Practice Questions
Last Two Digits
Find n mod 100. Use cyclicity mod 100 or binomial expansion for numbers ending in 1, 3, 7, 9.
Last Three Digits
Find n mod 1000. Useful for large powers and product problems in placement exams.
Techniques
- Work mod 100 or mod 1000 directly.
- Split the base into tens and units when squaring.
- Use Euler's theorem for advanced cases (φ(100)=40).
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Q1 EasyWhat will be the last two digits of 15^2 ?Explanation15² = 225. The last two digits are 25.
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Q2 EasyFind the last three digits of 7^3.Explanation7³ = 343. The last three digits are 343.
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Q3 EasyIf 13 is squared completely, which option correctly gives the tens and units digits of the answer?Explanation13² = 169. The last two digits are 69.
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Q4 EasyWhat are the last three digits of 12 cubed?Explanation12³ = 1728. The last three digits are 728.
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Q5 EasyWhich of the following represents the last two digits of 76 squared?Explanation76² = 5776. Any number ending in 76 keeps 76 as its last two digits when squared.
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Q6 EasyIf 19 is multiplied by itself once, what will be the last two digits of the result?Explanation19² = 361. The last two digits are 61.
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Q7 MediumWhat will be the last two digits of 7^4 ?Explanation7² = 49 and 7⁴ = 49² = 2401. The last two digits are 01.
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Q8 MediumFind the last two digits of 7^40.Explanation7⁴ ends in 01. Since 40 is a multiple of 4, 7^40 ends in 01.
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Q9 MediumFind the last two digits of 7^46.Explanation7⁴ ends in 01. 46 = 44 + 2, so 7^46 has the same last two digits as 7² = 49.
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Q10 MediumThe last two digits of the number 5^12345 isExplanationFor any power of 5 greater than 1, the last two digits are 25.
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Q11 MediumWhen 7 is raised to the power 53, what will be the last two digits of the resulting number?Explanation7⁴ ends in 01. 53 = 52 + 1, so the last two digits match 7¹ = 07.
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Q12 MediumWhen 2 is raised to the power 10, what digits appear in the tens and units places of the product?Explanation2¹⁰ = 1024. The last two digits are 24.
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Q13 MediumWhat are the last three digits of 5 raised to the power 10?Explanation5¹⁰ = 9765625. The last three digits are 625.
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Q14 HardWhat will be the last two digits of 3 raised to the power 20?Explanation3²⁰ = (3¹⁰)². 3¹⁰ ends in 49, and 49² = 2401, so the last two digits are 01.
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DS1 Easy
What are the last three digits of 5 raised to the power n, where n is at least 3?
Statement IThe exponent n is equal to 10.
Statement IIThe exponent n is greater than 2.
ExplanationFor n ≥ 3, 5ⁿ ends in 125 or 625. Statement I fixes n = 10, giving 625. Statement II alone does not determine whether the answer is 125 or 625. -
DS2 Easy
What are the last two digits of the positive integer N?
Statement IN is equal to 76 squared.
Statement IIN is a four-digit positive integer.
Explanation76² = 5776, so the last two digits are 76. Statement I alone is sufficient. Statement II alone only restricts the size of N and does not fix its last two digits.
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DS3 Medium
What are the last two digits of 7 raised to the power n?
Statement IThe value of n is 4.
Statement IIThe exponent n is an even positive integer.
Explanation7⁴ = 2401, so the last two digits are 01. Statement I alone is sufficient. Statement II alone only tells us n is even and does not fix the last two digits. -
DS4 Medium
What are the last two digits of the positive integer N?
Statement IN is the square of the two-digit prime number 13.
Statement IIThe integer N is a perfect square less than 200.
Explanation13² = 169, so the last two digits are 69. Statement I alone is sufficient. Statement II alone allows several squares such as 100, 121, and 144.