Entrance Exams
Govt. Exams
By Fermat's Little Theorem, since 13 is prime and gcd(5,13)=1, we have 5^12 ≡ 1 (mod 13). 100 = 12×8 + 4. So 5^100 ≡ 5^4 (mod 13). 5^4 = 625 = 48×13 + 1 ≡ 1 (mod 13).
Using Legendre's formula: floor(100/5) + floor(100/25) + floor(100/125) = 20 + 4 + 0 = 24.
First multiple of 7 ≥ 50: 56 (7×8). Last multiple of 7 ≤ 150: 147 (7×21). Count = 21 - 8 + 1 = 14. Hmm, should be 14 not 15. Let me verify: 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133, 140, 147. That's 14 numbers.
HCF takes minimum power of each prime: HCF = 2^min(4,3) × 3^min(3,2) × 5^min(1,2) = 2^3 × 3^2 × 5.
3^1 ≡ 3, 3^2 ≡ 9, 3^3 ≡ 27 ≡ 5, 3^4 ≡ 15 ≡ 4, 3^5 ≡ 12 ≡ 1 (mod 11). Wait: 3×4 = 12 ≡ 1. So 3^5 ≡ 1 (mod 11). Answer should be A.
Numbers divisible by 3: floor(100/3) = 33. Numbers divisible by both 3 and 5 (i.e., by 15): floor(100/15) = 6. Numbers divisible by 3 but not by 5 = 33 - 6 = 27. Wait, that's option B. Let me verify: 27 is correct.
Using the property: HCF × LCM = Product of two numbers. 6 × 60 = 12 × x. 360 = 12x. x = 30.
Unit digits of powers of 7 follow pattern: 7^1→7, 7^2→9, 7^3→3, 7^4→1, 7^5→7. Cycle = 4. Since 2019 = 504×4 + 3, unit digit of 7^2019 = unit digit of 7^3 = 3. Wait, let me recalculate: 7^3 = 343 (unit 3), but the cycle shows 7,9,3,1. For 2019 mod 4 = 3, so 7^3 has unit digit 3. Actually checking: option answer is B(7), but calculation shows 3. There may be a typo in options.
We need LCM(6, 8). 6 = 2×3, 8 = 2³. LCM = 2³×3 = 24. Therefore, 24 is the smallest number divisible by both 6 and 8.
Using Euclidean algorithm: 144 = 96×1 + 48, 96 = 48×2 + 0. Therefore HCF = 48. Alternatively, 144 = 2^4×3^2 and 96 = 2^5×3. HCF = 2^4×3 = 48.