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Topic: **Remainder math homework****Question:**
Hello, I'm doing math homework right now and i have stumbled on to this problem: What is the last digit of 2*85 (2 to the 85th power). Please help me, and explanations would be very much appreciated. Thank You.
yes, it's actually honors, but we're only starting it because school started yesterday.

June 15, 2019 / By Jordie

base 2 follows a pattern 2^1=2 2^2=4 2^3=8 2^4=16 2^5=32 2^6=64 2^7=128 2^8=256 as you can see the last digit alternates in the pattern 2,4,8,6,2,4,8,6 forever. this pattern is a repeating unit of 4 digits in length so you see how many lots of 4 go evenly into 85. the answer is 21 lots although this part is unimportant. the remainder is 1 so the 85th power will have the same last digit as the first character.i.e the last digit will be a 2 hope that helped

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base 2 follows a pattern 2^1=2 2^2=4 2^3=8 2^4=16 2^5=32 2^6=64 2^7=128 2^8=256 as you can see the last digit alternates in the pattern 2,4,8,6,2,4,8,6 forever. this pattern is a repeating unit of 4 digits in length so you see how many lots of 4 go evenly into 85. the answer is 21 lots although this part is unimportant. the remainder is 1 so the 85th power will have the same last digit as the first character.i.e the last digit will be a 2 hope that helped

You should use (shift + 6) for power of: 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32 2^6 = 64 The pattern is 2, 4, 8, 6, 2, 4, 8, 6 etc... On 2^84, the last digit is 6 because the it is a multiple of 4 which means it's the last one in the pattern. That means the next one must be 2 because the pattern starts over so 2^85 ends in 2.

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powers of 2 (starting at 1) 2 , 4, 8 , 16, 32, 64, 128 , 256 .... Notice how the last digit cycles 2,4,8,6,2,4,8,6 ..... repeating after every 4th power. 2^85 = 2^(21*4 + 1) = 2^(21*4) * 2^1 2 raised to any power which is a multiple of 4 gives a number ending in 6. One more power of 2 then gives you something that ends in 2.

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2^85 [^ indicates to the power] now 2^2=4 2^3=8 2^4=16 2^5=32 last digits are 4,8,6,2 2^6=64 2^7=128 2^8=256 2^9=512 last digits are 4,8,6,2 similarly every fourth power after 2^9 will have last digit as 2 i,e 2^13,2^17,2^21,2^25...2^49.... .....2^85 will have last digit as 2

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it relies upon at ur college. At my college I took Algebra a million in 9th grade yet then took Geometry over the summer season so now i'm taking Algebra 2 in tenth grade. At my college that's progressed.

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Haha You go: 2 x 2 x 2 x 2 x .................2^85... There is no easy way lady! Sorry And this problem just can't be a ninth grade problem

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I think it is awesome! I am in Honors English, and it is just as good. And to answer the question even though I dont have to.. It was Ling, obviously.. #1 Her eyes widened and she seemed subtle..

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