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Find the unit digit of expression (232)^...

Find the unit digit of expression
`(232)^(123!) × (353)^(124!) × (424)^(124!)`

A

3

B

4

C

6

D

8

Text Solution

AI Generated Solution

The correct Answer is:
To find the unit digit of the expression \( (232)^{123!} \times (353)^{124!} \times (424)^{124!} \), we will focus on the unit digits of each base number raised to their respective powers. ### Step 1: Identify the unit digits of the bases - The unit digit of \( 232 \) is \( 2 \). - The unit digit of \( 353 \) is \( 3 \). - The unit digit of \( 424 \) is \( 4 \). ### Step 2: Analyze the powers We need to find the unit digits of: - \( 2^{123!} \) - \( 3^{124!} \) - \( 4^{124!} \) ### Step 3: Determine the unit digit of \( 2^{123!} \) The unit digits of powers of \( 2 \) cycle every 4: - \( 2^1 = 2 \) - \( 2^2 = 4 \) - \( 2^3 = 8 \) - \( 2^4 = 6 \) - \( 2^5 = 2 \) (and so on...) To find the unit digit of \( 2^{123!} \), we need to calculate \( 123! \mod 4 \): - Since \( 123! \) contains all integers from \( 1 \) to \( 123 \), it includes \( 4 \), so \( 123! \equiv 0 \mod 4 \). - Therefore, \( 2^{123!} \equiv 2^0 \equiv 6 \) (from the cycle). ### Step 4: Determine the unit digit of \( 3^{124!} \) The unit digits of powers of \( 3 \) cycle every 4: - \( 3^1 = 3 \) - \( 3^2 = 9 \) - \( 3^3 = 7 \) - \( 3^4 = 1 \) - \( 3^5 = 3 \) (and so on...) Now, we calculate \( 124! \mod 4 \): - Similar to \( 123! \), \( 124! \) also contains \( 4 \), so \( 124! \equiv 0 \mod 4 \). - Therefore, \( 3^{124!} \equiv 3^0 \equiv 1 \). ### Step 5: Determine the unit digit of \( 4^{124!} \) The unit digits of powers of \( 4 \) cycle every 2: - \( 4^1 = 4 \) - \( 4^2 = 6 \) - \( 4^3 = 4 \) (and so on...) Now, we calculate \( 124! \mod 2 \): - Since \( 124! \) includes all even numbers, \( 124! \equiv 0 \mod 2 \). - Therefore, \( 4^{124!} \equiv 4^0 \equiv 6 \). ### Step 6: Combine the unit digits Now we have: - Unit digit of \( 2^{123!} \) is \( 6 \). - Unit digit of \( 3^{124!} \) is \( 1 \). - Unit digit of \( 4^{124!} \) is \( 6 \). Now, we multiply these unit digits together: - \( 6 \times 1 \times 6 = 36 \). ### Step 7: Find the unit digit of the product The unit digit of \( 36 \) is \( 6 \). ### Final Answer The unit digit of the expression \( (232)^{123!} \times (353)^{124!} \times (424)^{124!} \) is \( 6 \).
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ADVANCED MATHS BY ABHINAY MATHS ENGLISH-NUMBER SYSTEM -MULTIPLE CHOICE QUESTIONS
  1. Find the unit digit of expression (823)^(933!) × (777)^(223!) × (838)^...

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  2. Find the unit digit of expression 125^813 * 553^3703 * 4537^828?

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  3. Find the unit digit of expression (232)^(123!) × (353)^(124!) × (424)^...

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  4. Find the units digit in the expansion of (44)^44+(55)^55+(88)^88.

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  5. The last digit of the following expreesion is : (1!)^1 + (2!)^(2) + ...

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  6. Find the unit digit in the expression :(1!)^(1!) + (2!)^(2!) + (3!)^(3...

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  7. Find the unit digit in the expression : (3^57*6^41*7^63)

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  8. Find the units digit in the expansion of (44)^44+(55)^55+(88)^88.

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  9. If 100! divisible by 3^n then find the maximum value of n.

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  10. If 122! is divisible by 6^n then find the maximum value of n.

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  11. If 123! is divisible by 12^(n) then find the maximum value of n.

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  12. If 133! is divisible by 7^(n) then find the maximum value of n.

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  13. If 187! is divisible by 15^(n) then find the maximum value of n.

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  14. Find no of zeros in 100 !

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  15. Find the no. of zeros in expression : 1 × 2 × 3 × 4 .......... × 500

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  16. Find the no. of zeros in expression : (1 xx 3 xx 5 …….. 99) xx 100

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  17. Find the no. of zeros in expression : 1 × 3 × 5 × 7 .......... × 99

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  18. Find the no. of zeros in the product of (5 xx 10 xx 25 xx 40 xx 50 xx ...

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  19. Find the no. of zeros in expression : (1 xx 3 xx 5 …….. 99) xx 100

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  20. Find the no. of zeros in expression : 10 xx 20 xx 30 xx ……xx 1000.

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