The product of the last two digits of (1919)1919 is _____
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Step-by-Step Solution
Step 1: Find the last two digits of the base
To find the last two digits of (1919)1919, we only need to consider the last two digits of the base, which is 19. So, we need to find the last two digits of (19)1919.
Step 2: Find the Euler's totient function for 100
We use Euler's totient theorem, which states that if a and n are coprime, then aϕ(n)≡1(modn). Here, a=19 and n=100. First, we calculate ϕ(100). Since 100=22⋅52, ϕ(100)=100(1−1/2)(1−1/5)=40.
Step 3: Reduce the exponent modulo ϕ(100)
Now we need to reduce the exponent 1919 modulo ϕ(100)=40. Dividing 1919 by 40, we get 1919=40×47+39. So, 1919≡39(mod40). However, using 39 directly can be cumbersome. We can use 39≡−1(mod40) for easier calculation.
Step 4: Apply Euler's totient theorem
According to Euler's totient theorem, 191919≡1939(mod100). Since 39≡−1(mod40), we can write 1939≡19−1(mod100). This means we need to find the multiplicative inverse of 19 modulo 100.
Step 5: Find the multiplicative inverse of 19 modulo 100
We need to find an integer x such that 19x≡1(mod100). We can test values or use the extended Euclidean algorithm. By testing, we find that 19×79=1501, which leaves a remainder of 1 when divided by 100. Thus, 19−1≡79(mod100).
Step 6: Determine the last two digits and their product
The last two digits of (1919)1919 are 79. The last digit is 9 and the second to last digit is 7. The product of these digits is 7×9=63.