Question

the digit in the unit place of the product 127^7* 2^24 is

Answer 1

@satellite73

Answer 2

sorry the question is 127^79 * 2^24

Answer 3

no matter

Answer 4

sorry for the mistake in question

Answer 5

\(7^0=1,7^1=49,7^3=343,7^4=2401\) so choice of units digit for \(7^{79}\) is \[\{1,9,3\}\]

Answer 6

as you can see if the exponent is divisible by 4, you will get a 1, but in this case 79 is not divisible by 4, but 80 is, so the units digit of \((127)^{79}\) is 3

Answer 7

7^1=7 and 7^2=49 is it.....

Answer 8

yeah that was a typo

Answer 9

ok

Answer 10

the answer should be 8

Answer 11

damn let me start again
\[7^1=7,7^2=49,7^3=343,7^4=2401\] conclusion is still the same

Answer 12

yes, now we have to find units digit for powers of 2
that is easier, choices are \(2,4,8,6\) in that order

Answer 13

hope that part is clear,
\[2^1=2,2^2=4,2^3=8,2^4=16\] and then \(2^5=32\) so they start over

Answer 14

since 24 is divisible by 4 we know the units digit of \(2^{24}\) is 6

Answer 15

units digit of \((127)^{79}\) is 3, units digit of \(2^{24}\) is 6 and \(3\times 6=18\) so units digit of the product is 8

Answer 16

doesn't any one go to parties, shows, bars, dates, friends houses, something on saturday night instead of coming here to do math homework? damn!

Answer 17

how units digit of (127)^79 is 3

Answer 18

@experimentX

Answer 19

Let's look at this using modulo 10. \[ 127^7\cdot 2^{24} \pmod{10}\]We can simplify this problem significantly by notice that this is the same as \[ 7^7\cdot 2^{24} \pmod{10}\]Now, using Euler's theorem, and seeing that \(\phi(10)=4\) We can simplify this even further to \[7^3\cdot2^{24}\pmod{10}\]Since 2 isn't coprime to 10, we can't simplify that part. However, as satellite pointed out above, the last digit of \(2^n\) loops from \(2\rightarrow4\rightarrow6\rightarrow8\rightarrow2\), so \(2^{4n}\equiv 2\pmod{10}\) for integers n. Since \(24=4\cdot6\), we can simplify the problem to \[7^3\cdot2\pmod{10}\]This we can actually calculate to be\[3\cdot2\equiv 6\pmod{10}\]So the units digit is 6.

Answer 20

the answer should be 8

Answer 21

Right. I screwed up with the power of 2. It should be that \(2^{4n}\equiv 6\pmod{10}\) for all \(n>0\). So then you get \[7^3\cdot2^{24}\equiv3\cdot6\equiv8\pmod{10}\]My mistake.

Answer 22

did nt understand...can u make it clear

Answer 23

@shameer1 is it not clear from the pattern here
\[7^0=1,7^1=49,7^3=343,7^4=2401\] that the units digit of \(127)^{79}\)is 3 ?

Answer 24

no

Answer 25

forget the numbers not in the units place
we have \(7^0\to 1, 7^1\to 7,7^2\to 9, 7^3\to 3, 7^4\to 1\) and once we see the one in the units place, the digits will repeat

Answer 26

ok

Answer 27

that should be more or less clear because it doesn't matter what \(7^4\) is , only that is ends in a 1, so that \(7^5\) will end in a 7 , \(7^6\) will end in a 9, etc

Answer 28

so once again to sum up, \(7^{79}\) ends in a 3, \(2^{24}\) ends in a 6, and \(3\times 6=18\) so your number ends in an 8