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X ^ Y denotes X raised to
power Y, find the last two digits (2361 ^ 4507) + (1741 ^ 3581)
a. 02
b. 82
c. 42
d. 62
Read Solution (Total 6)
-
- its 62.
taking 2361^4507 last digit will be 1 and second last will be -> 2 (6*7=42 from 2361 and 4507) hence 21
similarly for 1741^3581 last digit=1 and 2nd last will be ->4 (4*1=4 from 1741 and 3581) hence 41
summing up 21+41= 62 (ANS)
- 12 years agoHelpfull: Yes(34) No(4)
- 62
1^7=1 nd 6*7=42 take 2 nd for 2nd no 1^1=1 nd 4*1=4
therefore 21+21=42
- 12 years agoHelpfull: Yes(3) No(5)
- 02 as 2361^4507 yield 61 last digit & 1741^3581 yield 41 last digit so 61+41=02 last two digit
- 12 years agoHelpfull: Yes(3) No(0)
- answer is 62
- 12 years agoHelpfull: Yes(1) No(3)
- d.)62 i will explain this later....very simple..try it using unit place digit and tens place digit's repitition..
- 12 years agoHelpfull: Yes(1) No(5)
- last two digits of(61^7 + 41 )
- 12 years agoHelpfull: Yes(0) No(5)
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