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There is no specific question as such. I am preparing for an aptitude exam, and am stuck at this particular point when we are required to find the last two digits/remainder when divided by 100 of an even number raised to a big power.
The above posted questions are just random ones, but are definitely part of my curriculum. the numbers cover even numbers ending in 2, 4, 6, 8. if anything else is required, please take any arbitrary example and help.
We look for example at $74^{53}$. This is congruent to $0$ modulo $4$. So we will know what it is modulo $100$ once we know what $74^{53}$ is modulo $25$.
Note that $74\equiv -1\pmod{25}$. Thus $74^{53}\equiv (-1)^{53}\equiv -1\pmod{25}$.
We therefore want to solve the system $x\equiv 0\pmod{4}$, $x\equiv -1\pmod{25}$. By inspection the answer is $x\equiv 24\pmod{100}$. The remainder is $24$.
Another one: The preceding problem could be done exceptionally quickly, because of the $74$. We look at $42^{70}$. Again, this is congruent to $0$ modulo $4$, so we work modulo $25$. By Euler's Theorem, $42^{80}\equiv 1\pmod{25}$. So $42^{79}$ is the inverse of $42$ modulo $25$. But $42\equiv -8\pmod{25}$. Since $(-8)(-3)\equiv -1\pmod{25}$, the inverse of $-8$ modulo $25$ is $3$. Thus our number is congruent to $3$ modulo $25$.
So we are looking for a number divisible by $4$ that has remainder $3$ on division by $25$. We conclude that the remainder on division by $100$ is $28$.
Hint: Use the binomial theorem:
$$48^{277} = (40 + 8)^{277} = \sum\limits_{k = 0}^{277} {277 \choose k}40^{277 - k} 8^k = (\text{things divisible by 100}) + {277 \choose 276} 40 \cdot 8^{276} + 8^{277}$$
Write out a few powers of $8$ and see if you can find a pattern to them.
For a slightly different way:
$$48^{277} = (2^4 3)^{277} = 2^{4 \cdot 277} 3^{277}$$
Try writing out some powers of $2$ and $3$ and see what happens; as a start, note that
$$2^{10} = 1024 \equiv 24 \mod 100$$
For another way, as suggested by Andre Nicolas, we can see that $48 \equiv 0 \mod 4$, and $48 \equiv (-2) \mod 25$. Hence,
$$48^{277} \equiv (-2)^{277} \equiv - (2^{277}) \mod 25$$
But $2^{10} \equiv -1 \mod 25$, so
$$2^{277} = (2^{10})^{27} 2^7 \equiv - 2^7 \mod 25$$
which is easily computed. Once we have two congruences, one mod $4$ and one mod $25$, it's straightforward to get a congruence mod $100$.
As you are preparing for aptitude I would prefer giving you some short cuts to find last 2 digits which will save you time
$(odd\ number)^{power}$ the last 2 digits only depends on last digit of power and the last 2 digits of the odd number given
CASE 1: if the odd number ends in 1 like $(y1)^{n}$ last 2 digits are (units digit of y*n)(1)
if odd number ends in 7 , 3 , 9 trick is first bring it to a form that ends in 1. keep in mind that $7^4$ , $9^2$ , $3^4$ ends in 1
CASE 2: if the number is like $(x7)^{4}$ then last two digits are (units digit of 2*x)(1)
CASE 3: if the odd number is like $(x3)^{4}$ then last two digits are (8 - [units digit of 2*x])(1)
CASE 4 : if the odd number is like $(x9)^{2}$ then last two digits are (8 - [units digit of 2*x])(1)
now having made numbers ending in 1 , then the process is as in case 1 with remaining of power after removing 4 for case 2 and case 3 or 2 for case 4
Now $(2)^{power}$ shortcut is as follows
CASE 5 : if ten's place of power is even the last 2 digits are just last 2 digits of $2^{last\ digit\ of\ power}$
CASE 6 : if ten's place of power is odd the last 2 digits are just last 2 digits of $3 \times 2^{last\ digit\ of\ power + 3}$
so case 1,2,3,4 is for number ending in odd number case 5,6 for power of 2
as any even number can be expressed as multiple of 2 and an odd number a combination ,of one of first 4 cases and one of case 5 or case 6 will fetch you the answer for $(even\ number ) ^{power}$
with the above short cut techniques I am able to calculate last 2 digits in a matter of 15 to 20 seconds without calculator
