what are the last two digits of 7^358 (the answer must be justified).?

Answer 1

You just need to calculate the last 2 -digits till 7^4
7^1=7
7*7=49
49*7=343
43*7=301 [Just consider the last two digits you don't need the earlier digits]

After this in the subsequent powers of 7 the last two digits will repeat in the above order ie 07,49,43,01 (becuase 01*7=07 and the series will repeat)

Since their are four possibilities divide the power in the question ie 358 by 4. The remainder is 2. So the last two digits will be given by the second member in the above series ie 49.

If the remainder turns out to be 0, we will consider the fourth member of the above series.

I know this technique might appear difficult in the first take but due practice will make it easy.

Mail me if you want any other justification..........

Answer 2

A fun problem in congruences:
You want 7^358 = x mod 100 and x is between 0 and 100.
So note: 7^4 = 2401 = 1 mod 100. (By Eulers formula this would also work).
and note: 358 = 4(89) + 2
so 7^358 = 7^(4(89)+2) = 1 * 7^2 = 49 as required.
If you need help on congruences, visit the site below and look up clock math.
Please THIS is the correct answers ignore the other two! the last 6 digits are 306449 by MAPLE so you know this is correct.

Answer 3

Looking only at last digits...

7^0 = 1
7^1 = 7 since 7 * 1 = 7
7^2 = 9 since 7 * 7 = 49
7^3 = 3 since 7 * 9 = 63

7^4 = 1 since 7 * 3 = 21
7^5 = 7 since 7 * 1 = 7
7^6 = 9 since 7 * 7 = 49
7^7 = 3 since 7 * 9 = 63

7^8 = 1 since 7 * 3 = 21
.
.
.

and this sequence will repeat for all successive powers of seven.

Answer 4

Since 7^4 mod 100 ≡ 1 mod 100, we have

7^358 mod 100
≡ 7^(2+4*89) mod 100
≡ 7^2 mod 100
≡ 49 mod 100

So, the last two digits are 49.