22-22-what-is-hte-last-digit-

Question Number 176058 by Shrinava last updated on 11/Sep/22

22^{22}

what is hte last digit ?

Answered by BaliramKumar last updated on 11/Sep/22

4

Commented by BaliramKumar last updated on 11/Sep/22

22^{22} = 22^{4 (5) + 2} [x^{4 (n) + r} = x^{r}] 0 < r \leq 4

2^{2} = 4 A n s w e r

e x a m p l e 39857^{2576}

39857^{4 (n) + 4}

7^{4} = 1 A n s w e r

e x a m p l e 2345678^{3566765!}

8^{4 (n) + 4} = 8^{4} = 6

I f x! t h e n r = 4 (x \geq 4)

Commented by Shrinava last updated on 11/Sep/22

solution please

Commented by Rasheed.Sindhi last updated on 11/Sep/22

Your rule is successful!

Actually I think :

In general

a \equiv b [10] \land m \equiv n [4] \Leftrightarrow a^{m} \equiv b^{n} [10]

Commented by Shrinava last updated on 13/Sep/22

cool dear sir thankyou

Answered by Rasheed.Sindhi last updated on 11/Sep/22

last digit = 22^{22} \mod 10

22 \equiv 2 (\mod 10)

22^{2} \equiv 2^{2} \equiv 4 (\mod 10) \dots \dots \dots \dots . (i)

22^{5} \equiv 2^{5} = 32 \equiv 2 (\mod 10)

{(22^{5})}^{4} \equiv 2^{4} = 16 \equiv 6 (\mod 10)

22^{20} \equiv 6 (\mod 10) \dots \dots \dots \dots \dots \dots (ii)

(i) \times (ii) :

22^{22} \equiv 24 \equiv 4 (\mod 10)

Answered by LordKazuma last updated on 11/Sep/22

22^{22} (m o d 10) = 2^{22} (m o d 10)

= {(2^{3})}^{7} \cdot 2^{1} (m o d 10)

= {(- 2)}^{7} \cdot 2 (m o d 10)

= {(2^{3})}^{2} \cdot (- 2) \cdot 2 (m o d 10)

= {(- 2)}^{2} \cdot (- 4) (m o d 10)

= 4 \cdot (- 4) (m o d 10)

= (- 16) (m o d 10)

= (4 - 2 \cdot 10) (m o d 10) = 4

s o t h e l a s t d i g i t o f 22^{22} i s 4