01Easy64×since 2002Q2897The remainder when (2021)2022+(2022)2021(2021)^{2022}+(2022)^{2021}(2021)2022+(2022)2021 is divided by 7 isA0B1C2D6Check answerSkip
02Easy64×since 2002Q2898The remainder when 72022+320227^{2022}+3^{2022}72022+32022 is divided by 5 is :A0B2C3D4Check answerSkip
03Easy64×since 2002Q2899Fractional part of the number 4202215\frac{4^{2022}}{15}1542022 is equal toA815\frac{8}{15}158B415\frac{4}{15}154C115\frac{1}{15}151D1415\frac{14}{15}1514Check answerSkip
04Medium64×since 2002Q2900Let the number (22)2022+(2022)22(22)^{2022}+(2022)^{22}(22)2022+(2022)22 leave the remainder α\alphaα when divided by 3 and β\betaβ when divided by 7. Then (α2+β2)\left(\alpha^{2}+\beta^{2}\right)(α2+β2) is equal to :A13B10C20D5Check answerSkip
05Easy64×since 2002Q290125190−19190−8190+219025^{190}-19^{190}-8^{190}+2^{190}25190−19190−8190+2190 is divisible by :A14 but not by 34Bneither 14 nor 34Cboth 14 and 34D34 but not by 14Check answerSkip
06Medium64×since 2002Q2902Among the statements : (S1) : 20232022−199920222023^{2022}-1999^{2022}20232022−19992022 is divisible by 8 (S2) : 13(13)n−12n−1313(13)^{n}-12 n-1313(13)n−12n−13 is divisible by 144 for infinitely many n∈Nn \in \mathbb{N}n∈NAboth (S1) and (S2) are incorrectBonly (S1) is correctConly (S2) is correctDboth (S1) and (S2) are correctCheck answerSkip
07Medium64×since 2002Q2903The coefficients of xp{x^p}xp and xq{x^q}xq in the expansion of (1+x)p+q{\left( {1 + x} \right)^{p + q}}(1+x)p+q areAequalBequal with opposite signsCreciprocals of each otherDnone of theseCheck answerSkip
08Medium64×since 2002Q2904The number of integral terms in the expansion of (3+\root8\of5)256{\left( {\sqrt 3 + \root 8 \of 5 } \right)^{256}}(3+\root8\of5)256 isA35B32C33D34Check answerSkip
09Medium64×since 2002Q2905The coefficient of xn{x^n}xn in expansion of (1+x)(1−x)n\left( {1 + x} \right){\left( {1 - x} \right)^n}(1+x)(1−x)n isA(−1)n−1n{\left( { - 1} \right)^{n - 1}}n(−1)n−1nB(−1)n(1−n){\left( { - 1} \right)^n}\left( {1 - n} \right)(−1)n(1−n)C(−1)n−1(n−1)2{\left( { - 1} \right)^{n - 1}}{\left( {n - 1} \right)^2}(−1)n−1(n−1)2D(n−1)\left( {n - 1} \right)(n−1)Check answerSkip
10Hard64×since 2002Q2906If the coefficient of x7{x^7}x7 in [ax2+(1bx)]11{\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}[ax2+(bx1)]11 equals the coefficient of x−7{x^{ - 7}}x−7 in [ax−(1bx2)]11{\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}[ax−(bx21)]11, then aaa and bbb satisfy the relationAa−b=1a - b = 1a−b=1Ba+b=1a + b = 1a+b=1Cab=1{a \over b} = 1ba=1Dab=1ab = 1ab=1Check answerSkip