Question

whats the last digit for (2^123!)*(7^76!)

Answer 1

The ones digit for powers of 2 repeats:
\[2^1=2\\
2^2=4\\
2^3=8\\
2^4=16\\
2^5=32\\
2^6=64\]
and so on, which gives a pattern of \(2-4-8-6-2-4-8-6-\cdots\).

The ones digit for powers of 7 also repeats:
\[7^1=7\\
7^2=49\\
7^3=343\\
7^4=2401\\
7^5=16807\\
7^6=117649\]
and so on, with a pattern of \(7-9-3-1-7-9-3-1-\cdots\).

Answer 2

The only foreseeable problem is checking the value of 123!... Most scientific calculators aren't equipped for numbers of that magnitude...

Answer 3

also u may work mod 10 directly :
last digit = (2^123!)*(7^76!) mod 10

Answer 4

this may help in reducing :
7^2 = 49 = -1 mod 10

Answer 5

2^(4x) MOD 10 = 6
7^(4y) MOD 10 = 1
So,
2^(4x) * 2^(4y) MOD 10 = 6

Answer 6

Since 123! and 76! can be expressed as 4k,
last digit of 2^123! * 7^76! is 6

Answer 7

ok thns