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07 Jan 2004, 14:16
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Find the remainder of the division 2^87/9
a. 8
b. 1
c. -1
d. None of these.
***I got answer choice (c) -1.
The official answer was (a). Whose correct? any idea
a. 8
b. 1
c. -1
d. None of these.
***I got answer choice (c) -1.
The official answer was (a). Whose correct? any idea
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A is the answer. Look at the pattern:
2^1 = 2 2^5 = 32
2^2 = 4 2^6 = 64
2^3 = 8 2^7 = 128
2^4 = 16 2^8 = 256
Look at the pattern of the units digits. It repeats after every 4. Therefore, the exponent, 87/4 gives 21 remainder 3. A remainder of 3 means that the unit digit will be an 8 at the end. When you divide that unit's digit by 9, you will get a remainder of 8. Hope it helps
2^1 = 2 2^5 = 32
2^2 = 4 2^6 = 64
2^3 = 8 2^7 = 128
2^4 = 16 2^8 = 256
Look at the pattern of the units digits. It repeats after every 4. Therefore, the exponent, 87/4 gives 21 remainder 3. A remainder of 3 means that the unit digit will be an 8 at the end. When you divide that unit's digit by 9, you will get a remainder of 8. Hope it helps
07 Jan 2004, 14:38
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Yep, 8 it is.
2^1 / 9 is 0 remainder 2
2^2 / 9 is 0 remainder 4
2^3 / 9 is 0 remainder 8 .....
for 2^1 to 2^6, the respective remainders are 2,4,8,7,5,1
after the sixth exponent, you find that you have yourself a cycle.
So, 87/6 is 14 remainder 3. Excluding 14 complete cycles, the third position is 8 which is the remainder.
Don't know if I've made sense but...
2^1 / 9 is 0 remainder 2
2^2 / 9 is 0 remainder 4
2^3 / 9 is 0 remainder 8 .....
for 2^1 to 2^6, the respective remainders are 2,4,8,7,5,1
after the sixth exponent, you find that you have yourself a cycle.
So, 87/6 is 14 remainder 3. Excluding 14 complete cycles, the third position is 8 which is the remainder.
Don't know if I've made sense but...
