Divide by 100 ——> the remainder after dividing by 100 will give the last 2 digits of the number
(1st way) we can use the Chinese Remainder Theorem and break the DEN into co-prime divisors of (25) and (4) [note: 25 * 4= 100 and 25 and 4 so not share any prime factors]
and then find the General form of the number “(7)^241 when divided by 100” by comparing the 2 results
Part 1:
(7)^41 / 25 ——-> remainder of = ?
(7)^2 = 49 ———> which is (-1) away from a multiple of 25
[ (49)^20 * (7) ] / 25 ———-> Rem of = ?
Excess remainder = (-1)^20 * (7) ———> after multiplying remainders, yields a remainder of 7
General Form 1:
N = 25a + 7 (eq 1)
Part 2:
(7)^41 / 4 ——-> remainder of = ?
(7)^2 = 49 ———> is (+1) more than a multiple of 4
[ (49)^20 * (7) ] / 4 ——-> excess remainder of = (1)^20 * 7 = 7
(7) / 4 ———> Rem of = 3
General form part 2:
N = 4b + 3 (eq 2)
Combining the 2 parts to find the General Form of “the dividend (7)^41 when divided by 100”:
N = 25a + 7 = 4b + 3
Where a and b are non negative integer quotients
4b = 25a + 4
b = (25a + 4) / 4
The first non negative value of a that will make (25a + 4) divisible by 4 ———> a = 0
B = (25* 0 + 4) / 4 = 1 ———-> so b = 1
N = 25a + 7 ——-> (25) (0) + 7 = 7
N = 4b + 3 ———-> (4) (1) + 3 = 7
Thus: the General form of “(7)^41 when divided by 100” will take the general form for any number positive integer N where:
N = 100k + 7
(7)^41 is just one of many N values that have the same remainder properties when divided by 100
Thus: (N / 100) ———-> (100k + 7) / 100 >>>>> yields a remainder of = 07
Thus the Tens Digit of the Term (7)^41 is
0 (A)
(Quicker method)
From memory, (7)^3 = 343
Thus, (7)^4 = (7) * (7)^3 = (7) * (343) = (7) (300 + 40 + 3) = (2100 + 280 + 21) =
2,401 ———-> which is (+1) more than a multiple of 100. Again, when we divide any dividend by 100, the remainder will give use the last 2 digits of the NUM
(7)^41 ———> (2,401)^ 10 * (7)
Dividing by 100, the excess remainder will be:
(1)^10 * (7) = 07
The tens digit or (7)^41 will be 0
(A) 0
Edit: is this an official question?
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