 7.

Multiplying the two units digits: 7 $$\times 1 = 7.$$​​

Units digit of $$1^{89} = 1$$​​

Final Product

Units digit =7 × 1 = 7

Alternate Method:

Units digit of $$7^{153}$$

​$$153-1 =152$$

152$$\div 4 = 38 \text{cycles}$$ remainder = 0 

we have subtracted 1 in the number thus now 0 + 1 = 1 

So, 7¹ = 7 unit digit  of $$7^{153}$$​​

Units digit of $$1^{89} = 1$$

Final Product

Units digit =7 × 1 = 7

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