if 16 to the power 123 what is the last two digits of the product?

Answer 1

The last digit is always 6 and the next-to-last goes in the pattern 1, 5, 9, 3, 7, 1.... So if 16^1 ends in 16, then so does 16^6 and every fifth one after it, so 16^121 ends in 16. Therefore 16^123 ends in 96.

Edit: Krystal, no calculator in the world could handle these kinds of numbers. The trick is to recognize the pattern early, because any attempt to actually find out the whole number will not end happily. I tried working this out in Excel and by 16^10 it had already resorted to scientific notation.

Answer 2

The last digit is easy since 6 times itself ends in 6.
The second to the last digit can be easily figured by using the calculator as follows. Enter "6", "*", "=". Every time you hit
"=" you're raising the power of 6 by 1. Notice that the 2nd to the last digit is in a repetitive cycle of 3,1,9,7,5 correponding to
powers of 6 which are 6mod1, 6mod2, 6mod3, 6mod4, 6mod5.
Since 123 is 6mod3 the second to the last digit is 9. So the last 2 digits are 96.

Answer 3

i'm going to anticipate you recommend the time-honored skill tower: 2^(4^(7^123)). (See link decrease than.) this is undemanding sufficient to coach that, modulo one hundred, the powers of two (from 2² onwards) repeat in cycles of 20. (*) So, as an occasion, 2? ends interior an identical 2 digits as 2²? or 2??; specifically, sixteen. Modulo 20, the powers of four repeat in cycles of two. So 4? ? 4 or sixteen (mod 20), in accordance as n is weird and wonderful or perhaps, respectively. needless to say 7^123 is weird and wonderful. subsequently 4^(7^123) ? 4 (mod 20). subsequently 2^4^(7^123)) ? sixteen (mod one hundred). it is, the final 2 digits of two^4^7^123 are sixteen. (*) extra: think a < b and a pair of^a ? 2^b (mod one hundred). Then 2^a ? 2^b (mod 2²) and a pair of^a ? 2^b (mod 5²). 2^a ? 2^b (mod 4) => a ? 2. (that's the place the requirement "from 2² onwards", above, comes from.) 2^a ? 2^b (mod 25) => 2^(b-a) ? a million (mod 25). Letting x = b-a, we look for the smallest x such that 2^x ? a million (mod 25). 2^x ? a million (mod 25) => 2^x ? a million (mod 5). via inspection, the smallest answer is x = 4, so the final answer mod 5 is x = 4k, the place ok is any helpful integer. Now we look for the smallest ok such that 2^(4k) ? a million (mod 25). we come across that ok = 5. So x = 20 is the smallest x such that 2^x ? a million (mod 25). putting the outcomes mod 4 and mod 25 mutually, we see that, modulo one hundred, the powers of two (from 2² onwards) repeat in cycles of 20.

Answer 4

considering only the last 2 digits
16 ^123 = 16*(16^2)^61 = 16*56^61
16*56^61 = 16*56*(56^2)^30 =96*36^30
96*36^30 = 96*(36^2)^15 =96*96^15
96*96^15 =96*96*(96^2)^7 = 16*16^7 =16^8
at this point my calculator can can figure 16^8 without scientific notation
16^8 =96

Answer 5

96

Answer 6

96

Answer 7

96

Answer 8

1.27 * 10 to the power of 148