The digit in the unit's place of the number represented by $(7^{95} - 3^{58})$ is:

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Solution

The correct option is B 4
If we successively multiply the unit digits only we shall see that the units numbers are repeated after a certain number of the power. Investigate that for 7 and 3.
$L e t u s f i n d o u t t h e u n i t s p l a c e n u m b e r f o r c e r t a i n p o w e r s o f 7 a n d 3. u n i t s n u m b e r u n i t s n u m b e r 7^{1} ⟶ 7 3^{1} ⟶ 3 7^{2} ⟶ 7 \times 7 \to 9 3^{2} ⟶ 9 7^{3} ⟶ 9 \times 7 \to 3 3^{3} ⟶ 3 \times 9 \to 7 7^{4} ⟶ 3 \times 7 \to 1 3^{4} ⟶ 3 \times 7 \to 1 7^{5} ⟶ 1 \times 7 \to 7 3^{5} ⟶ 3 \times 1 \to 3. S o, i n b o t h t h e c a s e s t h e u n i t s p l a c e n u m b e r s r e p e a t i n o r d e r a f t e r e a c h 4 t h p o w e r . T h e s e t i s {7, 9, 3, 1} f o r 7 a n d {3, 9, 7, 1} f o r 3. N o w i n 7^{95} ⟶ 95 = 4 \times 23 + 3. S o i t w i l l h a v e t h e 3 r d n u m b e r o f t h e s e t f o r 7. T h e r e f o r e u n i t p l a c e n u m b e r o f 7^{95} i s 3. A g a i n i n 3^{58} ⟶ 58 = 4 \times 14 + 2 T h e u n i t s p l a c e n u m b e r i s t h e 2 n d n u m b e r f o r 3. ∴ U n i t p l a c e n u m b e r o f 3^{58} i s 9. T h e d i f f e r e n c e \to u n i t s p l a c e 3 - u n i t p l a c e 9 = 13 - 9 (b y b o r r o w i n g 10 f r o m t e n s p l a c e) = 4 A n s w e r - O p t i o n B .$

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The digit in the unit's place of the number represented by (795−358) is:

The digit in the unit's place of the number represented by $(7^{95} - 3^{58})$ is: