if($_SERVER['REQUEST_URI']=='/' || $_SERVER['REQUEST_URI']=='/index.php'){?>
...for what may lead to a life altering association!
To find the remainder of a large exponent, reduce the base and look for a power that leaves +1 or –1 with the divisor. For example, 2³ = 8 leaves remainder +1 with 7, so 2300÷ 7 gives remainder 1.
Our goal in such questions is to find the remainder of a large power quickly and correctly. In your GMAT preparation course, you learn a simple routine. First reduce the base by the divisor. Then test small powers until one gives remainder +1 or remainder –1. Use that power as the cycle length. Shrink the exponent by this cycle. Replace the expression with 1 or –1. If the result is –1, add the divisor to make the remainder positive. Repeat the reduce and shrink steps for any extra factors. This keeps numbers small and avoids heavy multiplication. With steady work on GMAT practice tests, you apply the routine under time limits until it feels natural. The focus stays on clarity, not brute force.
Large powers become simple when you look for a small exponent that makes the base give a remainder of 1 or minus 1 with the divisor. Once you find that exponent, use it as the cycle length to shrink the power and finish in a few clean steps.
Question: What is the remainder when 9300 is divided by 7?
Remainder of 9 with 7 is 2. Or, 9 mod 7 is 2.
Reduce the base: 9 mod 7 = 2
Test small powers: 21 → remainder with 7 is 2
Test small powers: 22 → remainder with 7 is 4
Test small powers: 23 → remainder with 7 is 1
Cycle length = 3
Write the exponent: 300 = 3 x 100 + 0
Replace the power: 2300 has the same remainder as 1100
Final remainder: 1
Take the example of 2300, divided by 7.
Here, 2 itself is not close to +1 or -1 with 7.
But 2 cubed is 8, which leaves a remainder of +1 when divided by 7.
That insight transforms the expression into 8^100 with 7.
Since every 8 leaves a remainder of 1, the answer is simply 1.
Question: What is the remainder of 2303 with divisor 9?
Here, 2 cubed is 8, which leaves remainder -1 with 9.
Rewriting the problem gives (-1)^101, which is -1.
Since remainders cannot remain negative, we add the divisor 9, giving a final remainder of 8.
This small adjustment is critical and often tested.
The same approach applies to many other cases. If the base is larger than the divisor, always reduce the base first.
Here, 9260 with divisor 15 reduces to 260.
Since 24 is 16, which leaves remainder +1 with 15, the expression simplifies to 115.
The required remainder turns out to be 1.
Here, with divisor 17, 2^4 leaves remainder -1,
So, the answer becomes -1^15 = -1,
and after adding 17, the final remainder is 16.
The required remainder turns out to be 16.
Here, the base 100 gives remainder 2 with 7.
Thus, 100100 reduces to 2^100.
Since 23 = 8 leaves remainder +1 with 7,
the problem further reduces to 133 x 21,
The required remainder turns out to be 2.
Remainder problems test more than calculations. They test whether you can look past large numbers and recognize patterns. The ability to simplify, to use +1 and -1 intelligently, and to adjust negative remainders correctly reflects the kind of analytical clarity the GMAT seeks. These are not tricks but logical habits that serve well in problem-solving far beyond the test. Developing them not only helps in scoring high but also adds to the structured thinking that will help in your MBA admissions stage and in future professional challenges.
Remainder problems with large exponents reward a calm, structured approach. You learn to keep only what matters, work with cycles of four, and let checks guide results. Spotting when a power returns 1 or minus 1 trains your eye to see patterns quickly. That habit builds confidence, precision, and speed without strain. Practice the steps until they feel natural: take the last digit, divide the exponent by four, use the remainder, and handle special cases cleanly. Use GMAT simulation to rehearse under timed conditions. With steady practice, tough expressions shrink to a few clear moves, and clarity becomes your default.