3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Consider the expansion of (45+62)121 in powers of 45 and 62 by the binomial theorem, with all 122 terms written separately. How many of these 122 terms are irrational? Exactly one option is correct.
A112
B121
C122
D111
Show answer and solution
Answer: C
The general term is Tr+1=(r121)5(121−r)/42r/6, 0≤r≤121. It is rational exactly when 4∣(121−r) and 6∣r (since 5a and 2b with a,b non-negative rationals are rational only for integer exponents, and the product of a rational power of 5 and a rational power of 2 is rational only if both exponents are integers, by unique factorisation).
6∣r forces r even, while 4∣(121−r) forces r≡1(mod4), i.e. r odd. These are incompatible, so there is no rational term at all.
Hence all 122 terms are irrational.
Trap: counting multiples of lcm(4,6)=12 as for an even index gives 11 rational terms; the parity clash must be checked.
Difficulty: Hard
Question 2Level 2 · ExamSingle correct
In the expansion of (3−2x)20 in ascending powers of x, let Tk denote the k-th term from the beginning. At x=45, the term(s) of greatest absolute value is/are ______. Exactly one option is correct.
AT9 only
BT11 only
CT10 and T11 (equal in absolute value)
DT10 only
Show answer and solution
Answer: D
With x=45, Tr+1=(r20)320−r(−25)r, so the absolute values satisfy
∣Tr∣∣Tr+1∣=r21−r⋅35/2=6r5(21−r).
This exceeds 1 iff 105>11r, i.e. r≤9, and is never equal to 1 because 105/11 is not an integer. So the absolute values increase up to T10 and decrease afterwards; the unique numerically greatest term is T10 (the term with r=9).
Difficulty: Hard
Question 3Level 2 · ExamNumerical answer
Find the integer formed by the last three digits of 9123 (that is, the remainder when 9123 is divided by 1000). Give the answer as an integer.
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Answer: 929
9123=(10−1)123=−(1−10)123. Modulo 1000, only the terms with 100,101,102 matter:
(1−10)123≡1−123⋅10+(2123)⋅100=1−1230+7503⋅100.
7503⋅100=750300≡300, so (1−10)123≡1−1230+300=−929≡71(mod1000).
Therefore 9123≡−71≡929(mod1000). (Check: Python pow(9,123,1000)=929.)
Difficulty: Hard
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